text:
\section{Preface}\label{intr}

 Cirquent calculus  is a family of deep inference (cf. \cite{gug}) proof systems permitting various sorts of component-sharing between different parts of logical expressions.
The earlier article \cite{cl16} constructed a cirquent calculus   system $\lll$ for the elementary-base, recurrence-free propositional fragment of computability logic (the game-semantically conceived logic of computational problems introduced in \cite{Jap03})  and proved its   soundness and completeness. The present article takes that result to the first-order level, with the so called choice quantifiers. The resulting system  $\mmm$, in contrast to its classical counterpart, is a decidable predicate logic. While a variety of propositional systems have been built by now in cirquent calculus \cite{Cirq,Japdeep,taming1,taming2,cl16,XuIGPL,XuIf,XuLast}, $\mmm$ is the first cirquent calculus  system  with quantifiers. The formal semantics for its language (and beyond) was set up in \cite{lmcs}, but no axiomatizations had been attempted so far.

Being a continuation  of \cite{cl16} in the proper sense, the present article should  only be read in combination with its predecessor, as it relies on but does not reintroduce the main concepts from \cite{cl16}. Nor does it discuss related literature or the relevant motivations and philosophy underlying  cirquent calculus and computability logic, as this, again, is done in \cite{cl16}.

The language of $\mmm$ extends that of $\lll$ by augmenting its logical vocabulary with the {\em choice universal quantifier} $\ada$ (``{\em chall}'') and {\em choice existential quantifier} $\ade$ (``{\em chexists}''), and allowing atoms of any arities. Throughout this article, for simplicity, we assume that the universe of discourse is always the set $\mathbb{N}$ of natural numbers, by innocent abuse of concepts identified with the corresponding decimal numerals. If so, $\ada xG(x)$ is the game where, at the beginning of a play, the environment chooses one of $n\in\mathbb{N}$, after which the game continues as $G(n)$; if such a choice is never made, then the environment loses.  $\ade xG(x)$ is similar, only here it is the machine who can and must make an initial  choice. Thus, $\ada xG(x)$ is in fact nothing but the infinite choice conjunction $G(0)\adc G(1)\adc G(2)\adc\cdots$, and $\ade xG(x)$
is the infinite choice disjunction $G(0)\add G(1)\add G(2)\add\cdots$.

\section{Syntax}\label{Syntax}

The  language of $\mmm$ is the same as that of first-order classical logic, only with the quantifiers $\forall,\exists$ replaced by their choice (``constructive'') counterparts $\ada,\ade$, and additionally including the choice connectives $\adc,\add$ as well as the decimal numerals $0,1,2,\cdots$ referred to as {\bf constants}.
We will be typically using $x,y,z,\cdots$ as metavariables for the variables of the language, $p,q,r,\cdots$ for its predicate letters,  and $a,b,c\cdots$ for its constants.

We fix four pairwise disjoint infinite sets $\mathbb{C}(\add)$, $\mathbb{C}(\adc)$,  $\mathbb{C}(\ade)$ and $\mathbb{C}(\ada)$, whose elements will be referred to as   {\bf $\add$-clusters},  {\bf $\adc$-clusters}, {\bf $\ade$-clusters} and  {\bf $\ada$-clusters}, respectively. The Gothic letters $\mathfrak{a}, \mathfrak{b}, \mathfrak{c},\cdots$ will be used as metavariables for clusters.

A {\bf term} is either a variable  or a constant.  A {\bf nonlogical  atom} is $p(t_1,\cdots,t_n)$, where $p$ is an $n$-ary predicate letter and $t_1,\cdots,t_n$ are terms. A {\bf nonlogical literal} is either $A$  or $\neg A$, where $A$ is a nonlogical atom.  The expressions $\top$ and $\bot$ are said to be {\bf logical literals}.
 A (choice) {\bf universal quantor} (resp. {\bf existential quantor}) is the expression $\ada^\mathfrak{c}$ (resp. $\ade^\mathfrak{c}$), where $\mathfrak{c}$ is a $\ada$- (resp. $\ade$-) cluster.

\begin{definition}\label{basecir}
A {\bf cirquent} is defined inductively as follows:\vspace{-5pt}
\begin{itemize}
\item Each  (logical or nonlogical)  literal is a cirquent.\vspace{-7pt}
\item If $A$ and $B$ are cirquents, then $(A)\mlc(B)$ is a cirquent.\vspace{-7pt}
\item If $A$ and $B$ are cirquents, then $(A)\mld(B)$ is a cirquent.\vspace{-7pt}
\item If $A$ and $B$ are cirquents and $\mathfrak{c}$ is a $\adc$-cluster,  then $(A)\adc^\mathfrak{c}(B)$ is a cirquent.\vspace{-7pt}
\item If $A$ and $B$ are cirquents and $\mathfrak{c}$ is a $\add$-cluster,  then $(A)\add^\mathfrak{c}(B)$ is a cirquent.\vspace{-7pt}
\item If $A$ is a cirquent, $x$ is a variable and $\mathfrak{c}$ is a $\ada$-cluster,  then $\ada^\mathfrak{c}x(A)$ is a cirquent.\vspace{-7pt}
\item If $A$ is a cirquent, $x$ is a variable and $\mathfrak{c}$ is a $\ade$-cluster,  then $\ade^\mathfrak{c}x(A)$ is a cirquent.
\end{itemize}
\end{definition}

A cirquent of the form $(A)\mlc(B)$ is said to be {\bf $\mlc$-rooted}, a cirquent of the form $A\adc^\mathfrak{c}B$ is said to be $\adc^\mathfrak{c}$-rooted, and similarly for $\mld,\add^\mathfrak{c},\ada^\mathfrak{c},\ade^\mathfrak{c}$. If we simply say ``$\adc$-rooted'', it should be understood as ``$ \adc^\mathfrak{c}$-rooted for whatever $ \mathfrak{c}$''. Similarly for $\add,\ada,\ade$.

As in \cite{cl16}, negation is only allowed to be applied to atoms. $\neg \ada^\mathfrak{c} x F$ should be understood as an abbreviation of $\ade^\mathfrak{c} x\neg F$, and $\neg \ade^\mathfrak{c} xF$ as an abbreviation of $\ada^\mathfrak{c} x\neg F$. All other conventions of \cite{cl16} regarding the usage of $\neg$ and $\mli$ remain in force.

When omitting parentheses in cirquents,  our convention is that $\neg$, $\ada^\mathfrak{c}x$ and $\ade^\mathfrak{c}x$ have the highest precedence, then comes $\mli$, then $\adc^\mathfrak{c}$ and $\add^\mathfrak{c}$, and then $\mlc$ and $\mld$.

All (other) standard terminological and notational conventions of traditional logic also remain in force. This includes the concepts of free and bound occurrences of variables, or the practice of representing a cirquent as $A(x)$ when first mentioning it, and then writing $A(t)$ to mean the result of replacing in $A(x)$ all free occurrences of the variable $x$ by the term $t$. A cirquent is said to be {\bf closed} iff it has no free occurrences of variables.

\section{Semantics}

We (re)define {\em LegalRuns} as the set of all runs $\Gamma$ satisfying
the following conditions:\vspace{-5pt}
\begin{enumerate}
\item Every move of $\Gamma$ is the string $\mathfrak{c}.a$, where either (1) $\mathfrak{c}$ is a $\adc$- or $\add$-cluster and $a\in\{0,1\}$, or (2) $\mathfrak{c}$ is a $\ada$- or $\ade$-cluster and $a\in\mathbb{N}$.\vspace{-7pt}

\item Whenever $\Gamma$ contains a move $\mathfrak{c}.a$ where $\mathfrak{c}$ is a $\adc$- or $\ada$-cluster,  the move is $\bot$-labeled.\vspace{-7pt}

\item  Whenever $\Gamma$ contains a move $\mathfrak{c}.a$ where $\mathfrak{c}$ is a $\add$- or $\ade$-cluster, the move is $\top$-labeled.\vspace{-7pt}

\item  For any cluster $\mathfrak{c}$,  $\Gamma$ contains at most one move of the form $\mathfrak{c}.a$.\vspace{5pt}
 \end{enumerate}

The intuitive meaning of condition 1 is that every move signifies either a choice between ``left'' ($0$) and ``right'' ($1$) in a $\adc$- or $\add$-cluster, or a choice among the constants $ 0,1,2,\cdots $ in some $\ada$- or $\ade$-cluster. Conditions 2 and 3 say that the environment moves (chooses) only in $\adc$- or $\ada$-clusters, and the machine only in $\add$- or $\ade$-clusters. Finally,  condition 4 says that, in any given cluster, a choice can be made only once.

As in \cite{cl16}, given a run $\Gamma\in LegalRuns$, we  say that a cirquent of the form $A\add^\mathfrak{c} B$ or $A\adc^\mathfrak{c} B$ is {\bf $\Gamma$-resolved} iff
$\Gamma$ contains one of the moves $\mathfrak{c}.0$ or $\mathfrak{c}.1$; then by the {\bf $\Gamma$-resolvent} of the cirquent we mean $A$ if such a move is
$\mathfrak{c}.0$, and $B$ if it is $\mathfrak{c}.1$. Extending this terminology to quantifiers, we say that a cirquent of the form $\ade^\mathfrak{c}x A(x)$ or $\ada^\mathfrak{c} xA(x)$ is {\bf $\Gamma$-resolved} iff
$\Gamma$ contains the move $\mathfrak{c}.a$ for some constant $a$; then by the {\bf $\Gamma$-resolvent} of such a cirquent we mean $A(a)$. Sometimes, instead of saying that the cirquent $A\add^\mathfrak{c} B$ (or $A\adc^\mathfrak{c} B$, or $ \ada^\mathfrak{c}x A(x)$, or $\ade^\mathfrak{c}x A(x)$) is resolved, we may simply say that the cluster $\mathfrak{c}$ is resolved. In all cases, as expected, ``{\bf  unresolved}'' means ``not resolved''.

An {\bf interpretation}  is a function $^*$ that assigns
to every $n$-ary predicate letter $p$ a relation $p^*\subseteq\mathbb{N}^n$. When $(c_1,\cdots,c_n)\in p^*$, we say that $^*$ makes the atom $p(c_1,\cdots,c_n)$ {\bf true}, or simply that   $p^*(c_1,\cdots,c_n)$ is true. As usual, ``{\bf false}''   means ``not true''.   The concepts of truth and falsity extend to $\top$, $\bot$ and all $\neg,\mlc,\mld$-combinations of closed atoms in the standard way: $\top^*$ is always true; $\bot^*$ is always false; $(\neg A)^*$ is true iff $A^*$ is false; $(A\mlc B)^*$ is true iff both $A^*$ and $B^*$ are true;   $(A\mld B)^*$ is true iff at least one of $A^*,B^*$ is true.
When a cirquent is represented as $A(t)$, we usually write $A^*(t)$ instead of $\bigl(A(t)\bigr)^*$.

