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For A1, consider A(x) as passing in Phy, B(x) as passing in Chem.

A1 says that if there exists a student who passed in BOTH Phy and Chem, then there exists a student who passed in Phy and there exist a student who passed in Chem. This is true because we can be referring to the same student, who is mentioned in the LHS. Hence, it is a valid argument.

For A2, consider Q(x,y) as student x passes in subject y. It says that if there is a student who has passed in all subjects, then for all students there exists a student who has passed in them, which is also valid.

Hence, both the arguments are valid. However, it is interesting to note that the reverse is not valid in both the cases.
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What will happen if there is bi-implication in the first formula instead of implication?

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 not valid then because $\exists x [A(x)] \wedge \exists x [B(x)]$ can true even if there are two different x one true on A and other one on B but $\exists x [A(x) \wedge B(x)]$ can only be true if there exists some x which is true for both A and B.

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@goxul

Thanks

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$A1: \exists x [A(x) \wedge B(x)] \rightarrow \exists x [A(x)] \wedge \exists x [B(x)] $

It is valid 

You can try to create $T \rightarrow F$ result but you cannot because of there exist x which  is true for both A and B then that same x on RHS will also hold the property A and B.

$A2: \exists x  \forall y Q(x,y) \rightarrow \forall y \exists x Q(x,y)$

This is also valid.

LHS of A2 says that $\text{"There exist x for all y for which Q(x,y) holds"}$ 

Lets assume $Q(x,y)$ means x is a friend of y,

Then LHS of A2 says that there is someone x who is friends with everyone. 

and RHS of A2 says that $\text{"For every y there is some x for which Q(x,y) holds"}$ 

which can be interpreted as for every person y there is someone (x) such that x and y are friends. 

You can try to create $T \rightarrow F$ result but you cannot because if LHS is true that means there is someone who is friends with everyone and then RHS cannot be false. for RHS making false there must be someone who doesn't have any friend but LHS says that there is someone who is friends with everyone.

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Thanks .. :)
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