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Rosen 1.6

Proving Things About Sets. Rosen 1.6. Approaches to Proofs. Membership tables (similar to truth tables) Convert to a problem in propositional logic, prove, then convert back

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Rosen 1.6

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  1. Proving Things About Sets Rosen 1.6

  2. 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)

  3. Prove (AB)  (AB) = B A B (AB) (AB) (AB)(AB) 1 1 1 0 1 1 0 0 0 0 0 1 0 1 1 0 0 0 0 0

  4. Prove (AB)  (AB) = B (AB)  (AB) = {x | x(AB)(AB)} Set builder notation = {x | x(AB)  x(AB)} Def of  = {x | (xA  xB)  (xA  xB)} Def of  x2 and Def of complement = {x | (xB  xA )  (xB  xA )} Commutative x2 = {x | (xB  (xA  xA )} Distributive = {x | (xB  T } Or tautology = {x | (xB } Identity = B Set Builder notation

  5. A Ø = A AU = A AU = U A Ø = Ø AA = A A A = A (A) = A Identity Laws Domination Laws Idempotent Laws Complementation Law Set Identities (Rosen, p. 89)

  6. A  B = B  A A  B = B  A A(BC) = (AB)C A(BC) = (AB)C A(BC)=(AB)(AC) A(BC)=(AB)(AC) A  B = A  B A  B = A  B Commutative Laws Associative Laws Distributive Laws De Morgan’s Laws Set Identities (cont.)

  7. Prove (AB)  (AB) = B (AB)  (AB) = (BA)  (BA) Commutative Law x2 =B  (A  A) Distributive Law =B  U Definition of U =B Identity Law

  8. Prove (AB)  (AB) = B Proof: We must show that (AB)  (AB)  B and that B  (AB)  (AB) . First we will show that (AB)  (AB)  B. Let e be an arbitrary element of (AB)  (AB). Then either e (AB) or e (AB). If e (AB), then eB and eA. If e (AB), then eB and eA. In either case e B.

  9. Prove (AB)  (AB) = B Now we will show that B  (AB)  (AB). Let e be an arbitrary element of B. Then either e AB or e AB. Since e is in one or the other, then e  (AB)  (AB).

  10. Prove AB  A Proof: We must show that any element in AB is also in A. Let e be an element in AB. 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.

  11. AA = (AA) - (AA) = (A) - (A) = Ø Definition of  Idempotent Laws Definition of - Prove AA = 

  12. Prove A  B = AB Proof: To show that A  B = AB we must show that A  B  AB and AB  AB. First we will show that A  B  AB. Let e A  B. We must show that e is also  AB. Since e AB, then e AB. So either eA or eB. If eA then eA. If eB then eB. In either case eAB

  13. DeMorgan Proof (cont.) Next we will show that AB  AB. Let e  AB. Then e  A or e  B. Therefore, by definition e A or e B. Therefore e (AB) which implies that e  AB Since A  B  AB and AB  AB then A  B = AB.

  14. Prove A(BC)=(AB)(AC) Proof: To show that A(BC)=(AB)(AC) we must show that A(BC)  (AB)(AC) and (AB)(AC)  A(BC).

  15. Distributive Proof (cont.) First we will show that A(BC)  (AB)(AC). Let e be an arbitrary element of A(BC). Then e  A and e  (BC). Since e  (BC), then either eB or eC 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 (AB) or (AC). Therefore e must be in the union of (AB) and (AC).

  16. Distributive Proof (cont.) Next we will show that (AB)(AC)  A(BC). Let x  (AB)(AC). Then either x(AB) or x(AC) or x is in both. If x is in (AB), then xA and xB If xB, then x(BC). Therefore x  A(BC). By a similar argument if x(AC) then, again, x  A(BC). Since A(BC)  (AB)(AC) and (AB)(AC)  A(BC), then A(BC) = (AB)(AC).

  17. Prove: [AB  AB]  [A = B] Proof: We must show that when AB  AB is true then A=B is true. (Proof by contradiction) Assume that AB  AB is true but AB. If AB then this means that either  xA but xB, or  xB but xA. If  xA but xB, then x  AB but x  AB so AB is not a subset of AB and we have a contradiction to our original assumption. By a similar argument AB is not a subset of AB if  xB but xA. Therefore [AB  AB]  [A = B].

  18. Prove or Disprove [AB=AC]  [B=C] False! A= Ø, B={a}, C={b} [AB=AC]  [B=C] False! A={a}, B= Ø, C={a}

  19. Prove: A (B-A) = AB Proof: We must show that A(B-A)  AB and AB A (B-A). First we will show that A(B-A)  AB. Let e  A(B-A). Then either eA or e (B-A). If eA, then e AB. Note that e cannot be an element of both by the definition of (B-A). If e (B-A), then eB and e A by the definition of (B-A). In this case, too, e AB. Thus A(B-A)  AB.

  20. Prove: A (B-A) = AB (cont.) Next we will show that AB  A(B-A). Let e AB. Then either eA or eB or both. If eA or both, then e A(B-A). The other case is eB, eA. In this case e (B-A) by the definition of (B-A). Again, this means that e A(B-A). Thus AB A(B-A). Therefore A(B-A) = AB.

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