Remember from \cite{cl16}  that,  when  we say  ``{\bf won}'' without specifying a player (as in the following definition), it always means ``won by the machine''.  Similarly for ``{\bf lost}''.

\begin{definition} \label{semantics}
Every closed cirquent $C$ and interpretation $^*$ induces a unique game $C^{*}$, which we refer to as ``$C$ {\bf under} $^*$'', defined as follows. The
set $\legal{C^{*} }{}$ of legal runs of such a game is nothing but {\em LegalRuns}. Since  $\legal{C^*}{}$ does not depend on $C$ or $^*$, subsequently we shall simply say ``legal run'' rather than  ``legal run of $C^{*}$''.
The $\win{C^{*}}{}$ component of the game $C^{*}$ is defined by stipulating that a legal run  $\Gamma$ is a won   run of $C^{*}$ iff one of the following conditions is satisfied:

1.  $C$  is a  literal and $C^*$ is true.

2. $C$ has the form $A_0\mlc A_1$ (resp. $A_0\mld A_1$) and, for both (resp. at least one) $i\in\{0,1\}$,  $\Gamma$ is a won run of $A_{i}^{*}$.

3. $C$ is a $\Gamma$-resolved $\adc$-, $\add$-, $\ada$- or $\ade$-rooted cirquent and,  where $B$ is the resolvent, $\Gamma$ is a won run of $B^{*}$.

4. $C$ is a $\Gamma$-unresolved $\adc$- or $\ada$-rooted cirquent.

\end{definition}

When an interpretation $^*$ is fixed in the context or is irrelevant, by abuse of notation we may omit explicit references to it, and identify a cirquent $C$ with the game $C^*$. For instance, we may say that the machine wins $C$ instead of saying that the machine wins $C$ under $^*$.

\begin{definition} \label{krisha}
 Consider a closed cirquent $C$.

1. For an interpretation $^*$, a {\bf solution} of $C$ under $^*$, or simply a solution of $C^*$, is an HPM $\cal H$ such that ${\cal H}\models   C ^{*}$.
 We say that $C$ is {\bf computable} under $^*$, or simply that $C^*$ is computable, iff $C^*$ has a solution.

2. A {\bf logical}  {\bf solution} of $C$ is an HPM $\cal H$ such that, for any interpretation $^*$, $\cal H$ is a solution of $ C ^*$. We say that $C$ is (logically)  {\bf valid} if it has a logical solution; otherwise $C$ is  {\bf invalid}.
\end{definition}

\section{Axiomatics}

Just like $\lll$, our present system $\mmm$ has $\top$ as its only axiom.  The inference rules of  $\mmm$ are listed below,  where   all  notational conventions from \cite{cl16} remain  in force.  Note that all rules of $\lll$ are also rules of $\mmm$, even if ``somewhat'' renamed.

\begin{description}
\item[Por-commutativity:]  $X[B\mld A]\leadsto X[A\mld B]$.
\item[Pand-commutativity:] $X[B\mlc A]\leadsto X[A\mlc B]$.
\item[Por-associativity:]  $X[A\mld(B\mld C)]\leadsto  X[(A\mld B)\mld C ]$.
\item[Pand-associativity:] $X[A\mlc(B\mlc C)]\leadsto  X[(A\mlc B)\mlc C ]$.
\item[Por-identity:]  $X[A]\leadsto X[A\mld \bot]$.
\item[Pand-Identity:]  $X[A]\leadsto  X[A\mlc \top]$.
\item[Por-domination:]  $X[\top]\leadsto X[ A\mld \top]$.
\item[Pand-domination:]  $X[\bot]\leadsto  X[A\mlc \bot]$.
\item[Left chor-choosing:]  $X[A_1,\cdots,A_n]\leadsto  X[A_1 \add^\mathfrak{c} B_1,\cdots,A_n \add^\mathfrak{c} B_n]$, \ where $A_1\add^\mathfrak{c} B_1$, $\cdots$, $A_n \add^\mathfrak{c} B_n$ are all $\add^\mathfrak{c}$-rooted subcirquents of the conclusion.
\item[Right chor-choosing:] $X[B_1,\cdots,B_n]\leadsto  X[A_1 \add^\mathfrak{c} B_1,\cdots,A_n \add^\mathfrak{c} B_n]$,  \ where $A_1\add^\mathfrak{c} B_1$, $\cdots$, $A_n \add^\mathfrak{c} B_n$ are all $\add^\mathfrak{c}$-rooted subcirquents of the conclusion.
\item[Chexists-choosing:] \(X[A_1(a),\cdots,A_n(a)]\leadsto  X[\ade^\mathfrak{c} x_1A_1(x_1),\cdots,\ade^\mathfrak{c} x_nA_n(x_n)],\)  where $a$ is any constant and $\ade^\mathfrak{c} x_1A_1(x_1),\cdots,\ade^\mathfrak{c} x_nA_n(x_n)$ are all $\ade^\mathfrak{c}$-rooted subcirquents of the conclusion.
\item[Left chand-cleansing:] $X\bigl[Y[A]\adc^\mathfrak{c} C\bigr]\leadsto X\bigl[Y[A\adc^\mathfrak{c} B]\adc^\mathfrak{c} C\bigr]$.
\item[Right chand-cleansing:]  $X\bigl[C \adc^\mathfrak{c} Y[B]\bigr]\leadsto X\bigl[C \adc^\mathfrak{c} Y[A\adc^\mathfrak{c} B]\bigr]$.
\item[Chall-cleansing:] $X\bigl[\ada^\mathfrak{c} xY[A(x)]\bigr]\leadsto X\bigl[\ada^\mathfrak{c} xY[\ada^\mathfrak{c} yA(y)]\bigr]$.
\item[Pand-distribution:]     $X[(A\mld C)\mlc(B \mld C)]\leadsto  X[(A\mlc B)\mld C]$.
\item[Chand-distribution:]  $X[(A\mld C)\adc^\mathfrak{c} (B \mld C)]\leadsto  X[(A\adc^\mathfrak{c} B)\mld C]$.
\item[Chall-distribution:] $X [\ada^\mathfrak{c}x (A \mld B ) ]\leadsto  X[\ada^\mathfrak{c} xA \mld B]$, where $x$ has no free occurrences in $B$.
\item[Trivialization:] $X[\top]\leadsto  X[\neg A\mld A]$,  where $A$ is a nonlogical atom.
\item[Chandchotomy:] \[X \bigl[\bigl( (A\mlc  C \adc^\mathfrak{b} D  ) \adc^\mathfrak{a}  ( B\mlc  C \adc^\mathfrak{b} D  )\bigr)\adc^\mathfrak{c}  \bigl(( A\adc^\mathfrak{a} B \mlc C )
      \adc^\mathfrak{b} (  A\adc^\mathfrak{a} B \mlc D ) \bigr)\bigr] \leadsto  X[ A\adc^\mathfrak{a} B \mlc  C \adc^\mathfrak{b} D ],\]
where $\mathfrak{c}$ does not occur in the conclusion.
\item[Challchotomy:] \[X [\ada^\mathfrak{a}x  (A \mlc \ada^\mathfrak{b}yB   )\adc^\mathfrak{c} \ada^\mathfrak{b}y  (\ada^\mathfrak{a} xA \mlc B   ) ] \leadsto  X [\ada^\mathfrak{a}xA \mlc \ada^\mathfrak{b}yB  ],\]  where $\mathfrak{c}$ does not occur in the conclusion, $x$ has no free occurrences in $\ada^\mathfrak{b}yB $ and $y$ has no free occurrences in $\ada^\mathfrak{a}xA $.
\item[Chandallchotomy:] \[X \bigl[\bigl( (A\mlc \ada^\mathfrak{b}xC   )\adc^\mathfrak{a} (B\mlc \ada^\mathfrak{b}xC  )\bigr)\adc^\mathfrak{c} \ada^\mathfrak{b} x  ( A\adc^\mathfrak{a} B \mlc C   )\bigr] \leadsto  X [ A\adc^\mathfrak{a} B \mlc \ada^\mathfrak{b}xC  ],\]  where $\mathfrak{c}$ does not occur in the conclusion and $x$ has no free occurrences in $A\adc^\mathfrak{a} B$.
\item[Chand-splitting:] $A,B\leadsto  A\adc^\mathfrak{c} B$,  where neither $A$ nor $B$ has   occurrences of $\mathfrak{c}$.
\item[Chall-splitting:] $A(a)\leadsto  \ada^\mathfrak{c} xA(x)$,  where neither $\mathfrak{c}$ nor    $a$ has occurrences in $A(x)$.
\end{description}

We will be using the word ``{\bf Commutativity}'' as a common name of Por-commutativity and Pand-commutativity.  Similarly for all other rules. Throughout the rest of this article, ``$A$ is provable'' always means ``$A$ is provable in $\mmm$'', written $\mmm\vdash A$.

\begin{example}\label{exm1} The cirquent $\ade^\mathfrak{a}xp(x)\mld \ade^\mathfrak{b}xp(x)\mli \ade^\mathfrak{c}xp(x)$ can be shown to be unprovable. However, it becomes provable if the two clusters $\mathfrak{a}$ and $\mathfrak{b}$ are the same. Below is a proof of  $\ade^\mathfrak{a}xp(x)\mld \ade^\mathfrak{a}xp(x)\mli \ade^\mathfrak{c}xp(x)$, i.e., of $\bigl(\ada^\mathfrak{a}x\neg p(x)\mlc \ada^\mathfrak{a}x\neg p(x)\bigr)\mld \ade^\mathfrak{c}xp(x)$.

