Rosen 1.6

Approaches to Proofs

• Membership tables (similar to truth tables)• Convert to a problem in propositional logic, prove,

then convert back• Use set identities for a tabular proof (similar to

what we did for the propositional logic examples but using set identities)

• Do a logical (sentence-type) argument (similar to what we did for the number theory examples)

Prove (AB) (AB) = B

A B (AB) (AB) (AB)(AB)

1 1 1 0 1

1 0 0 0 0

0 1 0 1 1

0 0 0 0 0

Prove (AB) (AB) = B (AB) (AB)

= {x | x(AB)(AB)} Set builder notation

= {x | x(AB) x(AB)} Def of = {x | (xA xB) (xA xB)} Def of x2 and Def of

complement

= {x | (xB xA ) (xB xA )} Commutative x2

= {x | (xB (xA xA )} Distributive

= {x | (xB T } Or tautology

= {x | (xB } Identity

= B Set Builder notation

Set Identities (Rosen, p. 89)

A Ø = AAU = A

AU = UA Ø = Ø

AA = A A A = A

(A) = A

Identity Laws

Domination Laws

Idempotent Laws

Complementation Law

Set Identities (cont.)

A B = B AA B = B A

A(BC) = (AB)CA(BC) = (AB)C

A(BC)=(AB)(AC) A(BC)=(AB)(AC)

A B = A B A B = A B

Commutative Laws

Associative Laws

Distributive Laws

De Morgan’s Laws

Prove (AB) (AB) = B

(AB) (AB) =

(BA) (BA) Commutative Law x2

=B (A A) Distributive Law

=B U Definition of U

=B Identity Law

Prove (AB) (AB) = B

Proof: We must show that (AB) (AB) B and that B (AB) (AB) .

First we will show that (AB) (AB) B.

Let e be an arbitrary element of (AB) (AB). Then either e (AB) or e (AB). If e (AB), then eB and eA. If e (AB), then eB and eA. In either case e B.

Prove (AB) (AB) = B

Now we will show that B (AB) (AB).

Let e be an arbitrary element of B. Then either e AB or e AB. Since e is in one or the other, then e (AB) (AB).

Prove AB A

Proof: We must show that any element in AB is also in A. Let e be an element in AB. Since e is in the intersection of A and B, then e must be an element of A and e must be an element of B. Therefore e is in A.

Prove AA =

AA =

(AA) - (AA) =

(A) - (A) =

Ø

Definition of Idempotent Laws

Definition of -

Prove A B = AB

Proof: To show that A B = AB we must show that A B AB and AB AB.

First we will show that A B AB. Let e A B. We must show that e is also AB. Since e AB, then e AB. So either eA or eB. If eA then eA. If eB then eB. In either case eAB

DeMorgan Proof (cont.)

Next we will show that AB AB. Let e AB. Then e A or e B. Therefore, by definition e A or e B. Therefore e (AB) which implies that e AB

Since A B AB and AB AB then

A B = AB.

Prove A(BC)=(AB)(AC)

Proof: To show that A(BC)=(AB)(AC) we must show that A(BC) (AB)(AC) and (AB)(AC) A(BC).

Distributive Proof (cont.)

First we will show that A(BC) (AB)(AC). Let e be an arbitrary element of A(BC). Then e A and e (BC). Since e (BC), then either eB or eC or e is an element of both. Since e is in A and must be in at least one of B or C then e is an element of at least one of (AB) or (AC). Therefore e must be in the union of (AB) and (AC).

Distributive Proof (cont.)

Next we will show that (AB)(AC) A(BC). Let x (AB)(AC). Then either x(AB) or x(AC) or x is in both. If x is in (AB), then xA and xB If xB, then x(BC). Therefore x A(BC). By a similar argument if x(AC) then, again, x A(BC).

Since A(BC) (AB)(AC) and (AB)(AC) A(BC), then A(BC) = (AB)(AC).

Prove: [AB AB] [A = B]

Proof: We must show that when AB AB is true then A=B is true. (Proof by contradiction) Assume that AB AB is true but AB. If AB then this means that either xA but xB, or xB but xA. If xA but xB, then x AB but x AB so AB is not a subset of AB and we have a contradiction to our original assumption. By a similar argument AB is not a subset of AB if xB but xA.

Therefore [AB AB] [A = B].

Prove or Disprove

[AB=AC] [B=C]

False! A= Ø, B={a}, C={b}

[AB=AC] [B=C]

False! A={a}, B= Ø, C={a}

Prove: A (B-A) = AB

Proof: We must show that A(B-A) AB and AB A (B-A).

First we will show that A(B-A) AB. Let e A(B-A). Then either eA or e (B-A). If eA, then e AB. Note that e cannot be an element of both by the definition of (B-A). If e (B-A), then eB and e A by the definition of (B-A). In this case, too, e AB. Thus A(B-A) AB.

Prove: A (B-A) = AB (cont.)

Next we will show that AB A(B-A). Let e AB. Then either eA or eB or both. If eA or both, then e A(B-A). The other case is eB, eA. In this case e (B-A) by the definition of (B-A). Again, this means that e A(B-A). Thus AB A(B-A).

Therefore A(B-A) = AB.