1 / 20

Set Identities: Logical Approaches and Proofs

This educational text explores logical methods to prove set identities, including using membership and truth tables, propositional logic conversions, and applying set identities. It covers key principles such as intersection, union, and complement, demonstrating logical arguments for set equalities.

ssara
Download Presentation

Set Identities: Logical Approaches and Proofs

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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.

More Related