Verfiying satisfiability of formulas

Question

I have this question

And was wondering if someone could help improve my answer (I am learning English):

a) satisfiable as long P=True, Q=True, R= True. Then (P^Q^R) will be true. Also, (not P or R) will be (False or True), which is True. Then, True and True is True. So, (P and Q and R) and (not P or R) is also true.

b) not. P=True, Q=True, R=True, so not P is false, not R is false. Not P or not R is false. True and False is False. cannot find a true

c) will work as long as (if P then Q) is false and (if not P then not Q) is true, does not happen. so can be satisfiable

Answer 1

Your explanation for (c) is wrong.

$$A \rightarrow B$$ is satisfiable as long as you don't have both $A = \text{true}$ and $B = \text{false}$. Your answer says that the unsatisfied case is $A = \text{false}$ and $B = \text{true}$, which is incorrect.

In your formula $$\underbrace{(P \rightarrow Q)}_A \rightarrow \underbrace{(\lnot P \rightarrow \lnot Q)}_B$$

it is possible to create $A = \text{true}$ and $B= \text{false}$. This is the case of $P = \text{false}$ and $Q = \text{true}$. Therefore your claim that (c) is a tautology is incorrect.

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a) satisfiable as long P=True, Q=True, R= True. Then (P^Q^R) will be true. Also, (not P or R) will be (False or True), which is True. Then, True and True is True. So, (P and Q and R) and (not P or R) is also true.

Correct, but hard to read. Here is my suggestion:

• Both $(P \land Q \land R)$ and $(\lnot Q \lor R)$ must be true.
• $(P \land Q \land R)$ requires that $P=\text{true}$, $Q=\text{true}$, and $R=\text{true}$.
• That assignment makes $(\lnot Q \lor R)$ true, so it satisfies (a).

b) not. P=True, Q=True, R=True, so not P is false, not R is false. Not P or not R is false. True and False is False. cannot find a true

Again correct, but hard to read. Here is my suggestion.

• Both $(P \land Q \land R)$ and $(\lnot Q \lor \lnot R)$ must be true.
• $(P \land Q \land R)$ requires that $P=\text{true}$, $Q=\text{true}$, and $R=\text{true}$.
• That assignment makes $(\lnot Q \lor \lnot R)$ false, so nothing satisfies (b).