1. $\top$ \hspace{5pt} Axiom

2. $\top \mlc \top$ \hspace{5pt} Pand-identity: 1

3. $\bigl(\neg p(a)\mld p(a)\bigr)\mlc \bigl(\neg p(a) \mld p(a)\bigr)$ \hspace{5pt} Trivialization: 2 (twice)

4. $\bigl(\neg p(a)\mlc \neg p(a)\bigr)\mld p(a)$\hspace{5pt} Pand-distribution: 3

5. $\bigl(\neg p(a)\mlc \neg p(a)\bigr)\mld \ade^\mathfrak{c}xp(x)$ \hspace{5pt} Chexists-choosing: 4

6. $\ada^\mathfrak{a}x \Bigl(\bigl(\neg p(x)\mlc \neg p(x)\bigr)\mld \ade^\mathfrak{c}xp(x)\Bigr)$ \hspace{5pt} Chall-splitting: 5

7. $\ada^\mathfrak{a}x \bigl(\neg p(x)\mlc \neg p(x)\bigr)\mld \ade^\mathfrak{c}xp(x)$ \hspace{5pt} Chall-distribution: 6

8. $\Bigl(\ada^\mathfrak{a}x \bigl(\neg p(x)\mlc \neg p(x)\bigr)\mld \ade^\mathfrak{c}xp(x)\Bigr)\adc^\mathfrak{b}\Bigl(\ada^\mathfrak{a}x \bigl(\neg p(x)\mlc \neg p(x)\bigr) \mld \ade^\mathfrak{c}xp(x)\Bigr)$ \hspace{5pt} Chand-splitting: 7,7

9. $\ada^\mathfrak{a}x \bigl(\neg p(x)\mlc \neg p(x)\bigr)\adc^\mathfrak{b}\ada^\mathfrak{a}x \bigl(\neg p(x)\mlc \neg p(x)\bigr) \mld \ade^\mathfrak{c}xp(x)$ \hspace{5pt}Chand-distribution: 8

10. $\ada^\mathfrak{a}x \bigl(\neg p(x)\mlc \ada^\mathfrak{a}x\neg p(x)\bigr)\adc^\mathfrak{b}\ada^\mathfrak{a}x \bigl(\ada^\mathfrak{a}x \neg p(x)\mlc \neg p(x)\bigr) \mld \ade^\mathfrak{c}xp(x)$ \hspace{5pt} Chall-cleansing: 9 (twice)

11. $\bigl(\ada^\mathfrak{a}x\neg p(x)\mlc \ada^\mathfrak{a}x\neg p(x)\bigr)\mld \ade^\mathfrak{c}xp(x)$ \hspace{5pt} Challchotomy: 10
\end{example}

\section{The preservation lemma}

A {\bf surface occurrence} of a subcirquent or an operator  in a given cirquent is an occurrence which is not in the scope of a choice operator (i.e., of $\adc^\mathfrak{c}$, $\add^\mathfrak{c}$, $\ada^\mathfrak{c}$ or $\ade^\mathfrak{c}$ for whatever $\mathfrak{c}$).

\begin{definition}\label{residue}
Given a closed cirquent $C$ and a legal run $\Gamma$, the {\bf $\Gamma$-residue} of $C$ is the  $\adc,\add,\ada,\ade$-free cirquent obtained from $C$ as a result of repeatedly replacing  until no longer possible:

1. every surface occurrence of every $\Gamma$-resolved $\adc$-, $\add$-, $\ada$- or $\ade$-rooted subcirquent by its resolvent;

2. every surface occurrence of every $\Gamma$-unresolved $\adc$- or $\ada$-rooted subcirquent by $\top$;

3. every surface occurrence of every $\Gamma$-unresolved $\add$- or $\ade$-rooted subcirquent by $\bot$.

\end{definition}

\begin{lemma}\label{f1}
Consider any closed cirquent $C$, interpretation $^*$  and legal run $\Gamma$. Let $R_C$ be the $\Gamma$-residue of $C$. Then  $\Gamma$ is a won (by the machine) run of $C^*$ iff $R_{C}^{*}$ is true.
\end{lemma}
\begin{proof} Let $C$, $^*$, $\Gamma$, $R_{C}$ be as above. We proceed by induction on the complexity of $C$.

If $C$ is a literal, then $R_C=C$. Hence, from clause 1 of Definition \ref{semantics} (and the fact that all other clauses of that definition are about non-literal cases), $\Gamma$ is a won run of $C^*$  iff $R_{C}^{*}$ is true.

If $C$ is of the form $A\mlc B$, then obviously $R_{C} =R_A\mlc R_B$, where $R_A$ is the residue of $A$ and $R_B$ is the residue of $B$. By Definition \ref{semantics},  $\Gamma$ is a won run of  $(A\mlc B)^*$ iff it is a won run of both $ {A}^{*}$ and $ {B}^{*}$. But, by the induction hypothesis, $\Gamma$ is a won run of both $A^*$ and $B^*$   iff both $R_{A}^{*}$ and $R_{B}^{*}$ are true. ``Both $R_{A}^{*}$ and $R_{B}^{*}$ are true'', in turn, means nothing but that $R_{A}^{*}\mlc R_{B}^{*}$, i.e., $R_{C}^{*}$, is true.

The case of $C$ having the form $A\mld B$ will be handled similarly.

Suppose $C$ is of the form $A\adc^\mathfrak{c} B$. If $C$ is unresolved, then, by clause 4 of Definition \ref{semantics}, $\Gamma$ is a won run of $C^*$. But then, as desired, $R_{C}^{*}$ is true because $R_{C}=\top$. Now suppose $C$ is resolved. Without loss of generality we may assume that the resolvent is $A$. Notice that then $R_{C}$ is the residue of not only $C$ but also of $A$. We have: $\Gamma$ is a won run of $C^*$ iff  (by clause 3 of Definition \ref{semantics}) it is a won run of $A^*$ iff (by the induction hypothesis) $R_{C}^{*}$ is true, as desired.

The remaining cases of $C$ having the form $A\add^\mathfrak{c} B$, $\ada^\mathfrak{c} x A(x)$ or  $\ade^\mathfrak{c} x A(x)$ will be handled similarly.
\end{proof}

\begin{lemma} \label{pres}
\

1. Each application of any of the rules of $\mmm$ preserves logical validity in the premises-to-conclusion direction, i.e., if all premises are   valid, then so is the conclusion.

2. Each application of any of the rules of $\mmm$ other than (the three versions of) Choosing also preserves logical validity in the conclusion-to-premises direction, i.e., if the conclusion is   valid, then so are all premises.
\end{lemma}

\begin{proof} As in \cite{cl16}, we will implicitly rely on the clean environment assumption, allowing us to rule out the possibility that the environment ever makes any illegal moves. We shall also implicitly rely on the straightforward fact that    if, for every   interpretation $^*$,  computability of a cirquent $A$ under $^*$ implies computability   of a cirquent $B$ under the same $^*$,    then  logical validity  of $A$ implies logical   validity of $B$.

If $E\leadsto F$ is an application of any of the rules other than Splitting, Chotomy or Choosing, it is not hard to see that, for any interpretation $^*$,  $E^*$  and $F^*$ are identical as games. So, a logical solution of $E$ is automatically a logical solution of $F$, and vice versa. Let us just look at Chall-cleansing as an   example. Consider an application $E\leadsto F$ of this rule, where $E=X\bigl[\ada^\mathfrak{c} xY(x)[A(x)]\bigr]$ and $F=X\bigl[\ada^\mathfrak{c} xY(x)[\ada^\mathfrak{c} yA(y)]\bigr]$. Fix some arbitrary interpretation for the present context, and let $\Gamma$ be an arbitrary legal run.  We want to show that $\Gamma$ is a won run of $E $ iff it is a won run of $F $. If $\mathfrak{c}$ is $\Gamma$-unresolved, then  $\Gamma$ is a won run of the  $ \ada^\mathfrak{c} xY(x)[\ada^\mathfrak{c} yA(y)] $ component of the conclusion  just as it is a won run of  the $ \ada^\mathfrak{c} xY(x)[A(x)] $ component of the premise. Then, since $E$ and $F$ only differ in that one has $\ada^\mathfrak{c} xY(x)[\ada^\mathfrak{c} yA(y)]$ where the other has $\ada^\mathfrak{c} xY(x)[A(x)]$, we find that $\Gamma$ is a won run of both games $E ,F $ or neither. Now assume $\mathfrak{c}$ is resolved, i.e., $\Gamma$ contains the move $\mathfrak{c}.a$ for some constant $a$. Then  $\Gamma$ is a won run of  $ \ada^\mathfrak{c} xY(x)[\ada^\mathfrak{c} yA(y)] $ iff it is a won run  of $ Y(a)[A(a)] $  iff it is a won run of   $ \ada^\mathfrak{c} xY(x)[A(x)] $. This, again,    implies that $\Gamma$ is a won run of both games $ E , F $ or neither.

{\em Chor-choosing} is taken care of in the proof of Lemma 6.1 of \cite{cl16}, which shows that this rule preserves computability under any given interpretation.  The same can be said about either direction of
{\em Chand-splitting}.

Consider an application $X[A_1(a),\cdots,A_n(a)]\leadsto  X[\ade^\mathfrak{c} x_1A_1(x_1),\cdots,\ade^\mathfrak{c} x_nA_n(x_n)]$ of {\em Chexists-choosing}, and assume $\cal M$ is a logical
solution of the premise. Let ${\cal N}$ be an HPM that, at the beginning of the play, makes the move $\mathfrak{c}.a$, after which it plays exactly as
$\cal M$ would. Obviously $\cal N$
is a logical solution of  the conclusion.

Of the three {\em Chotomy} rules, let us just consider Challchotomy, with the remaining two rules being similar. Consider an application
\[X [\ada^\mathfrak{a}x  (A \mlc \ada^\mathfrak{b}yB   )\adc^\mathfrak{c} \ada^\mathfrak{b}y  (\ada^\mathfrak{a} xA \mlc B   ) ] \leadsto  X [\ada^\mathfrak{a}xA \mlc \ada^\mathfrak{b}yB  ] \]
of Challchotomy.  It is not hard to see that, under any interpretation,  any won run of the conclusion  is also (``even more so'') a won run of the premise, meaning that a   solution of the conclusion  would automatically also be a   solution of the premise. This takes care of the conclusion-to-premise direction. For the premise-to-conclusion direction, fix an arbitrary interpretation and assume that $\cal M$ is a solution of the premise. Let $\cal N$ be an HPM that, until  it sees that its environment has resolved either $\mathfrak{a}$ or $\mathfrak{b}$, plays just as $\cal M$ would play in the scenario where $\mathfrak{c}$ is not (yet) resolved but otherwise $\cal M$'s imaginary environment is making the same moves as $\cal N$'s real environment is making. If and when it sees that a move $\mathfrak{a}.i$ (resp. $\mathfrak{b}.i$) has been made by its environment, $\cal N$  imagines that $\cal M$'s environment has correspondingly made not only the same move $\mathfrak{a}.i$ (resp. $\mathfrak{b}.i$) but also  $\mathfrak{c}.0$  (resp. $\mathfrak{c}.1$), and  continues playing exactly as $\cal M$ would continue playing in that case. With a little thought, $\cal N$ can be seen to be a solution of the conclusion.

Finally, consider an application $A(a)\leadsto  \ada^\mathfrak{c} xA(x)$ of {\em Chall-splitting}.

For the conclusion-to-premise direction, assume $\cal M$ is a logical solution of $ \ada^\mathfrak{c} xA(x)$. Let $\cal N$ be the HPM that plays exactly as $\cal M$ would play in the scenario where, at the very beginning of the play, the environment makes the move $\mathfrak{c}.a$. It is not hard to see that $\cal N$ is a logical solution of $A(a)$.

For the premise-to-conclusion direction,
assume $\cal M$ is a logical solution of $A(a)$.
Let $\cal N$ be an HPM that plays $\ada^\mathfrak{c} xA(x)$ as follows. While continuously polling its run tape, $\cal N$ maintains a list $L$ of  moves, initially empty. This is just to keep track of which moves made by the environment have already been ``processed'' by $\cal N$. Call the moves that are not in $L$ {\em unprocessed}. $\cal N$ further maintains a partial function   $f:\mathbb{N}\rightarrow \mathbb{N}$. Call the value of $f(c)$ the {\em image} of $c$, and call the constants at which $f$ is not (yet) defined {\em imageless}. Initially, the image of each constant occurring in $A(x)$ is that constant itself, and all other constants are imageless.

