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

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The discussion centers on proving that if the set difference \( c \setminus b \) is a subset of \( c \setminus a \), then it follows that \( a \subset b \). The proof involves assuming \( x \in c \) but \( x \notin b \), leading to a contradiction when \( x \) is also shown to be in \( a \). The participants clarify that negating the hypothesis is not appropriate in this context, emphasizing the importance of maintaining logical consistency throughout the proof process.

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

Let's negate the conclusion: [itex]a\not\subset b[/itex], which means that [itex]\exists x\in a[/itex], so that [itex]x\not\in b[/itex]. But by hypothesis, [itex]\forall x\not\in b, x\not\in a[/itex]. Contradiction, right ?
 
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You are correct. I see the difference. Thank you for the help.
 

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