To check whether this statement is always true
$(\forall x (P(x)) \implies \forall x (Q(x))) \implies (\forall x(P(x) \implies Q(x)))$
Ans:
Let $(\forall x (P(x)) \implies \forall x (Q(x)))$ be A and
$(\forall x(P(x) \implies Q(x)))$ be B
So for the above statement to be always true
A $\implies$ B can never be False. That is,
T F = F, this case can never arise.
Now to prove it by contradiction we'll try to make A: T and B: F
Lets first make B: F
$(\forall x(P(x) \implies Q(x)))$
Again for the B to be F,
$P(x)$ should be T and $Q(x)$ should be F.
The following table shows for different values of X the Predicate values of P and Q drawn for B to be false.
| $X$ |
$P(x)$ |
$Q(x)$ |
| Tuhin |
True |
False |
| Dutta |
True |
False |
Now according to these values we'll have to check whether A is always True then we can prove that A: T and B: F case holds and thus the statement given in question cannot be valid or always true or tautology.
$\forall x (P(x)) \implies \forall x (Q(x))$
or, $\neg \forall x (P(x)) \lor \forall x (Q(x))$
or, $\exists x (P(x)) \lor \forall x (Q(x))$
or, $ T \lor F$ gives True
Thus A: T and B: F, hence the statement given in question cannot be valid or always true or tautology.
Is my approach correct? Is this a correct solution?
