given $X\subseteq A$ prove $X\cap B \subseteq A$

Question

It looks that trivial for me that I can't be 100% sure I do the proof correctly. would appreciate your attention and critique.

this is what I did:

• given: $\forall {x\in X}: x\in A$
• Need to prove: $\forall y\in X\cap B: y\in A$

So $y\in X \cap B$ implies $y\in X$ and $y\in B$. Then from the statment '$p\land q \to p$' it follows that $y \in X$, but we also know that $X\subseteq A$. Hence $y \in A$. Therefore we conclude that: $X\cap B \subseteq A$.

Is everything correct??

Answer 1

Using your notation and a natural deduction style,

• Given: ∀x∈X:x∈A
• Need to prove: ∀y∈X∩B:y∈A
• --Assume y∈X∩B
• --Infer y∈X and y∈B
• --Instantiate ∀x∈X:x∈A with x=y to obtain (if $y\in X$ then $y\in A$)
• --Conclude y∈A using modus ponens
• Discharge the assumption y∈X∩B and generalize to recover ∀y∈X∩B:y∈A