Question

For any sets A and B prove that:
1) A-B =A intersection B

Answer 1

Question:-

For any two sets $\sf{A}$ and $\sf{B}$ prove that $\sf{A-B=A-(A\cap B).}$

Solution:-

Let,

$\longrightarrow\sf{x\in A-B}$

$\longrightarrow\sf{x\in A\land x\notin B}$

$\longrightarrow\sf{x\in A\land x\notin[(A\cap B)\cup(B-A)]}$

$\longrightarrow\sf{x\in A\land x\in[(A\cap B)\cup(B-A)]'}$

$\longrightarrow\sf{x\in (A\cap A)\land x\in [(A\cap B)'\cap (B-A)']}$

$\longrightarrow\sf{x\in A\land x\in A\land x\in (A\cap B)'\land x\in (B-A)'}$

$\longrightarrow\sf{[x\in A\land x\notin(A\cap B)]\land [x\in A\land\lnot[x\in B\land x\notin A]}$

$\longrightarrow\sf{x\in [A-(A\cap B)]\land [x\in A\land (x\notin B\lor x\in A)]}$

$\longrightarrow\sf{x\in[A-(A\cap B)]\land [(x\in A\land x\notin B)\lor (x\in A\land x\in A)]}$

$\longrightarrow\sf{x\in [A-(A\cap B)]\land x\in [(A-B)\cup A]}$

$\longrightarrow\sf{x\in A\land x\notin(A\cap B)\land x\in A\quad\quad[\because\ (A-B)\subseteq A]}$

$\longrightarrow\sf{x\in [A-(A\cap B)]}$

This implies,

$\longrightarrow\sf{[A-(A\cap B)]\subseteq (A-B)\quad\quad\dots (1)}$

Let,

$\longrightarrow\sf{x\in [A-(A\cap B)]}$

$\longrightarrow\sf{x\in A\land x\notin (A\cap B)}$

$\longrightarrow\sf{x\in A\land\lnot[x\in A\land x\in B]}$

$\longrightarrow\sf{x\in A\land (x\notin A\lor x\notin B)}$

$\longrightarrow\sf{(x\in A\land x\notin A)\lor(x\in A\land x\notin B)}$

$\longrightarrow\sf{x\in [(A\cap A')\cup (A-B)]}$

$\longrightarrow\sf{x\in (A-B)}$

This implies,

$\longrightarrow\sf{(A-B)\subseteq[A-(A\cap B)]\quad\quad\dots (2)}$

Then, from (1) and (2),

$\longrightarrow\sf{\underline {\underline {A-B=A-(A\cap B)}}}$

Hence Proved!