At the beginning of the play, $\cal N$ waits till the environment makes the move $\mathfrak{c}.b$ for some constant $b$.
If and when this happens (and if not, $\cal N$ wins $\ada^\mathfrak{c} xA(x)$ by doing nothing), $\cal N$ adds $\mathfrak{c}.b$ to $L$, declares $b$ to be the image of (the so far imageless) $a$, and  starts simulating an imaginary play of $A(a)$ by $\cal M$. In this simulation:
\begin{itemize}
  \item  Whenever $\cal N$ sees an unprocessed move $\mathfrak{a}.i$ on its run tape where $\mathfrak{a}$ is a $\adc$-cluster, it adds this move to $L$  and appends $\bot \mathfrak{a}.i$  to the content of the imaginary run tape of $\cal M$.
  \item  Whenever $\cal N$ sees an unprocessed move $\mathfrak{a}.c$ on its run tape where $\mathfrak{a}$ is a $\ada$-cluster, it adds this move to $L$    and appends  $\bot \mathfrak{a}.d$  to the content of the imaginary run tape of $\cal M$, where $d$ is an (say, the smallest) imageless constant; after that $\cal M$ declares $c$ to be the image of $d$.
\item Whenever $\cal N$ sees that the simulated $\cal M$ made a move $\mathfrak{a}.i$
where $\mathfrak{a}$ is a $\add$-cluster,  it makes the same move  $\mathfrak{a}.i$ in its real play.
\item Whenever $\cal N$ sees that the simulated $\cal M$ made a move $\mathfrak{a}.c$ where $\mathfrak{a}$ is a $\ade$-cluster and $c$ is imageless,  $\cal N$  makes the  same move  $\mathfrak{a}.c$ in its real play,  and declares $c$ to be its own image.
\item Whenever $\cal N$ sees that the simulated $\cal M$ made a move $\mathfrak{a}.c$ where $\mathfrak{a}$ is a $\ade$-cluster and $c$ is not imageless,  $\cal N$  makes the   move  $\mathfrak{a}.d$ in its real play, where $d$ is the image of $c$.
\end{itemize}
We claim that $\cal N$ is a logical solution of  $\ada^\mathfrak{c} xA(x)$. To see this, consider an arbitrary interpretation $^*$ and an arbitrary ``real'' play by $\cal N$. Let $\Gamma_{\cal N}$ be the run that has taken place in that play, and $\Gamma_{\cal M}$ be the run that has correspondingly taken place in $\cal M$'s  play as imagined by $\cal N$. Our goal is to show that $\Gamma_{\cal N}$ is a won run of $\bigl(\ada^\mathfrak{c} xA(x)\bigr)^*$. Let $R_{\cal M}$ be the $\Gamma_{\cal M}$-residue of $A(a)$, and let $c_1,\cdots,c_n$ be all constants occurring in $R_{\cal M}$. Notice that each such constant has acquired  an image at some time during the work of $\cal N$ (and never lost or changed it afterwards).   Let $R_{\cal N}$ be the result of replacing in $R_{\cal M}$ each occurrence of each $c_i\in\{c_1,\cdots,c_n\} $ by the image of $c_i$. Let $^\circ$ be an interpretation such that, for any atom $X$ of $R_{\cal M}$, $X^\circ$ is true iff so is $Y^*$, where $Y$ is the result of replacing all constants in $X$ by their images.

An analysis of the work of $\cal N$, details of which are left to the reader, reveals  that   $R_{\cal N}$ is the $\Gamma_{\cal N}$-residue of $\ada^\mathfrak{c} x A(x)$, and that $R_{\cal N}^{*}$ is true iff  so is $R_{\cal M}^{\circ}$. Since $\cal M$ is a logical solution of $A(a)$, $\Gamma_{\cal M}$ is a won run of $A^\circ(a)$. By Lemma \ref{f1}, this implies that $R_{\cal M}^{\circ}$ is true. Hence so is  $R_{\cal N}^{*}$, which, again by Lemma \ref{f1}, implies that $\Gamma$ is a won run of $\bigl(\ada^\mathfrak{c} xA(x)\bigr)^*$, as desired.
\end{proof}

\section{Rank  and purification}
Our proofs in this section closely  follow those given in Section 7 of \cite{cl16}.
As in \cite{cl16}, $^na$ means ``tower of $a$'s of height $n$'' (tetration), defined inductively by $^1 a= a$ and $^{n+1}a=a^{( ^na)}$.
\begin{definition}\label{rankdef}
The {\bf rank} $\Rank(C)$ of a cirquent $C$ is the number defined as follows:

1. If $C$ is  a literal, then its rank is $1$.

2. If $C$ is  $A\add^\mathfrak{c} B$ or $A\adc^\mathfrak{c} B$, then its rank is $\Rank(A)+\Rank(B)$.

3. If $C$ is  $\ade^\mathfrak{c} xA$ or $\ada^\mathfrak{c}x A$, then its rank is $\Rank(A)+1$.

4. If $C$ is  $A\mlc B$, then its rank is $5^k$, where $k=\Rank(A)+\Rank(B)$.\footnote{In fact, a smaller number can be taken here and below instead of $5$, but why bother.}

5. If $C$ is $A\mld B$, then its rank is \ $^k5$, where $k=\Rank(A)+\Rank(B)$.
\end{definition}

Due to due to the monotonicity of the functions $x+y$, $x+1$, $5^x$ and $^x5$, we have:
\begin{lemma}\label{monot}
The rank function is {\bf monotone} in the following sense. Consider a cirquent $A$ with a subcirquent $B$. Assume $B'$ is a cirquent with {\em $\Rank(B')< \Rank(B)$}, and $A'$ is the result of replacing an occurrence of $B$ by $B'$ in $A$. Then {\em $\Rank(A')< \Rank(A)$}.
\end{lemma}

\begin{definition}\label{pd}
 We say that a cirquent $E$ is {\bf pure} iff the following conditions are satisfied:

1. $E$ has no surface occurrence  of $\bot$ unless $E$ itself is $\bot$.

2. $E$ has no surface occurrence of $\mlc$ which is in the scope of $\mld$.

3. $E$ has no surface occurrence of $\adc^\mathfrak{c}$ or $\ada^\mathfrak{c}$ (whatever cluster $\mathfrak{c}$) which is in the scope of $\mld$.

4. $E$ has no surface occurrence of the form $A_1\mld\cdots\mld A_n$ such that, for some atom $A$, both
$A$ and $\neg A$ are among $A_1,\cdots,A_n$.

5. $E$ has no surface occurrences of $\top$ unless $E$ itself is $\top$.

6. If $E$ is of the form $A_1\mlc\cdots\mlc A_n$ ($n\geq 2$), then at least one $A_i$ ($1\leq i\leq n$) is neither $\adc$- nor $\ada$-rooted.

7. If $E$ is of the form $A\adc^\mathfrak{c} B$, then neither $A$ nor $B$ contains the cluster $\mathfrak{c}$.

8. If $E$ is of the form $\ada^\mathfrak{c} xA$, then   $A$ does not contain  the cluster $\mathfrak{c}$.
\end{definition}

Below we describe a procedure which takes a cirquent $E$ and applies to it a series of modifications. Each modification changes the value of $E$ so that the old value of $E$ follows from the new value by one of the  rules  of $\mmm$ other than Choosing and Splitting. The procedure is divided into eight stages, and the purpose of each stage $i\in\{1,\cdots,8\}$ is to make $E$ satisfy the corresponding condition $\#i$ of Definition \ref{pd}.\vspace{5pt}

{\bf Procedure Purification} applied to a cirquent $E$:
Starting from Stage 1, each  of the following eight stages is a loop that should be iterated until it no longer modifies (the current value of) $E$; then the procedure goes to the next stage, unless the current stage was Stage 8, in which case the procedure terminates and returns (the then-current value of) $E$.

{\em Stage 1}:   If $E$ has a surface occurrence of the form $\bot\mld A$ or  $A\mld\bot$, change the latter to $A$
using Por-identity perhaps in combination with Por-commutativity.  Next, if $E$ has a surface occurrence of the form $\bot\mlc A$  or $A\mlc\bot$, change it to $\bot$  using Pand-domination   perhaps in combination with Pand-commutativity.

{\em Stage 2}: If $E$ has a surface occurrence of the form $(A\mlc B)\mld C$ or $C\mld(A\mlc B)$, change it to $(A\mld C)\mlc (B\mld C)$ using Pand-distribution perhaps in combination with Por-commutativity.

{\em Stage 3}:   (a) If $E$ has a surface occurrence of the form  $A\adc^\mathfrak{c} B \mld C$ or $C\mld A\adc^\mathfrak{c} B $, change it to $(A\mld C)\adc^\mathfrak{c} (B \mld C)$ using Chand-distribution perhaps in combination with Por-commutativity. (b) Next, if $E$ has a surface occurrence of the form  $\ada^\mathfrak{c}xA  \mld B$ or $B\mld\ada^\mathfrak{c}xA$, change it to $\ada^\mathfrak{c}x (A \mld B )$ using Chall-distribution perhaps in combination with Por-commutativity.

{\em Stage 4}: If $E$ has a surface occurrence of the form $A_1\mld \cdots\mld A_n$ and, for some atom $A$, both
$A$ and $\neg A$ are among $A_1,\cdots,A_n$, change $A_1\mld \cdots\mld A_n$ to $\top$ using Trivialization, perhaps in combination with Por-domination, Por-commutativity and Por-associativity.

{\em Stage 5}: If $E$ has a surface occurrence of the form $\top\mld A$ or $A\mld \top$, change it to  $\top$  using Por-domination   perhaps in combination with Por-commutativity. Next, if $E$ has a surface occurrence of the form
$\top \mlc  A$ or $A\mlc\top$,  change it to $A$
using Pand-identity perhaps in combination with Pand-commutativity.

{\em Stage 6}: In all three cases below, $\mathfrak{c}$ is a $\adc$-cluster not occurring in $E$. (a) If $E$ has a surface occurrence of the form $ A \adc^\mathfrak{a} B \mlc  C \adc^\mathfrak{b} D $, change it to
$\bigl( (A\mlc  C \adc^\mathfrak{b} D  )\adc^\mathfrak{a}  (B\mlc  C \adc^\mathfrak{b} D  )\bigr)
\adc^\mathfrak{c} \bigl( ( A\adc^\mathfrak{a} B \mlc C )\adc^\mathfrak{b} ( A\adc^\mathfrak{a} B \mlc D )\bigr)$  using Chandchotomy.
  (b) Next, if $E$ has a surface occurrence of the form $\ada^\mathfrak{a}xA \mlc \ada^\mathfrak{b}yB $, change it to
$\ada^\mathfrak{a}x (A \mlc \ada^\mathfrak{b}yB  )\adc^\mathfrak{c} \ada^\mathfrak{b}y  (\ada^\mathfrak{a} xA \mlc B   )$ using Challchotomy.  (c) Next, if $E$ has a surface occurrence of the form $ A\adc^\mathfrak{a} B \mlc \ada^\mathfrak{b}xC $ or
$ \ada^\mathfrak{b}xC\mlc A\adc^\mathfrak{a} B $,
  change it to $\bigl( (A\mlc \ada^\mathfrak{b}xC   )\adc^\mathfrak{a} (B\mlc \ada^\mathfrak{b}xC  )\bigr)\adc^\mathfrak{c} \ada^\mathfrak{b} x  ( A\adc^\mathfrak{a} B \mlc C   )$ using Chandallchotomy perhaps in combination with Pand-commutativity.

