Verfiying satisfiability of formulas 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
 A: 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.


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).

