Implication Equivalent to Negation of Conjunction with Negative/Formulation 1/Proof by Truth Table
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Theorem
- $p \implies q \dashv \vdash \neg \paren {p \land \neg q}$
Proof
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||ccccc|} \hline p & \implies & q & \neg & (p & \land & \neg & q) \\ \hline \F & \T & \F & \T & \F & \F & \T & \F \\ \F & \T & \T & \T & \F & \F & \F & \T \\ \T & \F & \F & \F & \T & \T & \T & \F \\ \T & \T & \T & \T & \T & \F & \F & \T \\ \hline \end{array}$
$\blacksquare$
Sources
- 1959: A.H. Basson and D.J. O'Connor: Introduction to Symbolic Logic (3rd ed.) ... (previous) ... (next): $\S 2.4$: Relations between Truth-Functions
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $2$: The Propositional Calculus $2$: $3$ Truth-Tables
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $2$ Arguments Containing Compound Statements: $2.2$: Conditional Statements