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š: Show that the equivalence p ā§Ā¬p ā” F can be derived
using resolution together with the fact that a condi
tional statement withafalsehypothesisistrue.[Hint: Let
q =r =Finresolution.]
š¤: Sure! To derive the equation $p \land \neg p \equiv F$ using resolution, we need to find a resolvent rule that allows us to transform $p \land \neg p$ into $F$.
Recall that in resolution, we use the following rules:
- Resolution: If $A \lor B$, then $A \to C \lor D \to E \equiv (A \lor B) \to C \lor D \lor E$.
- Modus Ponens: From $A \to B$ and $B$, infer $A$.
- Conditional Elimination: From $A \to B$ and $not(B)$, infer $not(A)$.
Using these rules, we can derive the desired equation as follows:
\begin{align\*} p \land \neg p &= p \to \neg p \\ &\to (\neg p \lor p) \lor \neg p \\ &= F \lor \neg p \\ &\equiv F \end{align\*})
The second line uses modus ponens to infer $p \to \neg p$ from $p$ and $\neg p$. The third line uses conditional elimination to infer $\neg p \lor p$ from $\neg p$ and $p$. Finally, the fourth line uses resolution to conclude that $F \lor \neg p \equiv F$.
Note that we have used the fact that a conditional statement with a false hypothesis is true in the last step of the derivation. This allows us to conclude that $\neg p \lor p \equiv T$, which completes the proof.