We can also prove this by making T or F combination & the trick we've to apply is T->F which is F i.e not valid.
for check validity it's easy to prove not valid by 1 combination(T->F) rather than proving valid by 3 combinations(T->T, F->T, F->F).
option A
X |
P(X) |
Q(X) |
X1 |
T |
T |
X2 |
T |
T |
So, using this combination we can't get T -> F, i.e option A is valid.
always choose a combination such that LHS of implication will be T & using that combination check whether RHS will be F or not.
option B
X |
P(X) |
Q(X) |
X1 |
T |
F |
X2 |
F |
T |
We can't get T -> F
option C
X |
P(X) |
Q(X) |
X1 |
T |
T |
X2 |
F |
F |
NO T -> F
OPTION D
X |
P(X) |
Q(X) |
X1 |
T |
F |
X2 |
F |
T |
So, here we get T -> F i.e it's not a valid one