Implication Equivalent to Negation of Conjunction with Negative/Formulation 2/Proof by Truth Table
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Theorem
- $\vdash \paren {p \implies q} \iff \paren {\neg \paren {p \land \neg q} }$
Proof
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.
$\begin{array}{|ccc|c|ccccc|} \hline (p & \implies & q) & \iff & (\neg & (p & \land & \neg & q)) \\ \hline \F & \T & \F & \T & \T & \F & \F & \T & \F \\ \F & \T & \T & \T & \T & \F & \F & \F & \T \\ \T & \F & \F & \T & \F & \T & \T & \T & \F \\ \T & \T & \T & \T & \T & \T & \F & \F & \T \\ \hline \end{array}$
$\blacksquare$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 6$
- 1980: D.J. O'Connor and Betty Powell: Elementary Logic ... (previous) ... (next): $\S \text{I}: 13$: Logical Equivalences: Exercise $\text{(A)} \ 1$