Let’s make a truth table this problem: P<->~Q 
~P<->Q.
|
P |
~P |
Q |
~Q |
P<->~Q |
~P<->Q |
|
T |
F |
T |
F |
F |
F |
|
T |
F |
F |
T |
T |
T |
|
F |
T |
T |
F |
T |
T |
|
F |
T |
F |
T |
F |
F |
To show that a sequent is invalid, you need to find an assignment
where ALL the premises are true and the conclusion is false. In this case,
wherever the antecendent (P<->~Q) is true, the succedent
(~P<->Q ) is true. Therefore, we can say that the following sequent is
valid: P<->~Q ~P<->Q.