text: # Prove P from B ∨ P when we have proven ¬B
  
  I have the following proof so far:
  
  In step 9 I'm not sure how to prove P from the steps I have before. I thought that I could use ∨ Elim but I don't think I can now.
  
  • $(B\lor P)\equiv (\lnot(\lnot B\land\lnot P))\equiv(\lnot B\implies P)$ – trying Aug 8 '17 at 7:52
  • @trying The first $\equiv$ isn't constructively valid from left to right. – DanielV Aug 8 '17 at 8:16
  • @DanielV He is using DeMorgan's Law and the definition of implication. How is that invalid? – Dan Christensen Aug 8 '17 at 14:44
  • @DanChristensen Many classical laws aren't constructive. For DeMorgan's, $\lnot (\lnot A \lor \lnot B) \to (A \lor B)$, just take $B = \lnot A$ and you get $\lnot \lnot A \to A$, double negation elimination which is not taken for granted. – DanielV Aug 8 '17 at 14:58
  • @DanielV So, it is valid in what you call "classical" logic. Thanks. – Dan Christensen Aug 8 '17 at 16:57
  
  The trick is to use $\bot \ Elim$. Now, to continue from what you have, you can do:
  Although conceptually, it may be helpful to keep the original subproof, and (since it shows that $B$ leads to a contradiction $\bot$) derive $\neg B$, and to then combine $\neg B$ with $B \lor P$ (the latter is a super common patterns, so remember that one!!):
  Finally, you can set this up as a proof by cases within a conditional proof, i.e. derive your goal $P \land D$ from each of the cases $B$ and $P$:
  I've actually seen some texts define Or-Elimination as $$\frac{\lnot B,~ B \lor P}{P}$$ which is actually constructively valid. Anyway, the way to get there from regular Or-Elimination (proof by cases) is just to show that in both cases of $B$ and case $P$ that $P$ follows:
  $$\begin{array} {rll} (1) & \lnot B & \text{Given} \\ \\ % (2) & \quad B & \text{assumption, Case 1} \\ % (3) & \quad \bot & \text{Contradiction} \\ % (4) & \quad P & \text{Explosion} \\ \\ % (5) & \quad P & \text{assumption, Case 2} \\ \\ % (6) & B \lor P & \text{Given} \\ % (7) & P & \text{Proof by cases (Or Elim) of 6, 2 through 4, 5 through 5} \\ % \end{array}$$
explanation: The document contains a math question related to a proof in logic. The user is asking for help in completing a proof to derive \( P \) from \( B \lor P \) given that \( \neg B \) has been proven. The document also includes a detailed response providing multiple methods to complete the proof.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Prove P from B ∨ P when we have proven ¬B
  
  I have the following proof so far:
  
  In step 9 I'm not sure how to prove P from the steps I have before. I thought that I could use ∨ Elim but I don't think I can now.
  
  • $(B\lor P)\equiv (\lnot(\lnot B\land\lnot P))\equiv(\lnot B\implies P)$ – trying Aug 8 '17 at 7:52
  • @trying The first $\equiv$ isn't constructively valid from left to right. – DanielV Aug 8 '17 at 8:16
  • @DanielV He is using DeMorgan's Law and the definition of implication. How is that invalid? – Dan Christensen Aug 8 '17 at 14:44
  • @DanChristensen Many classical laws aren't constructive. For DeMorgan's, $\lnot (\lnot A \lor \lnot B) \to (A \lor B)$, just take $B = \lnot A$ and you get $\lnot \lnot A \to A$, double negation elimination which is not taken for granted. – DanielV Aug 8 '17 at 14:58
  • @DanielV So, it is valid in what you call "classical" logic. Thanks. – Dan Christensen Aug 8 '17 at 16:57
  
  The trick is to use $\bot \ Elim$. Now, to continue from what you have, you can do:
  Although conceptually, it may be helpful to keep the original subproof, and (since it shows that $B$ leads to a contradiction $\bot$) derive $\neg B$, and to then combine $\neg B$ with $B \lor P$ (the latter is a super common patterns, so remember that one!!):
  Finally, you can set this up as a proof by cases within a conditional proof, i.e. derive your goal $P \land D$ from each of the cases $B$ and $P$:
  I've actually seen some texts define Or-Elimination as $$\frac{\lnot B,~ B \lor P}{P}$$ which is actually constructively valid. Anyway, the way to get there from regular Or-Elimination (proof by cases) is just to show that in both cases of $B$ and case $P$ that $P$ follows:
  $$\begin{array} {rll} (1) & \lnot B & \text{Given} \\ \\ % (2) & \quad B & \text{assumption, Case 1} \\ % (3) & \quad \bot & \text{Contradiction} \\ % (4) & \quad P & \text{Explosion} \\ \\ % (5) & \quad P & \text{assumption, Case 2} \\ \\ % (6) & B \lor P & \text{Given} \\ % (7) & P & \text{Proof by cases (Or Elim) of 6, 2 through 4, 5 through 5} \\ % \end{array}$$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question related to a proof in logic. The user is asking for help in completing a proof to derive \( P \) from \( B \lor P \) given that \( \neg B \) has been proven. The document also includes a detailed response providing multiple methods to complete the proof.
  
  Label: question-and-answer