Syllogism proof using set notation

Question

$$B\subseteq C \wedge B\nsubseteq A^c \Rightarrow A\nsubseteq C^c$$

So I need to prove that $$\exists x \in A\,\ x\notin C^c$$

so far I'm up to the point where I can say that $$\exists x \in C , x \in A$$

I've noticed that if I swap the A and C around then I have my answer but what steps do I take to be able to swap it around or is the ending statement simply false?

Answer 1

$$B\subseteq C \wedge B\nsubseteq A^c$$

by the second part we have $\exists x\in B,x\notin A^c$, we can change $\notin A^c$ into $\in A$ to get $$\exists x\in B,x\in A$$from the first part of the original statement we can change $B$ to $C$ to get$$\exists x\in C,x\in A$$change the order we are looking at it to get $$\exists x\in A,x\in C$$ changing the order is okay since those statements are simply properties of $x$, there is no order to those properties, they all are just... exists you can say. it is like saying: i am a human and i am standing VS i am standing and i am a human.

we can now change $\in C$ into $\notin C^c$ to get $$\exists x\in A,x\notin C^c$$hence we get $$A\nsubseteq C^c$$

Answer 2

$B \subseteq C$ means: $\forall x \ (x \in B \to x \in C)$.

And $B \nsubseteq A^c$ means: $\exists x \ (x \in B \land x \notin A^c)$, i.e: $\exists x \ (x \in B \land x \in A)$.

For $(\exists x∈A)x∉C^c$, see Restricted quantifiers; it is an abbreviation for: $\exists x \ ( x∈A \land x∉C^c)$ that in turn means: $\exists x \ (x∈A \land x∈C)$.

Now we can complete the syllogism:

$\exists x \ (x \in A \land x \in B)$,

$\forall x \ (x \in B \to x \in C)$;

therefore:

$\exists x \ (x∈A \land x∈C)$.