If ∀x∈A (x≠0) and A⊆B, then ∀x∈B (x≠0).

Which line of the following proof is wrong. Change the conclusion of the theorem so that the theorem is true.

Let x be an arbitrary element of A. Since ∀x∈A, (x≠0), we can conclude that x≠0. Also, since A⊆B, x∈B. Since x∈B, x≠0, and x was arbitrary, we can conclude that ∀x∈B, x≠0.

Question

If ∀x∈A (x≠0) and A⊆B, then ∀x∈B (x≠0).

Which line of the following proof is wrong. Change the conclusion of the theorem so that the theorem is true. 

Let x be an arbitrary element of A. Since ∀x∈A, (x≠0), we can conclude that x≠0. Also, since A⊆B, x∈B. Since x∈B, x≠0, and x was arbitrary, we can conclude that ∀x∈B, x≠0.

anonymous · Answer

We can say $x \in B$ but we can't say $\forall x \in B$.

anonymous · Answer

We could say $\forall x \in A\land x\in B (x\neq 0)$

zzr0ck3r · Answer

if B was included in A it would work.

anonymous · Answer

$x$ wasn't arbitrary since it was limited to elements in $A$.

zzr0ck3r · Answer

look at {1,2} in {0,1,2,3}