Usually operator , (comma) is used to replace braces in logical expressions
Let s consider the domain of x as {1,2}.
We will invalidate options one by one by invalidating the given conditional.
(A) $\{ \beta \rightarrow (\exists x , \alpha(x) ) \} \rightarrow \{ \forall x , \beta \rightarrow \alpha(x) \}$
The above expression can be re-written as :
$\{ \beta \rightarrow (\exists x ( \alpha(x) ) ) \} \rightarrow \{ \forall x ( \beta \rightarrow \alpha(x) ) \}$
Notice comma has been replaced by ()
Now we don't know what variable input does $\beta$ take so let us assume it to be a simple propositional variable.
The L.H.S. can be re-written considering domain of x as {1,2} as
$\beta \rightarrow (\alpha_{1}+ \alpha_{2})$
This can be further written as : $\sim\beta + (\alpha_{1}+ \alpha_{2})$
R.H.S. can be re-written as : $(\beta \rightarrow \alpha_{1}) \land (\beta \rightarrow \alpha_{2})$
which evaluates as $\sim\beta + \alpha_{1}.\alpha_{2}$
Now we can clearly see that L.H.S $\rightarrow$ R.H.S. will be false if $\beta$ is true, $\alpha_{1}$ is true and $\alpha_{2}$ is false. Implication became invalid so we eliminate this option.
(B) $\{ \exists x , \beta \rightarrow \alpha(x) \} \rightarrow \{ \beta \rightarrow (\forall x, \alpha(x)) \}$
after replacing comma it becomes
$\{ \exists x ( \beta \rightarrow \alpha(x) ) \} \rightarrow \{ \beta \rightarrow (\forall x (\alpha(x))) \}$
Now re-writing L.H.S. and R.H.S. considering domain of x
L.H.S. : $(\beta \rightarrow \alpha_{1})+(\beta \rightarrow \alpha_{2})$ and this evaluates to
$\sim\beta + \alpha_{1} + \alpha_{2}$
R.H.S. : $\beta \rightarrow (\alpha_{1}.\alpha_{2})$
and this evaluates to $\sim \beta + \alpha_{1}.\alpha_{2}$
This implication can also become false, following same reasoning from option (A).
(C) $\{ (\exists x , \alpha(x) ) \rightarrow \beta \} \rightarrow \{ \forall x, \alpha(x) \rightarrow \beta \}$
After replacing comma it becomes
$\{ (\exists x ( \alpha(x)) ) \rightarrow \beta \} \rightarrow \{ \forall x (\alpha(x) \rightarrow \beta) \}$
L.H.S. : $(\alpha_{1} + \alpha_{2}) \rightarrow \beta$
which evaluates to
$\sim\alpha_{1}.\sim\alpha_{2} + \beta$
R.H.S. : $(\alpha_{1}\rightarrow\beta).(\alpha_{2}\rightarrow\beta)$
which evaluates to $\sim\alpha_{1}.\sim\alpha_{2} + \beta$
which is same as L.H.S. and hence this implication is valid.
(D) $\{ (\forall x, \alpha(x)) \rightarrow \beta \} \rightarrow \{ \forall x, \alpha(x) \rightarrow \beta \}$
after replacing comma it becomes
$\{ (\forall x (\alpha(x))) \rightarrow \beta \} \rightarrow \{ \forall x (\alpha(x) \rightarrow \beta )\}$
Considering domain of x to be {1,2} we re-write L.H.S. and R.H.S.
L.H.S. : $(\alpha_{1}.\alpha_{2})\rightarrow \beta$
and this evaluates to
$\sim \alpha_{1} + \sim \alpha_{2} + \beta$
R.H.S. : $(\alpha_{1}\rightarrow \beta)(\alpha_{2}\rightarrow\beta)$
Which evaluates to $\sim\alpha_{1}.\sim\alpha_{2}+\beta$
Now this implication can be set false, when $\sim\alpha_{1} $ is true , $\beta$ is false and $\sim\alpha_{2}$ is false.
Hence, this implication is invalid.
Answer (c)