Question

if A is subset of B and x not belongs to B, then x not belongs to A is true then prove it
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Answer 1

Since $A\subseteq B,$ we have,

$\longrightarrow B=A\cup(B-A)\quad\quad\dots(1)$

Given,

$\longrightarrow x\notin B$

From (1),

$\longrightarrow x\notin [A\cup(B-A)]$

This also means,

$\longrightarrow x\in [A\cup(B-A)]'$

By De Morgan's Theorem,

$\longrightarrow x\in [A'\cap(B-A)']$

Expanding it we get,

$\longrightarrow x\in A'\ \land\ x\in(B-A)'$

Or,

$\longrightarrow x\notin A\ \land\ x\notin(B-A)$

The statements implies that,

$\longrightarrow\underline{\underline{x\notin A}}$

else it makes the statement $x\notin B$ false.

Hence Proved!