(Verification) If $A \subsetneq B$, and $B \subsetneq C$, then $A \subsetneq C$

Question

Statement to be proved: If $A \subsetneq B$, and $B \subsetneq C$, then $A \subsetneq C$.

First method: Inequality: If $A \subsetneq B$ then there exists an $x\in B$ such that $x\notin A$. Let $z:z\in B \wedge z\notin A$

Similarly, if $x\in B\implies x\in C$, then $z\in C$, but if $z\notin A$, then there is an element in $C$ not in $A$, so $A\ne C$.

$(x\in A\implies x\in B, x\in B \implies x\in C) \implies (x\in A \implies x\in C)$. Thus at best $x\subset C$. But since $A\ne C$, then $A\subsetneq C$. Q.E.D.

Second method (I couldn't use this in the exercise as cardinality came later):

$A\subsetneq B$ means that all elements in $A$ are in $B$, but some elements in $B$ are not in $A$. So $n(A)<n(B)$. Same can be said about $B$ and $C$: $n(B)<n(C)$. Combining the two inequalities: $n(A)<n(C)$ so it follows that $A\neq C$.

The fact that all elements in $A$ are in $C$ is proven the same way as in the first method.

I think this is correct, but I'm not entirely sure as to the logical soundess. I also think there is a faster way of doing it (but I mainly want to know if it is correct).

EDIT: I use $\subsetneq$ to mean "A is a subset of B, but A is not equal to B"

Answer 1

I really don't think the second solution is correct because is not a bimplication. It works only one way. If A is a subset of C, then n(A)<n(C). Not the other way round.

Coming to the first one, I think this one is correct. Perhaps you don't even need to prove using an element which doesn't exist in A. It is simply shown by first few lines of your solutions that for all elements which are present in A, they are also present in C. Thus A is a subset of C.