Answers.Body: <p><span class="math-container">$f_1 \vdash f_2$</span>, if and only if <span class="math-container">$f_2$</span> must be true if we assume <span class="math-container">$f_1$</span> to be true.</p>
  <p>Similarly, <span class="math-container">$f_2 \vdash  f_1$</span>, if and only if <span class="math-container">$f_1$</span> must be true if we assume <span class="math-container">$f_2$</span> to be true.</p>
  <p>Logically, by taking any value for <span class="math-container">$A$</span> or <span class="math-container">$B$</span>, from the domain <span class="math-container">$\{1, 0 \}$</span>, one could verify that <span class="math-container">$f_1 \vdash f_2$</span>, because <span class="math-container">$f_2$</span> is true whenever <span class="math-container">$f_1$</span> is true (for example, when both <span class="math-container">$A = 0$</span> and <span class="math-container">$B = 0$</span>).</p>
  <p>However, <span class="math-container">$f_2 \vdash  f_1$</span> is not true. As, in two cases, <span class="math-container">$f_2$</span> is true (e.g. <span class="math-container">$A = 0$</span> and <span class="math-container">$B=1$</span>, or vice-versa), but <span class="math-container">$f_1$</span> is not true.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-01-21T19:38:28.980Z
Answers.Text: This is an old question and answer, but when you use $\vdash$, didn't you want to use $\rightarrow$?
Answers.UserDisplayName: nbro
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2017-01-28T19:59:37.583Z
Answers.LastActivityDate: 2021-01-21T21:06:40.843Z
Answers.LastEditDate: 2021-01-21T21:06:40.843Z
Answers.UserDisplayName: Coder
Answers.Body: <p>One way of verifying whether two boolean expressions are equivalent is to assign all possibilities to all variables, and comparing all results.</p>
  <div class="s-table-container">
  <table class="s-table">
  <thead>
  <tr>
  <th>A</th>
  <th>B</th>
  <th>f1</th>
  <th>f2</th>
  </tr>
  </thead>
  <tbody>
  <tr>
  <td>T</td>
  <td>T</td>
  <td>F</td>
  <td>F</td>
  </tr>
  <tr>
  <td>T</td>
  <td>F</td>
  <td>F</td>
  <td>T</td>
  </tr>
  <tr>
  <td>F</td>
  <td>T</td>
  <td>F</td>
  <td>T</td>
  </tr>
  <tr>
  <td>F</td>
  <td>F</td>
  <td>T</td>
  <td>T</td>
  </tr>
  </tbody>
  </table>
  </div>
  <p>We can see <code>(F, F, F, T)</code> does not equal <code>(F, T, T, T)</code>, for example for the assignment <code>(A, B) = (T, F)</code> we get result <code>(f1, f2) = (F, T)</code> , meaning <code>f1</code> <span class="math-container">$\ne$</span> <code>f2</code>.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-01-21T20:19:14.593Z
Answers.LastActivityDate: 2021-05-30T09:03:15.330Z
Answers.LastEditDate: 2021-05-30T09:03:15.330Z
Answers.UserDisplayName: Gulzar
Answers.Body: <p>I find it easy to get a quick intuition of the truth value of logical statements involving negations by converting them wherever possible.</p>
  <p>So assume by way of contradiction that <span class="math-container">$\textit{f}_1 \iff \textit{f}_2$</span>, then the two-way contrapositive <span class="math-container">$\neg \textit{f}_1 \iff \neg \textit{f}_2$</span> also holds, hence <span class="math-container">$A \lor B \iff A \land B$</span> (De Morgan's law). Since <span class="math-container">$A \lor B \implies A \land B$</span> is easily confirmed false (by plugging in A=True and B=False) this is a contradiction. <span class="math-container">$\Box$</span></p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-05-30T16:08:24.003Z
Answers.LastActivityDate: 2021-05-30T16:08:24.003Z
Answers.UserDisplayName: Tcfkaj
Body: <p>If I have 2 statements, say <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, from which I formed 2 formulae:</p>
  <ol>
  <li><p><span class="math-container">$f_1: (\lnot A) \land (\lnot B)$</span></p>
  </li>
  <li><p><span class="math-container">$f_2: (\lnot A) \lor (\lnot B)$</span></p>
  </li>
  </ol>
  <p>Are <span class="math-container">$f_1$</span> and <span class="math-container">$f_2$</span> equivalent?</p>
  
Tags: <logic><knowledge-representation>
Title: Given statements $A$ and $B$, are the formulae $(\lnot A) \land (\lnot B)$ and $(\lnot A) \lor (\lnot B)$ equivalent?
UserDisplayName: Coder
ThreadText: # Given statements $A$ and $B$, are the formulae $(\lnot A) \land (\lnot B)$ and $(\lnot A) \lor (\lnot B)$ equivalent?
  
  (Asked by: username_0 on 2017-01-27)
  
  If I have 2 statements, say $A$ and $B$, from which I formed 2 formulae:
  
  1. $f\_1: (\lnot A) \land (\lnot B)$
  2. $f\_2: (\lnot A) \lor (\lnot B)$
  
  Are $f\_1$ and $f\_2$ equivalent?
  
  **Tags:** logic, knowledge-representation
  
  ---
  
  ## 3 answers
  
  ### Answer by username_0 on 2017-01-28. Score: 0 | Accepted answer
  
  $f\_1 \vdash f\_2$, if and only if $f\_2$ must be true if we assume $f\_1$ to be true.
  
  Similarly, $f\_2 \vdash f\_1$, if and only if $f\_1$ must be true if we assume $f\_2$ to be true.
  
  Logically, by taking any value for $A$ or $B$, from the domain $\{1, 0 \}$, one could verify that $f\_1 \vdash f\_2$, because $f\_2$ is true whenever $f\_1$ is true (for example, when both $A = 0$ and $B = 0$).
  
  However, $f\_2 \vdash f\_1$ is not true. As, in two cases, $f\_2$ is true (e.g. $A = 0$ and $B=1$, or vice-versa), but $f\_1$ is not true.
  
  > **Comments:**
  > **username_1**: This is an old question and answer, but when you use $\vdash$, didn't you want to use $\rightarrow$?  
  
  ---
  
  ### Answer by username_2 on 2021-01-21. Score: 3
  
  One way of verifying whether two boolean expressions are equivalent is to assign all possibilities to all variables, and comparing all results.
  
  | A | B | f1 | f2 |
  | --- | --- | --- | --- |
  | T | T | F | F |
  | T | F | F | T |
  | F | T | F | T |
  | F | F | T | T |
  
  We can see `(F, F, F, T)` does not equal `(F, T, T, T)`, for example for the assignment `(A, B) = (T, F)` we get result `(f1, f2) = (F, T)` , meaning `f1` $\ne$ `f2`.
  
  
  ---
  
  ### Answer by username_3 on 2021-05-30. Score: 0
  
  I find it easy to get a quick intuition of the truth value of logical statements involving negations by converting them wherever possible.
  
  So assume by way of contradiction that $\textit{f}\_1 \iff \textit{f}\_2$, then the two-way contrapositive $\neg \textit{f}\_1 \iff \neg \textit{f}\_2$ also holds, hence $A \lor B \iff A \land B$ (De Morgan's law). Since $A \lor B \implies A \land B$ is easily confirmed false (by plugging in A=True and B=False) this is a contradiction. $\Box$
  
  
  ---
  
  