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1 votes

In my opinion, all combinations make sense,

Implication:

$\forall{x}(P \to C)$: For all $x$, if $P$ is true then $C$ is also true.

$\exists{x}(P \to C)$: There exists atleast a $x$ such that if $P$ is true then $C$ is also true.

Conjunction:

$\forall{x}(P \wedge C)$: For all $x$, Both $P$ and $C$ are true.

$\exists{x}(P \wedge C)$: There exists atleast a $x$, such that both $P$ and $C$ are true.

However, if you will post specific example where you find problem/confusion, question will be more clear.

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