{\em Stage 7}: If $E$ is  of the form $X[A\adc^\mathfrak{c} B]\adc^\mathfrak{c} C$ (resp. $C\adc^\mathfrak{c} X[A\adc^\mathfrak{c} B]$), change it to $X[A]\adc^\mathfrak{c}C$ (resp. $C\adc^\mathfrak{c} X[B]$) using Left (resp. Right) chand-cleansing.

{\em Stage 8}: If $E$ is of the form $ \ada^\mathfrak{c} xX[\ada^\mathfrak{c} yA(y)] $, change it to $ \ada^\mathfrak{c} xX[A(x)] $  using Chall-cleansing.

\

\begin{lemma}\label{terminate}
Each stage of the Purification procedure strictly reduces the rank of $E$.
\end{lemma}

\begin{proof} Each stage replaces an occurrence of a subcirquent $A$ of $E$ by some cirquent $B$. In view of Lemma \ref{monot}, in order to show that such a replacement reduces the rank of $E$, it is sufficient to show that $\Rank(B)<\Rank(A)$.   Here we shall only consider Stages 3(b), 6(b,c) and 8, as all other stages or cases are covered in the proof of Lemma 7.4 of \cite{cl16}.

{\em Stage 3(b)}: With $k$ abbreviating $\Rank(A)+\Rank(B)$, the rank of $\ada^\mathfrak{c}xA  \mld B$ or $B\mld\ada^\mathfrak{c}xA$ is $^{k+1}5$, and the rank of   $\ada^\mathfrak{c}x (A \mld B )$ is $^{k }5+1$. The latter is clearly smaller than the former.

{\em Stage 6(b)}:   With $k$ abbreviating $\Rank(A)+\Rank(B)$, the rank of $\ada^\mathfrak{a}xA \mlc \ada^\mathfrak{b}yB $   is $5^{k +2}$, and the rank of
$\ada^\mathfrak{a}x (A \mlc \ada^\mathfrak{b}yB  )\adc^\mathfrak{c} \ada^\mathfrak{b}y  (\ada^\mathfrak{a} xA \mlc B   )$
is $5^{k+1 }+5^{ k+1 }   $. Of course the latter is smaller than the former.

{\em Stage 6(c)}:  With $a$, $b$ and $c$ standing for the ranks of $A$, $B$ and $C$, respectively, the rank of $ A \adc^\mathfrak{a} B \mlc \ada^\mathfrak{b}xC$ or  $\ada^\mathfrak{b}xC\mlc  A \adc^\mathfrak{a} B $ is $5^{a+b+c+1}$, and the rank of
$\bigl( (A\mlc \ada^\mathfrak{b}xC   )\adc^\mathfrak{a}  (B\mlc \ada^\mathfrak{b}xC )\bigr)\adc^\mathfrak{c} \ada^\mathfrak{b}x ( A \adc^\mathfrak{a} B \mlc C  ) $
is $5^{ a+c+1 }+5^{b+c+1 }+5^{ a+b+c }  +1$. Obviously the latter is smaller than the former.

{\em Stage 8}: Each iteration of this stage replaces a subcirquent $\ada^\mathfrak{c} yA(y)$ by $A(x)$. Since the rank of  $ A(x)$ is obviously the same as the rank of  $A(y)$, the rank $\Rank\bigl(A(x)\bigr)+1$ of $\ada^\mathfrak{c} xA(x)$ is greater than the rank $\Rank\bigl(A(y)\bigr)$ of $A(y)$.
 \end{proof}

Where $A$ is the initial value of $E$ in the Purification procedure and $B$ is its final value (which exists by Lemma \ref{terminate}), we call $B$ the {\bf purification} of $A$.

\begin{lemma}\label{pl}
 For any closed cirquent $E$ and its purification $F$, we have:

1. If $F$ is provable in $\mmm$, then so is $E$.

2. $E$ is   valid iff so is $F$.

3. $F$ is pure.

4. The rank of $F$ does not exceed the rank of $E$.
\end{lemma}

\begin{proof} {\em Clause 1}: When obtaining $F$ from $E$, each transformation performed during the Purification procedure applies, in the conclusion-to-premise direction, one of the inference rules of $\mmm$. Reversing the order of those transformations, we get a derivation of $E$ from $F$ in $\mmm$. Appending that derivation to a proof of $F$
(if one exists) yields a proof of $E$.

{\em Clause 2}: Immediate from the two clauses of Lemma \ref{pres} and the fact that, when obtaining $F$ from $E$
using the Purification procedure, Choosing is never used.

{\em Clause 3}: One by one, Stage 1 eliminates all surface occurrences of $\bot$ in $E$ (unless $E$ itself is $\bot$). So, at the end of the stage, $E$ satisfies condition 1 of Definition \ref{pd}. None of the subsequent steps make $E$ violate that condition, so $F$, too, satisfies that condition.
 Similarly, a routine examination of the situation reveals that Stage 2 (resp. 3, \ldots, resp. 8) of the Purification procedure makes $E$ satisfy condition 2 (resp. 3, \ldots, resp. 8) of Definition \ref{pd}, and $E$ continues to satisfy that condition throughout the rest of the stages. So, $F$ is pure.

{\em Case 4}:  Immediate from Lemma  \ref{terminate}. \end{proof}

\section{The soundness and completeness of $\mmm$}
\begin{theorem}\label{theo}
A closed cirquent $E$ is   valid if (soundness) and only if (completeness) $\mmm\vdash E$.
\end{theorem}

\begin{proof} The soundness part is immediate from clause 1 of Lemma \ref{pres} and the fact that the axiom $\top$ is   valid. The rest of this section is devoted to a proof of the completeness part. Pick an arbitrary closed cirquent $E$, and let $F$ be its purification.   We proceed by induction on the
rank of $E$.

By clauses 3-4 of Lemma \ref{pl}, $F$ is a pure  cirquent whose rank does not exceed that of
$E$. We shall implicitly rely on this fact below. In view of $F$'s being pure, after some analysis it is clear that   one of the following conditions should be satisfied:

\begin{description}
  \item[Condition 0.] $F$ is either $\bot$ or a nonlogical literal.\vspace{-7pt}
  \item[Condition 1.] $F$ is $\top$.\vspace{-7pt}
  \item[Condition 2.] $F$ is of the form $A_0 \add^\mathfrak{c}  A_1$.\vspace{-7pt}
  \item[Condition 3.] $F$ is of the form $\ade^\mathfrak{c} xA(x)$.\vspace{-7pt}
  \item[Condition 4.] $F$ is of the form $A_0 \adc^\mathfrak{c} A_1$, and neither $A_0$ nor $A_1$ contains the cluster $\mathfrak{c}$.\vspace{-7pt}
  \item[Condition 5.] $F$ is of the form $\ada^\mathfrak{c} xA(x)$, and $A(x)$ does not contain the cluster $\mathfrak{c}$.\vspace{-7pt}
  \item[Condition 6.] $F$ is of the form $A_1\mld\cdots\mld A_n$ ($n\geq 2$),  where each disjunct  is either    a nonlogical literal, or $\add$-rooted, or $\ade$-rooted; besides, for no atom $A$ do we have that both $A$ and $\neg A$ are among $A_1,\cdots,A_n$.\vspace{-7pt}
  \item[Condition 7.] $F$ is of the form $B_1\mlc\cdots\mlc B_m$ ($m\geq 2$), where at least one conjunct $B_e$ is  either   (0) a nonlogical literal, or (1) $\add$-rooted, or  (2) $\ade$-rooted, or (3) satisfies   Condition 6 in the role of $F$.
\end{description}

Assume $E$ is valid. We want to show that then $E$ is provable. For this, in view clause 1 of Lemma \ref{pl}, it is sufficient to show that $F$ is provable. Keep in mind that, by clause 2  of Lemma \ref{pl},  $F$ is   valid.   This immediately rules out Condition 0, because, of course, neither $\bot$ nor nonlogical literals are  valid. So, we only need to show that $F$ is provable in each of the following seven cases:

{\em Case 1}:  $F$ is $\top$  as in Condition 1. Then $F$ is an axiom and hence provable.

{\em Case 2}: $F$ is $A_0 \add^\mathfrak{c} A_1$ as in Condition 2. Let $\cal M$ be a logical solution of $F$.
Consider the work of $\cal M$ in the scenario where the environment does not move until $\cal M$ makes the move $\mathfrak{c}.i$ for one of $i\in\{0,1\}$. Sooner or later $\cal M$ has to make such a move, for otherwise $F$ would be lost
due to being $\add^\mathfrak{c}$-rooted.  Since in the games that we deal with the order of moves is irrelevant, without loss of  generality we may assume that the move $\mathfrak{c}.i$ is made by $\cal M$ before any other moves. Let $D$ be the result of replacing in $F$ all subcirquents of the form $X_0\add^\mathfrak{c} X_1$ by $X_i$.  Observe that, after the move $\mathfrak{c}.i$ is made, in any scenario that may follow, $\cal M$ has to continue and win $D$. This means that  $\cal M$ is a logical solution of (not only $F$ but also) $D$. Thus, $D$ is valid. The rank of $D$ is of course smaller than that of $F$. Hence, by the induction hypothesis, $D $  is provable. Then so is $F$ because it follows from $D $ by Chor-choosing.

{\em Case 3}:  $F$ is $\ade^\mathfrak{c} xA(x)$ as in Condition 3. This case is rather similar to the preceding one. Let  $\cal M$ be a logical solution of $F$. Consider the work of $\cal M$ in the scenario where the environment does not move until $\cal M$ makes the move $\mathfrak{c}.a$ for some constant $a$.
 Sooner or later $\cal M$ has to make such a move, for otherwise $ F$ would be lost  due to   being $\ade^\mathfrak{c}$-rooted. As in case 2, we may assume that the move  $\mathfrak{c}.a$  is made  before any other  moves. Let $D$ be the result of replacing in $F $ all subcirquents of the form $\ade^\mathfrak{c} yX(y)$ by $X(a)$.  After the move $\mathfrak{c}.a$ is made, in any scenario that may follow, $\cal M$ has to continue and win $D$. This means that  $\cal M$ is a logical solution of  $D$. The rank of  $D$ is smaller than that of $F$. Hence, by the induction hypothesis, $D$  is provable. Then so is $F$, as it follows from $D$ by Chexists-choosing.

