If $P, Q, \text{and} R$ are propositions, show that (without using the truth tables):
i) $(P \land Q) \implies R$ is equivalent to $(P \land \lnot R) \implies \lnot Q$
ii) $(P \iff Q)$ is equivalent to $ (\lnot P \iff \lnot Q)$
iii) $P \implies (\lnot Q \implies \lnot R)$ is equivalent to $(P \lor \lnot Q) \land (P \lor R)$
My attempt:
i) $((P \land Q) \implies R) \iff \lnot (P \land Q) \lor R$
$\iff (\lnot P \lor \lnot Q) \lor R$
$\iff \lnot P \lor (\lnot Q \lor R)$
$\iff \lnot P \lor (R \lor \lnot Q )$
$\iff (\lnot P \lor R) \lor \lnot Q $
$ \iff \lnot (\lnot P \lor R) \implies \lnot Q $
$\iff (P \land \lnot R) \implies \lnot Q$
ii) $(P \iff Q) \iff ((P \implies Q) \land (Q \implies P))$
$\iff ((\lnot P \lor Q) \land (\lnot Q \lor P))$
$\iff ((\lnot Q \lor P) \land (\lnot P \lor Q))$
$\iff ((P \lor \lnot Q) \land ( Q \lor \lnot P))$
$\iff ((\lnot P \implies \lnot Q) \land (\lnot Q \implies \lnot P))$
$\iff (\lnot P \iff \lnot Q)$
Is that true? And what about (iii) please.
Best Answer
i & ii are correct, just make sure you are not assuming what you are trying to proof. Listing your axiom can help. iii is not equivalent. It's false, for example, when P and R are both false. below is the list of your assumptions roughly on order of appearance.
$\begin{align} A\rightarrow B \iff \neg A \vee B && &\text{definition of implication}\\ \neg(A\wedge B) \iff \neg A \vee \neg B && &\text{De Morgan's law} \\ (A \vee B) \vee C \iff A \vee (B \vee C) && &\text{associative law} \\ \neg (\neg A) \iff A && &\text{double negation}\\ (A \iff B) \iff (\neg A \iff \neg B) && & \text{sub-expression negative equivalence *(ii)}\\ (A \iff B) \implies ((A \vee C) \iff (B \vee C )) && &\text{sub-expression disjunctive equivalence} \\ ((A\iff B) \wedge (B \iff C)) \iff (A \iff C) && &\text{transitivity of equivalence}\\ (A \iff B) \implies ((A \wedge C) \iff (B \wedge C)) && &\text{sub-expression conjunctive equivalence}\\ \end{align}$
Typically, you only need the $\implies$ part of ii for dealing with sub-expressions.