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I was doing this question, but it tells me one of the option I selected was wrong.

The second option I thought it means "all of the things in A is not in B" indicates that "there exists something in A that is not in B"

The last option I thought means "not all things in A are in B" indicates that "there exists something in A that is not in B"

Both make sense to me, please help explain which one I selected is wrong. Also is there any that I haven't selected is correct?

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    $\begingroup$ What if $A$ is always false? $\endgroup$ – Dan Doel Feb 23 at 5:40
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Only the last implication is correct.

As you have observed, you can think of the predicates $A$ and $B$ as sets. So, knowing the first three don't hold, we should try to construct a counter-model.

Both the first and second fail on the model $A = \emptyset, B = \emptyset$, and the third fails on the model $A = \{o_1,o_2\}, B = \{o_2\}$.

Where you went wrong is assuming $A$ will contain an element. Indeed $$\exists x( A(x) ), \forall x( A(x) \Rightarrow B(x) ) \vDash \exists x( A(x) \wedge \neg B(x) )$$ actually holds.

So, beware cases where you just have $\forall$ assumptions, because they don't guarantee that objects will actually exist in your set.

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