Non-Equivalence as Disjunction of Negated Conditionals
Theorem
- $\neg \paren {p \iff q} \dashv \vdash \neg \paren {p \implies q} \lor \neg \paren {q \implies p}$
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 | $\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }$ | Sequent Introduction | 1 | Rule of Material Equivalence | |
| 3 | 1 | $\neg \paren {p \implies q} \lor \neg \paren {q \implies p}$ | Sequent Introduction | 2 | De Morgan's Laws: Disjunction of Negations |
$\Box$
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\neg \paren {p \implies q} \lor \neg \paren {q \implies p}$ | Premise | (None) | ||
| 2 | 1 | $\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }$ | Sequent Introduction | 1 | De Morgan's Laws: Disjunction of Negations | |
| 3 | 1 | $\neg \paren {p \iff q}$ | Sequent Introduction | 2 | Rule of Material Equivalence |
$\blacksquare$
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.
$\begin{array}{|cccc||ccccccccc|} \hline
\neg & (p & \iff & q) & \neg & (p & \implies & q) & \lor & \neg & (q & \implies & p) \\
\hline
\F & \F & \T & \F & \F & \F & \T & \F & \F & \F & \F & \T & \F \\
\T & \F & \F & \T & \F & \F & \T & \T & \T & \T & \T & \F & \F \\
\T & \T & \F & \F & \T & \T & \F & \F & \T & \F & \F & \T & \T \\
\F & \T & \T & \T & \F & \T & \T & \T & \F & \F & \T & \T & \T \\
\hline
\end{array}$
$\blacksquare$
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- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$