{\em Case 4}: $F=A_0 \adc^\mathfrak{c} A_1$ is as in Condition 4. By clause 2 of Lemma \ref{pres}, both $A_0$ and $A_1$ are valid, because $F$ follows from either one by Chand-splitting. Both $\Rank(A_0)$ and $\Rank(A_1)$ are smaller  than  $\Rank(F)$. Hence, by the induction hypothesis, both $A_0$ and $A_1$ are provable. Therefore, by Chand-splitting, so is $F$.

{\em Case 5}: $F=\ada^\mathfrak{c} xA(x)$ is as in condition 5.   Let $c$ be a constant not occurring in $F$. By clause 2 of Lemma \ref{pres},  $A(c)$ is   valid, because $F$ follows from it by Chall-splitting. The rank of $A(c)$ is smaller  than that of $F$. Hence, by the induction hypothesis,   $A(c)$ is provable. Therefore, by Chall-splitting, so is $F$.

{\em Case 6}:  $F=A_1\mld\cdots\mld A_n$ is as in Condition 6.    Not all of the cirquents $A_1,\ldots,A_n$ can be literals, for otherwise $F$ would be automatically lost under an interpretation that makes   all those literals false, contrary to our assumption that $F$  is   valid. With this observation in mind, without loss of generality, we may assume that, for some
$k,m$ with $1\leq k+m \leq n$, the first $k$ cirquents $A_1,\cdots,A_k$ are of the form $B_{0}^{1}\add^{\mathfrak{b}_1} B_{1}^{1}$, \ldots,
$B_{0}^{k}\add^{\mathfrak{b}_k} B_{1}^{k}$, the next $m$ cirquents are of the form $\ade^{\mathfrak{c}_1}x_1C_1(x_1), \cdots, \ade^{\mathfrak{c}_m}x_mC_m(x_m) $, and the remaining $n-k-m$ cirquents  are literals. Let $\cal M$ be a
logical solution of $F$. Consider the work of $\cal M$ in the scenario where the environment does not move until $\cal M$ makes either (a)   the move $\mathfrak{b}_j.i$  for some $j\in\{1,\cdots,k\}$ and  $i\in\{0,1\}$,  or (b)
the move $\mathfrak{c}_j.a$  for some $j\in\{1,\cdots, m\}$ and  $a\in\mathbb{N}$. At some point, $\cal M$ should indeed make such a move, for otherwise $F$
would be lost under an(y) interpretation which makes all of the literal cirquents $A_{k+m+1},\cdots,A_n$ false.
  In case (a), let $D$ be the result of replacing in $F$ every subcirquent of the form $X_0 \add^{\mathfrak{b}_j} X_1$  by $X_i$; in case (b), let $D$ be the result of replacing in $F$ every subcirquent of the form $ \ada^{\mathfrak{c}_j} y X (y)$  by $X (a)$.  With some analysis left to the reader, $\cal M$ can be seen to be a logical solution of $D$. Thus, $D$ is   valid. The rank of $D$ is smaller than that of $F$ and hence, by the induction hypothesis, $D$ is provable. But then so is $F$, because it follows from $D$ by Choosing.

{\em Case 7}: $F=B_1\mlc\cdots\mlc B_m$ and $B_e$ are is as in Condition 7.
 The  validity of $F$, of course, implies that
$B_e$, as one of its $\mlc$-conjuncts, is also  valid. This rules out the possibility that $B_e$ is a nonlogical literal, because, as we observed earlier, a nonlogical literal cannot be   valid. Therefore  we are left with one of the following three possible subcases, corresponding to subconditions (1), (2) and (3) of Condition 7:

{\em Subcase 7.1}: $B_e$ is of the form $C_0\add^\mathfrak{c} C_1$. The argument given for Subcase 5.1 in the proof of Theorem 7.6 of \cite{cl16} goes through without any changes.

{\em Subcase 7.2}: $B_e$ is of the form $\ade^\mathfrak{c} xC(x)$. Let $\cal M$ be a logical solution of $F$.  Consider the work of $\cal M$ in the scenario where the environment does not move until $\cal M$ makes the move   $\mathfrak{c}.a$ for some constant $a$ (otherwise $F$ would be lost).
  Let $D$ be the result of replacing  in $F$  every subcirquent of the form $ \ade^\mathfrak{c} y X(y)$
 by $X(a)$. Then, as in  Case 3, $\cal M$ can be seen to be a logical solution of $D$, meaning that  $D$ is   valid. The rank of $D$ is smaller than that of $F$ and hence, by the induction hypothesis, $D $ is provable. But then so is $F$, because it follows from $D $ by Chexists-choosing.

{\em Subcase 7.3}: $B_e$ satisfies   Condition 6 in the role of $F$.  This
case is very similar to Case 6 and, almost literally repeating our reasoning in the latter, we find that $F$
is provable. \end{proof}

\section{The decidability of $\mmm$}
\begin{theorem}\label{theor}
$\mmm$   is decidable. Namely,  the algorithm   DECISION   described below accepts a closed cirquent $E$ if $\mmm\vdash E$ and rejects if $\mmm\not\vdash E$.
\end{theorem}

\begin{proof} Consider an arbitrary closed cirquent $E$. The algorithm  DECISION  given below is a recursive one. It terminates because  every recursive call strictly decreases the rank of the cirquent that is being processed. This is how the algorithm acts on input $E$:

First, using the Purification algorithm,  DECISION   constructs the purification $F$ of $E$. Note   that, in  view of Theorem \ref{theo}, ``provable'' and ``valid'' can (and will) be used interchangeably; additionally, by clause 2 of Lemma \ref{pl}, so can be ``$E$'' and ``$F$'' when we talk about their provability or computability.
 As pointed out in the proof of Theorem \ref{theo}, $F$ should satisfy one of the eight Conditions listed  in that proof.

If $F$ is $\bot$ or a nonlogical literal as in Condition 0, it is  invalid, and we let  DECISION  reject $E$.

If $F$ is $\top$ as in Condition 1, then it is valid, and we let  DECISION  accept  $E$.

Assume $F$ is $A_0 \add^\mathfrak{c}A_1$   as in Condition 2. Let $D_{i}$ ($i\in\{0,1\}$) be the result of  replacing in $F$ all subcirquents of the form $X_0\add^\mathfrak{c} X_1$ by $X_i$.    DECISION recursively calls itself  on $D_0$ and then on $D_1$ to figure out whether these cirquents are valid/provable. The ranks of both $D_0$ and $D_1$ are  smaller than the rank of  $F$ and hence, in view of clause 4 of Lemma \ref{pl}, smaller than the rank of $E$, as promised in the first paragraph of the present proof.
 If at least one of $D_0,D_1$ turns out to be provable,    we let   DECISION   accept $E$,    because $F$ follows from either cirquent by (one or more applications of) Chor-choosing. Otherwise, if    both $D_0,D_1$ turn out to be invalid,  we let  DECISION  reject $E$ because,    as (in fact) observed within Case 2 of the proof of Theorem \ref{theo}, if $F$ was valid, then so would be either $D_0$ or $D_1$.

For the subsequent cases, we merely state how   DECISION  acts. A verification of the adequacy of the corresponding acceptance/rejection decisions is left to  the reader.

Assume $F$ is $\ade^\mathfrak{c} xA(x)$ as in Condition 3. Let $\{a_1,\ldots,a_s\}$ be all constants occurring in $F$, and let $a_{s+1}$ be a constant not occurring in $F$.\footnote{In fact, there is no need for considering $a_{s+1}$ unless $s=0$, but let us be generous. The same comment applies to our treatment of the cases of $F$ being as in Condition 6 or Condition 7(2).}   DECISION runs itself on each of the  cirquents $A(a_1),\cdots, A(a_{s+1})$. If all of these are rejected, then DECISION rejects $E$, otherwise it accepts $E$.

Assume $F= A_0 \adc^\mathfrak{c} A_1$ is as in Condition 4. DECISION  runs itself on $A_0$ and then on $A_1$. If both are accepted, DECISION accepts $E$, otherwise it rejects $E$.

Assume $F=\ada^\mathfrak{c} xA(x)$ is as in Condition 5. Let $c$ be a constant not occurring in $F$. DECISION  calls itself on $A(c)$ and generates the same acceptance/rejection decision for $E$ as the call does for $A(c)$.

Assume $F=A_1\mld\cdots\mld A_n$    is as in Condition 6.  Let   $B_{0}^{1}\add^{\mathfrak{b}_1} B_{1}^{1}$, \ldots,
$B_{0}^{k}\add^{\mathfrak{b}_k} B_{1}^{k}$   and   $\ade^{\mathfrak{c}_1}x_1C_1(x_1)$, \ldots, $\ade^{\mathfrak{c}_m}x_mC_1(x_m) $  be as in Case 6 of the proof of Theorem \ref{theo}. Let $a_1,\ldots,a_s$ be all constants occurring in $F$, and let $a_{s+1}$ be a constant not found in $F$.
  For $i\in\{1,\cdots,k\}$ and $j\in\{0,1\}$, let $D_{i}^{j}$  be the result of replacing  in $F$  all subcirquents of the form $X_0\add^{\mathfrak{b}_i} X_1$ by $X_j$. Further, for $i\in\{1,\cdots,m\}$ and $j\in\{ 1,\cdots, s+1 \}$, let $G_{i}^{j} $ be the result of replacing  in  $F$  all subcirquents of the form $\ada^{\mathfrak{c}_i}yX (y) $ by $X(a_j)$. DECISION calls itself on each of the
cirquents  $D_{1}^{0},  \cdots,  D_{k}^{0}$, $D_{1}^{1},  \cdots,  D_{k}^{ 1}$,  \  $G_{1}^{1}, \cdots, G_{m}^{ 1}$, \ $\cdots$, \ $G_{1}^{s+1}, \cdots, G_{m}^{s+1}$.
If all calls reject their arguments, then   DECISION rejects $E$, otherwise it accepts $E$.

Finally, assume $F=B_1\mlc\cdots\mlc B_m$ and $B_e$ are   as in   Condition 7.

Assume $A_e$ is of the form $C_0\add^\mathfrak{c} C_1$ as in Condition 7(1).  For $i\in\{0,1\}$, let $D_i$   be the result of replacing  in $F$  all subcirquents of the form  $X_0\add^\mathfrak{c} X_1$ by $X_i$.  DECISION runs itself on $D_0$ and   $D_1$. If both cirquents are rejected, then DECISION rejects $E$, otherwise it accepts $E$.

Assume $A_e$ is of the form  $\ade^\mathfrak{c} xC(x)$ as in Condition 7(2). Let $a_1,\ldots,a_s$ be all constants occurring in $F$, and let $a_{s+1}$ be a constant not found in $F$.  For each $j\in\{1,\ldots,s+1\}$, let $D_j$ be the result of replacing in $F$ all subcirquents of the form $\ade^\mathfrak{c}yX(y)$ by $X(a_j)$.  DECISION runs itself on each of the arguments $D_1,\ldots,D_{s+1}$. If all $s+1$ cirquents are rejected,   DECISION rejects $E$, otherwise it accepts $E$.

