Could someone check this proof? If c\b subset c\a, then prove a subset b

[IntroAnalysis]

Assume c\b is a subset of c\a. This means if x Є c Λ (Not Є) b, then it is an Є c Λ (Not Є) a.

Assume x Є c Λ (Not Є) b, but is Not Є c Λ (Not Є) a. Then x Є c Λ a. But this contradicts,
c\b is a subset of c\a. Therefore, a must be subset of b.

 

[dextercioby]

It doesn't look right to me. Why would you negate the hypothesis ? This is not reductio ad absurdum.

Let's negate the conclusion: $a\not\subset b$, which means that $\exists x\in a$, so that $x\not\in b$. But by hypothesis, $\forall x\not\in b, x\not\in a$. Contradiction, right ?

 

[IntroAnalysis]

You are correct. I see the difference. Thank you for the help.