Just wondering if someone could help me with this proof from a practice exam I'm studying for.

Let A and B be sets. Prove or disprove:

(B-A) union (A-B) = (A union B) - (A intersection B)

Thanks for any help.

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- September 13th 2006, 10:37 AMOntarioStudSet Theory Proof
Just wondering if someone could help me with this proof from a practice exam I'm studying for.

Let A and B be sets. Prove or disprove:

(B-A) union (A-B) = (A union B) - (A intersection B)

Thanks for any help. - September 13th 2006, 10:45 AMtopsquark
There are probably some set rules to break this down, but look at it this way: there are only 4 possibilities that you need to check.

1) A and B are disjoint.

2) A and B intersect, but do not contain one or the other.

3) A is a subset of B.

4) B is a subset of A.

(I suppose you ought to include the possibilities were either A or B or both are the empty set, but these should be easy.)

-Dan - September 13th 2006, 01:47 PMPlato
(A-B)U(B-A)

=(A^B’)U(B^A’)

=(AUB)^(B’UB)^(AUB’)^(B’UA’)

=(AUB)^(B’UA’) [(AUA’)^(B’UB) is space.]

=(AUB)^(B^A)’

=(AUB)-(A^B)

Too bad TeX is down. - September 13th 2006, 01:54 PMSoroban
Hello, OntarioStud!

I will use**A'**for "complement of A" and**0**for the empty set.

Quote:

Let A and B be sets.

Prove or disprove: .(B - A) U (A - B) = (A U B) - (A ∩ B)

Defintion of set subtraction: .A - B .= .A ∩ B'

Axiom #1: .A ∩ A' = 0

Axiom #2: .A U 0 = A

Distributive Laws: .A U (B ∩ C) = (A U B) ∩ (A U C)

. . . . . . . . . . . . . .A ∩ (B U C) = (A ∩ B) U (A ∩ C)

The right side is: .(A U B) - (A ∩ B)

. . . . . . . . . . .= .(A U B) ∩ (A ∩ B)' . Def. of Subtraction

. . . . . . . . . . .= .(A U B) ∩ (A' U B') . DeMorgan's Law

. . . . . . . . . . .= .[(A U B) ∩ A'] U [(A U B) ∩ B'] . Distr.Law

. . . . . . . . . . .= .[(A ∩ A') U (B ∩ A')] U [(A ∩ B') U (B ∩ B')] . Distr.Law

. . . . . . . . . . .= .[0 U (B ∩ A')] U [(A ∩ B') U 0] . Axiom #1

. . . . . . . . . . .= .(B ∩ A') U (A ∩ B') . Axiom #2

. . . . . . . . . . .= .(B - A) U (A - B) . Def. of Subtraction