00:01
Hi here for the given question.
00:03
We need to prove that here in our case the value of a minus b union b minus a is equal to a union b minus a intersection b.
00:15
So here in our case this value we need to prove that for all set a and b.
00:20
So here first part we need to prove is a minus b union b minus a is contained in a union b minus a intersection b and the second part which we can prove easily is a union b minus a intersection c a intersection b is contained in a minus b union b minus a.
00:48
So here in our case now here first we will start with the first part.
00:53
So here let x be any arbitrary element which belongs in a minus b union b minus a which means x either is in a minus b or b minus a.
01:11
So here in our case first let us assume that x belongs to a minus b.
01:17
So here as we know that this implies x belongs to a but not in b.
01:26
Now here since x belongs to a which implies x also belongs to a union b.
01:34
However, x does not belongs to a intersection b.
01:41
So here in our case, we can prove that x belongs to a union b minus a intersection b.
01:50
So this contains x element.
01:52
So here we can say that for the first case we have proved that it is contained...