Showing a first-order predicate formula is valid.

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There is first-order predicate formula

$$A = \forall a (\neg P(a,a))$$

$$B = \forall a \forall b \forall c (P(a,b)\wedge P(b,c)\Rightarrow P(a,c))$$

$$B'= \forall a \forall b \forall c (P(a,b)\wedge P(b,c)\Rightarrow \neg P(a,c))$$

$$C = \forall a \forall b (P(a,b)\Rightarrow \neg P(b,a))$$

Question: "(A ∧ B) ⇒ C" is valid.

Valid means there is no truth assignment makes it false.

What I got understood are truth assignment and resolving $A\Rightarrow B$ to $\neg A \vee B$. But I can't understand how to implement '$\forall $' Universal notation into formula. And what happened when we use $P(a,b)$ instead of just $a$ or $b$.

So, can anyone solve above question with explaination step by step .

Many thanks. You're all legends!

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Hint

Assume not, i.e. that $A \land B$ is True and $C$ is False.

$C$ False means that $P(a,b)$ and $P(b,a)$ both hold, for some $a,b$.

Thus, from $B$ : $P(a,b) ∧ P(b,a) \to P(a,a)$ we get : $P(a,a)$.