Assume $A_e$ is as in Condition 7(3). With $A_e$ in the role of $F$,   DECISION acts exactly as in the above case of $F$ satisfying Condition 6.
\end{proof}

redpajama_set_name: RedPajamaArXiv


A collection that explores the relationship between People of color and america. This collection tells the story of how people of color have influenced many iconic american concepts , specifically the most american symbol of all: The Cowboy.
The iconic Image of the American cowboy was created by hollywood, however the true cowboys were heavily made up of black americans, Native americans, & Mexican Americans. The term "cowboy" was a term used for black men who worked with the cattle, since black men were not allowed to be called "men". They were often given the toughest jobs and were responsible to taming the horses. One of the most notable of these "Cowboys" was Bill pickett. Pickett actually went on to invent the rodeo.
this Goal of this collection is to push the viewer to think about how american identity is portrayed, and think about the stories that have been left out. this collection was created to celebrate the history of people of color in America, as we look toward their future.
Performance Look designed for artist Vic Mensa during his time on the JAY- Z "4:44 TOUR".
The SILHOUETTE draws upon vic mensa's love for punk aesthetic while seamlessly blending with his hip hop roots.
EACH OUTFIT IS TO BE WORN FOR MULTIPLE DAYS UNTIL THE AUDIENCE IS FULLY AWARE OF THE IDENTITY. THE SILHOUETTE CONSISTS OF A DIAGONAL BOXY OUTERWEAR PIECE THAT IS REMINISCENT OF WHITE BLUE COLLAR AMERICA. THE SILHOUETTE ALSO INCLUDES PATCHWORK LEATHER PANTS IN MULTIPLE ITERATIONS. THE T-SHIRTS ARE INTERCHANGEABLE BUT WE SUGGEST MESH SO VIC CAN CAPITALIZE OFF OF USING HIS TATTOOS AS MERCH GRApHICS IN THE IMMEDIATE FUTURE. WE'VE ALSO INCLUDED MESH T-SHIRTS THAT HAVE PANELS THAT ALLOWS VIC TO CHANGE HIS MESSAGING EVERY NIGHT LIKE A TWITTER FEED.


text:
#pragma once

enum class ConvertType { GRAY, HSL, HSL2, THERMOGRAPHIC };

redpajama_set_name: RedPajamaGithub


text:
FOOD INSECURITY IN RIVERSIDE
History of Black film in Toronto
NAVIGATE LIFE IN THE BIG CITY
Like a trusted friend, The Green Line's got your back with our handy guides to Toronto's many diverse neighbourhoods and local hot topics.
Intersectional Feminism
Hana (Frizz Kid) Shafi, an illustrator and writer, breaks down the complexities of modern feminism and explains how Torontonians can better understand the movement through an intersectional lens.
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THROUGH THE WAY WE LIVE
We want to acknowledge that the land on which The Green Line operates is the traditional territory of many nations, including the Mississaugas of the Credit, the Anishnabeg, the Chippewa, the Haudenosaunee and the Wendat peoples. Today, it is home to many diverse First Nations, Inuit and Métis peoples from across Turtle Island. We also acknowledge that Toronto is covered by Treaty 13 with the Mississaugas of the Credit. Our team is grateful to have the opportunity to work on this land.
© 2023 The Green Line. All Rights Reserved.
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redpajama_set_name: RedPajamaCommonCrawl


The Sumaco Ñahui Advisory Team is a group of professionals from all different fields and backgrounds. Their advice and knowledge will greatly contribute to the success of this vision. We will add more advisors….
Travel to the tropics is a transformative experience. To bring this level of experiential learning to the students of Colorado Mesa University, I have led more than 10 Tropical Field Biology courses to Ecuador where students have had the opportunity to live and work in the field alongside professional biologists to expand their understanding of their own potential in biology.
Experienced Field Biologist with special interest in Herpetofauna.
Experienced Ecuador Nature Guide, with special interest in Wildlife Photography.
Experienced Ecuador Nature Guide, with special interest in Wildlife Natural History.


As of today Stumbleupon is my top referrer, followed by twitter and so on… but also, it is because I haven't updated my other social networks yet. Though I am working on it now. I wanted to be active with my other social networks that helps me get going.


text:
Home » News » NYHS Members Edith & Geoff Geduld Honoured with Ontario Historical Society Cruikshank Medal
OHS Executive Director Rob Leverty (left) and Chair of the OHS Honours and Awards Committee Ian Radforth (far right) present the 2016 OHS Cruikshank Gold Medal to Geoff Geduld (centre-right) and Edith Geduld.
NYHS Members Edith & Geoff Geduld Honoured with Ontario Historical Society Cruikshank Medal
The Ontario Historical Society's Cruikshank Medal is presented on rare occasions to individuals who have performed with distinction on behalf of the Society. This year, the Ontario Historical Society is proud to recognize Geoff and Edith Geduld with this honour.
The Gedulds have worked in close partnership with the Ontario Historical Society for over thirty years. Geoff served as a Director on the Ontario Historical Society's board and both Edith and Geoff have served as Presidents of the North York Historical Society, an organization that has worked closely with the OHS for decades. Since 1978, they have been involved in virtually all of the efforts to preserve the significant heritage structures of North York, including Milne House, Gibson House, Green Meadows, and of course, the John McKenzie House.
Please see the Ontario Historical Society press release below for more information.
Congratulations, Edith & Geoff!
 Longtime OHS Volunteers Honoured for Outstanding Dedication to Heritage (opens 1-page PDF)
NYHS News
NYHS Remembers Barbara Kane, August 5, 1921 – April 4, 2017
NYHS Remembers Donalda Ewart Badone (Née Hastie), 1927-2016


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On May 8th, the Somerset Triad held a fund raiser featuring comedian Steve Lopes, known as "The Portuguese Fireman." Here are some highlights of Steve's performance, along with Joe Shepard, who talks about some of the projects that Somerset Triad is invloved with.
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Ireland could see the balance of its taxation system reshuffled, creating new taxes and diminishing those already established.
Ireland may soon see a restructuring of its taxation system, if suggestions by Ireland's Commission on Taxation to the Government are accepted. The Commission, which was created in 2008 to examine Ireland's taxation structure, is expected to release a report by the 28th of August with its recommended taxation changes.
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ABHES is an independent non-profit agency unrelated to any trade or membership organization. To learn more about ABHES, please visit www.abhes.org.


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Маури́сио Андре́с Пельегри́но (; ) — аргентинский футболист, защитник, который провёл большую часть своей карьеры, выступая за «Велес Сарсфилд» и испанскую «Валенсию». В настоящее время — футбольный тренер. Является старшим братом Максимилиано Пельегрино.

Карьера игрока
Пельегрино был важнейшей частью аргентинского клуба «Велес Сарсфилд», когда команда смогла добиться величайших успехов в своей истории, трижды выиграв чемпионат Клаусуры, один раз — чемпионат Апертуры и взяв пять международных трофеев, в числе которых Кубок и Суперкубок Либертадорес.

В 1999 году Маурисио перешёл в «Валенсию», с которой выиграл два титула чемпиона Испании, взял в 2004 году Кубок УЕФА и два года подряд играл в финале Лиги чемпионов. Этот турнир так и не покорился ему — в финале 2001 года именно он не забил решающий одиннадцатиметровый в серии послематчевых пенальти, и титул достался мюнхенской «Баварии».

В 2005 году 33-летний Пельегрино покинул «Валенсию», не найдя общего языка с новым тренером клуба Клаудио Раньери и перешёл в «Ливерпуль», который тренировал его бывший наставник Рафаэль Бенитес, подписав контракт на шесть месяцев. Однако в новой команде возрастной защитник закрепиться не смог и отправился заканчивать карьеру в испанский «Алавес».

Карьера тренера
После завершения карьеры игрока Маурисио вошёл в тренерский штаб «Валенсии», где он стал тренером юношеской команды. В июне 2008 года Рафаэль Бенитес в интервью официальному сайту «Ливерпуля» сообщил, что Пельегрино возвращается в Мерсисайд, чтобы занять пост тренера первой команды клуба. 7 мая 2012 года подписал контракт с «Валенсией» сроком на два года. Для аргентинца работа в левантийском клубе станет первым самостоятельным опытом — ранее он работал исключительно помощником. 1 декабря этого же 2012 года в связи с неудачным ходом сезона и разгромным поражением 2:5 от клуба «Реал Сосьедад» уволен с поста тренера «Валенсии».

В апреле 2013 года принял «Эстудиантес», но через два года был уволен из-за неудовлетворительных результатов команды. Летом 2015 года был назначен тренером «Индепендьенте», победив в первых шести матчах, он стал самым успешным тренером команды за последние годы, но в итоге тоже был уволен из-за плохих результатов. В 2016 году возглавил «Алавес».

23 июня 2017 года официально был назначен тренером английского «Саутгемптона», подписав с клубом трёхлетний контракт. Уже 12 марта 2018 года уволен с должности за плохие результаты. При нём команда выиграла лишь один из 17 последних матчей и балансирует в одном очке от зоны вылета из АПЛ.

2 июня 2018 года был назначен новым тренером испанского клуба «Леганес».

16 апреля 2020 года назначен главным тренером «Велес Сарсфилд». Контракт подписан до 30 июня 2021 года.

