# Help with Equivalence Statements and Tautologies

I apologize for this bulky, cumbersome post but in reviewing for a proofs midterm I have tomorrow I was asked some questions on logical equivalence and tautologies that I want to make sure I understand.

We're asked to consider the following statements:

a) $P \land Q$

b) $P\land\neg Q$

c) $\neg P\land Q$

d) $\neg P\land\neg Q$

e) $P\land\neg P$

f) $P\lor Q$

g) $P\lor\neg Q$

h) $\neg P\lor Q$

i) $\neg P \lor \neg Q$

j) $P\lor \neg P$

k) $P\Rightarrow Q$

l) $P\Rightarrow \neg Q$

m) $\neg Q\Rightarrow P$

n) $\neg P \Rightarrow \neg Q$

o) $\neg Q \Rightarrow \neg P$

p) $[P\land(P\Rightarrow Q)]\Rightarrow Q$

q) $[Q\land(P\Rightarrow Q)]\Rightarrow P$

1. Find all statements logically equivalent to $P\Rightarrow Q$.

While $P\Rightarrow Q$ and itself are obviously equivalent, its contrapositive version $\neg Q \Rightarrow \neg P$ also works. However, I'm unsure if $P\land Q$ and $[P\land(P\Rightarrow Q)]\Rightarrow Q$ should also be included.

1. Find all statements logically equivalent to $\neg(P\Rightarrow Q)$.

I want to say $P\Rightarrow\neg Q$ and $P\land\neg Q$ are equivalence statements but I feel like there may be more.

1. Find all statements logically equivalent to $P\lor Q$.

I thought about using a negation of negations via DeMorgan's Laws but no such thing was listed as a possibility, so I can't think of any above statements being equivalent to this one (except itself).

1. Which of these are tautologies?

$P\lor\neg P$ is an obvious tautology since $P$ is either true or false.

$[P\land(P\Rightarrow Q)]\Rightarrow Q$ is a tautology because if $P$ implies $Q$ and $Q$ is true, then $P$ must be true.

Again, I'm sorry for how bulky this question is but I need to assure myself that I'm not overlooking any equivalences or tautologies. Thanks for taking the time to read this.

1. While the contrapositive $\neg Q \Rightarrow \neg P$ is clearly equivalent to $P \Rightarrow Q$, neither $P \wedge Q$ nor $[P \wedge (P \Rightarrow Q)] \Rightarrow Q$ is equivalent. The only case in which $P \Rightarrow Q$ is false is when $P$ is true and $Q$ is false. The only case when $P \wedge Q$ is true is when both $P$ and $Q$ are true. Finally, $[P \wedge (P \Rightarrow Q)] \Rightarrow Q$ is always true. On the other hand, $\neg P \vee Q$ is equivalent to $P \Rightarrow Q$.
2. The statement $\neg(P \Rightarrow Q)$ is equivalent to $\neg(\neg P \vee Q)$, which is equivalent to $P \wedge \neg Q$. It is not equivalent to $P \Rightarrow \neg Q$, which is equivalent to $\neg P \vee \neg Q$.
3. $P \vee Q$ is, however, equivalent to $\neg P \Rightarrow Q$ as well as $\neg Q \Rightarrow P$.
The one rule that will likely help you the most is the one that establishes the equivalence of $x \Rightarrow y$ and $\neg x \vee y$.