Conjunction Equivalent to Negation of Implication of Negative/Formulation 2
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Theorem
- $\vdash \paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} }$
Proof 1
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \land q$ | Assumption | (None) | ||
| 2 | 1 | $\neg \paren {p \implies \neg q}$ | Sequent Introduction | 1 | Conjunction Equivalent to Negation of Implication of Negative: Formulation 1: Forward Implication | |
| 3 | $\paren {p \land q} \implies \paren {\neg \paren {p \implies \neg q} }$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged | ||
| 4 | 4 | $\neg \paren {p \implies \neg q}$ | Assumption | (None) | ||
| 5 | 4 | $p \land q$ | Sequent Introduction | 4 | Conjunction Equivalent to Negation of Implication of Negative: Formulation 1: Reverse Implication | |
| 6 | $\paren {\neg \paren {p \implies \neg q} \implies \paren {p \land q} }$ | Rule of Implication: $\implies \II$ | 4 – 5 | Assumption 4 has been discharged | ||
| 7 | $\paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} }$ | Biconditional Introduction: $\iff \II$ | 3, 6 |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.
$\begin{array}{|ccc|c|ccccc|} \hline (p & \land & q) & \iff & (\neg & (p & \implies & \neg & q)) \\ \hline \F & \F & \F & \T & \F & \F & \T & \T & \F \\ \F & \F & \T & \T & \F & \F & \T & \F & \T \\ \T & \F & \F & \T & \F & \T & \T & \T & \F \\ \T & \T & \T & \T & \T & \T & \F & \F & \T \\ \hline \end{array}$
$\blacksquare$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T38}$