Достижения
  «Велес Сарсфилд»
 Чемпион Аргентины: Кл. 1993, Aп. 1995, Кл. 1996, Кл. 1998
 Обладатель Кубка Либертадорес: 1994
 Обладатель Межконтинентального кубка: 1994
 Обладатель Межамериканского кубка: 1995
 Обладатель Суперкубка Южной Америки: 1996
 Обладатель Рекопы Южной Америки: 1997

  «Барселона»
 Чемпион Испании: 1998/99

  «Валенсия»
 Чемпион Испании: 2001/02, 2003/04
 Обладатель Суперкубка Испании: 1999
 Обладатель Кубка УЕФА: 2003/04
 Обладатель Суперкубка УЕФА: 2004
 Финалист Лиги чемпионов: 1999/00, 2000/01

Примечания

Ссылки
 Профиль игрока на Liverbird.ru
 Профиль игрока на официальном сайте «Ливерпуля»
 Профиль игрока на LFCHistory.net
 Статистика игрока в чемпионате Аргентины

Футболисты Аргентины
Игроки сборной Аргентины по футболу
Игроки ФК «Велес Сарсфилд»
Игроки ФК «Барселона»
Игроки ФК «Валенсия»
Игроки ФК «Ливерпуль»
Игроки ФК «Алавес»
Футбольные тренеры Аргентины
Тренерский штаб ФК «Ливерпуль»
Тренеры ФК «Алавес»
Тренеры ФК «Валенсия»
Тренеры ФК «Индепендьенте»
Тренеры ФК «Эстудиантес»
Тренеры ФК «Саутгемптон»
Тренеры ФК «Леганес»
Тренеры ФК «Велес Сарсфилд»

redpajama_set_name: RedPajamaWikipedia


The Episode starts with Aayat punishing Mariam. She asks her to get potatoes from the hot water. Mariam gets her hand burnt and goes back. She apologizes and says I won't call dad again. Aayat asks servant to go if he is pitying Mariam. She goes to Farhaan and gives his phone. She asks him not to interfere in house matters. She goes. Farhaan messages his leader. Servant offers help to Mariam. She says no, Aayat will beat you, I will never talk to dad now. She gets the potato and peels it. Madiha and Meher go to see Mahira and don't see her in room.
her and asks what were you doing. She says I lost you once, I can't lose you again, I can't live without you, I love you. She hugs him. Rifat looks on shocked and says you are cheating my innocent son, Mahira's family should be blamed for this. Mahira says I just married Zain, I just love you. Rifat says there will be a real twist in their love story.
Madiha says Rihaan is here because of you. They see Rihaan sleeping in his room. They go and see Mahira talking on phone to Zain. Madiha smiles. Mariam gets peeled potatoes. She steps on potatoes to mash them. Aayat comes there. Mariam says I mashed all potatoes. Aayat says you ruoned them, you don't even deserve to be punished, you will sleep on the ground today, I don't want to see you in party tomorrow, get lost. She goes. Junaid and Hamdam look on.
Everyone parties with Jibraan. Junaid and Hamdan come to Mariam and try to instigate her to go to party. Mariam says my dad is coming, he will feed me tasty food, but I can hide and see the party. Aayat blesses Jibraan. Aye dil….plays….. Mariam recalls her family. She misses them. Jibraan sees Mariam hiding. Junaid pays a man and gets a box of honeybees. Mariam imagines her friends. Mariam comes after Junaid and asks about the gift. He says don't know who got this. She says I will keep it with all gifts. He says Aayat got this for Jibraan. She asks how is this noise coming. He lies and asks her to come with him to play. Mariam goes to Jibraan. He says Mariam won't play with me. Junaid says its good chance to trouble her. Jibraan agrees. Aayat calls Rifat and says Majaaz is coming, I know a way to stop him. Rifat says I m worried, you help yourself, don't think I will stop Majaaz. She says my problems are increasing, no…. I need to do something for my son and his happiness. Jibraan goes to hide in storeroom. Some honey falls on him, he sees a gift and thinks of Aayat's words. He says this might be mum's secret gift.
Jibraan opens the gifts. Honeybees attack him. Mariam tries to save him.


text:
CREATE TABLE cfg_bam (
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  state_source int default NULL,

  description varchar(254) default NULL,
  level_w float default NULL,
  level_c float default NULL,
  sla_month_percent_w float default NULL,
  sla_month_percent_c float default NULL,
  sla_month_duration_w int default NULL,
  sla_month_duration_c int default NULL,
  current_level float default NULL,
  downtime float default NULL,
  acknowledged float default NULL,
  activate enum('1','0') NOT NULL default '0',
  last_state_change int default NULL,
  current_status smallint default NULL,
  in_downtime boolean default NULL,
  must_be_rebuild enum('0', '1', '2') NOT NULL default '0',
  id_reporting_period int default NULL,
  ba_type_id int NOT NULL,
  organization_id int NOT NULL,

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  FOREIGN KEY (id_reporting_period) REFERENCES timeperiod (tp_id)
    ON DELETE SET NULL,
  FOREIGN KEY (ba_type_id) REFERENCES cfg_bam_ba_types (ba_type_id)
    ON DELETE RESTRICT,
  FOREIGN KEY (organization_id) REFERENCES cfg_organizations (organization_id)
    ON DELETE CASCADE
);


Dual Standby Dual Sim Samsung C6112 to access the SIM card easily and conveniently. sleek and slim design of Samsung C6112 is equipped with a brushed metal finish, the slim slider measuring just 16.5 mm. Dedicated button to switch SIM on the side of the Samsung C6112, it is easy to manually switch between SIM cards. Its 2 MP camera takes pictures that you need. One Click buttons SNS Samsung C6112 connects you to the most popular SNS sites.


View first unread post Post The Deed. Post It.
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text:
Bucs hire 3 assistant coaches
Eagles finalize coaching staff
Pacers burned by Heat 117-112
BY foxsports • February 8, 2011
By TIM REYNOLDS
AP Sports Writer
MIAMI (AP) -- With LeBron James on the bench, the Indiana Pacers quickly took command.
Unfortunately for them, he returned.
The league's two-time reigning MVP had 41 points, 13 rebounds and eight assists, Chris Bosh added 19 points and the Miami Heat rallied from a 14-point deficit to beat the Indiana Pacers 117-112 on Tuesday night for their seventh straight victory.
Miami held Indiana to 28 percent shooting in the fourth quarter, giving up only 15 points after yielding a 97 in the first three periods, and saw Mario Chalmers force Dahntay Jones into a key 5-second violation with 8.9 seconds remaining.
James took care of the rest, and Miami moved within a half-game of idle Boston for the Eastern Conference's best record.
"We figured it out in the fourth quarter," James said. "Held them to 15 points and we executed."
In his last 20 games, James is averaging 29.6 points, 8.6 rebounds and 7.5 assists. Not coincidentally, the Heat are 17-3 in those contests.
"I'll tell you what, he's playing MVP-like basketball on both ends of the court," Heat coach Erik Spoelstra said. "He had a motor tonight that did not fatigue. It was amazing."
James had 26 points in the first half, sending one dunk through the net and accidentally off Dwyane Wade's head -- "My bad," James said later -- then going on a tear in the final 2 minutes where he turned what was a 66-52 deficit into a 66-63 game going into halftime.
James went on the attack -- outrunning Danny Granger in transition for one dunk, flashing out of the high post and taking a pass from Chalmers for another, then converting a spinning layup while getting fouled.
"We would have been up by 40 if it wasn't for him," Granger said. "He was unconscious."
In the end, Miami needed every bit of his heroics.
"The best player in the world played at his best in his own building," said Pacers interim coach Frank Vogel, who fell to 4-1 since taking over on the Indiana sideline. "It's tough to beat."
Wade scored 17 and Chalmers added 16 for Miami, which moved five games ahead of second-place Atlanta in the Southeast Division.
"You don't have to be awesome every night," Wade said. "You have to find a way to help your ballclub win and we were able to do that."
Roy Hibbert scored 20 points and added 10 rebounds for Indiana. Granger scored 19 and Tyler Hansbrough added 18 for the Pacers.
Indiana was shooting better than 60 percent at one point in the second half, and was at 57 percent entering the final quarter. But the Pacers shot 5 for 18 in the fourth, including one dreadful stretch that gave the Heat life.
Indiana missed nine straight shots over a span of nearly 6 minutes, but Granger's layup with 51.9 seconds remaining -- which snapped the drought -- got the Pacers within 111-110.
They needed a stop, and couldn't deliver.
Mike Miller missed a 3-pointer, but Bosh grabbed the offensive rebound to extend the possession and James eventually wound up connecting from 17 feet with 14.1 seconds left for a 113-110 Heat lead.
"A shootout for three quarters, both teams scoring the ball, making highlight plays ... and scoring with relative ease," Spoelstra said. "Finally we made a stand in the fourth quarter."
Darren Collison scored 15 points, Josh McRoberts added 12 and Jones scored 11 for Indiana, which was trying to join Boston and Dallas as teams that beat Miami twice this season.
The Heat spoke both Monday and pregame Tuesday about wanting to avenge a 93-77 home loss to the Pacers on Nov. 22. It easily was their worst home defeat of the season -- considering the 16-point margin in that game exceeded the 13-point margin of their other four losses in Miami combined.
Words didn't translate into play. At least, not for much of the night, they didn't.
With James on the court in the opening two quarters, Miami outscored Indiana by 11. Without him, the Heat were outscored by 14 -- and it happened quickly.
James only got one break in the opening half, a 3-minute, 35-second rest at the start of the second quarter. And Indiana outscored Miami 18-4 in that stretch to turn a 33-29 deficit into a 45-37 lead by the time James came back into the game.
It took until 4:34 remained in the fourth for Miami to get back on top, coming when Erick Dampier slammed in the rebound of a miss by Wade for a 106-105 edge -- the first Heat lead since the opening 2 minutes of the second quarter.
The Pacers didn't lead again.
"If we want greatness, greatness is consistency," Spoelstra said. "And we're not there yet."
NOTES: Pacers President Larry Bird was in attendance, and shook hands with a
few fans. Tuesday marked the 25th anniversary of his victory at the
inaugural 3-point shootout, now a staple of All-Star weekend. ...
Miller gave referee David Jones a frustrated look after the veteran
official nearly ran in the way of a pass coming his way with 5:20
remaining. ... Pacers guard Brandon Rush (ankle) was active, but did
not play. Rush hasn't played since Jan. 26.


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There has never been a better time to be an entrepreneur. The barriers to entry have been removed and anyone who is willing to put forth the effort can build a business and create the life of their dreams. This book is written to inspire entrepreneurs to create joy, passion and profits in their business by becoming the best entrepreneur they can possibly be.
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Commissioners investigating the Parkland school shooting said Thursday that law enforcement school resource officers, or SROs, should play a larger role at schools.
Florida Sen. Lauren Book, who sits on the commission, asked the group to recommend that SROs be given access to students' educational and disciplinary records.
The state's public safety commission is meeting for its second day of meetings this month, at the BB&T Center in Sunrise.
On Wednesday the commission watched a surveillance video showing former Marjory Stoneman Douglas High School SRO Scot Peterson during the February shooting. He did not enter the building where the shooting was taking place. Commissioners announced they had summoned Peterson to testify at next month's meeting.
On Thursday the commission discussed standardizing and automizing the threat assessment process for schools across the state. That's the way schools document and keep track of student incidents, including any behavioral issues that could be considered threats.


2. Under Proclamation Date select "September 28, 2018".
3. Complete remaining fields with your personal/organization's information.
4. In the box beneath Proposed Text copy/paste the Draft Proclamation (below) in its entirety, including the References.
6. Click here to send us an email and let us know you've submitted a proclamation request for GA!
I, NATHAN DEAL, Governor of the State of Georgia, do hereby proclaim September 28, 2018, as NATIONAL PENICILLIN ALLERGY DAY in Georgia.


The uppersides of this handsome butterfly are velvety vlack with a red stripe through the centre of the forewing and a red margin to the hindwing. There are also several white spots near the tip of the forewing. The undersides are mottled greyish brown, marked with red, white and blue on the forewing. After a short hibernation in the south (Mediterranean) he arrives in The Netherlands in May and stays till October. The new generation will fly to the south when temperatures become low. This wide-ranging migratory butterfly can be found almost anywhere with flowers and ripe fruit. They often drink sap oozing from wounded trees and are also found of over-ripe and fermenting fruit, often they become a little 'drunk'. The Red Admiral is often common in gardens. The larval foodplant is Stinging Nettle.
This link will lead you to the metamorphosis of the Red Admiral.