Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1
Jump to navigation
Jump to search
Theorem
- $p \implies q \dashv \vdash \neg \paren {p \land \neg q}$
This can be expressed as two separate theorems:
Forward Implication
- $p \implies q \vdash \neg \paren {p \land \neg q}$
Reverse Implication
- $\neg \paren {p \land \neg q} \vdash p \implies q$
Proof by Truth Table
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
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Theorems $35$
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $2$ Arguments Containing Compound Statements: $2.4$: Statement Forms
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.10$: Exercise $2.3$