Problem taken from here.
Use Logical Equivalences to prove that $[(p \land \lnot(\lnot p \lor q)) \lor (p \land q)] \implies p$ is a tautology.
- implication law…
$\lnot[(p \land \lnot(\lnot p \lor q)) \lor (p \land q)] \lor p $ - demorgans $[\lnot(p \land \lnot(\lnot p \lor q)) \land \lnot(p \land q)] \lor p$
- demorgans $[\lnot(p \land \lnot(\lnot p \lor q)) \land (\lnot p \lor \lnot q)] \lor p$
-
demorgans $[(\lnot p \lor (\lnot p \lor q)) \land (\lnot p \lor \lnot q)] \lor p$
-
distribution… $[(\lnot p \lor q) \land (\lnot p \lor \lnot q)] \lor p$
- distribution… $[(\lnot p \lor (q \land \lnot q)] \lor p$
- negation… $[\lnot p \lor F] \lor p$
- associativity $(p \lor \lnot p) \lor f$
- domination $T \lor F$
- $T$
does this make sense? i just want to make sure there are multiple ways to go about it. obviously implication would be better saved till the end but on an exam etc. just want to make sure I know multiple ways to get there.
Best Answer
$\color{blue}{\checkmark \mathrm A{-}}$
Your steps are correct, although some of their justifications don't match them.
The first 'distribution' is actually 'association and idempotence'.
The 'associativity' should be 'association and commutation'.
Other than that, spot on.