Non-Equivalence as Conjunction of Disjunction with Disjunction of Negations
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Theorem
- $\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \paren {\neg p \lor \neg q}$
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\neg \paren {p \iff q}$ | Premise | (None) | ||
2 | 1 | $\paren {p \lor q} \land \neg \paren {p \land q}$ | Sequent Introduction | 1 | Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction | |
3 | 1 | $\paren {p \lor q} \land \paren {\neg p \lor \neg q}$ | Sequent Introduction | 2 | De Morgan's Laws: Disjunction of Negations |
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \lor q} \land \paren {\neg p \lor \neg q}$ | Premise | (None) | ||
2 | 1 | $\paren {p \lor q} \land \neg \paren {p \land q}$ | Sequent Introduction | 1 | De Morgan's Laws: Disjunction of Negations | |
3 | 1 | $\neg \paren {p \iff q}$ | Sequent Introduction | 2 | Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction |
$\blacksquare$
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.
$\begin{array}{|cccc||ccccccccc|} \hline
\neg & (p & \iff & q) & (p & \lor & q) & \land & (\neg & p & \lor & \neg & q) \\
\hline
\F & \F & \T & \F & \F & \F & \F & \F & \T & \F & \T & \T & \F \\
\T & \F & \F & \T & \F & \T & \T & \T & \T & \F & \T & \F & \T \\
\T & \T & \F & \F & \T & \T & \F & \T & \F & \T & \T & \T & \F \\
\F & \T & \T & \T & \T & \T & \T & \F & \F & \T & \F & \F & \T \\
\hline
\end{array}$
$\blacksquare$