Rosen 1'6

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
Set Identities (Rosen, p. 89)

• 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

6
Set Identities (cont.)

• 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 Morgans Laws

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
Prove A?A ?

• A?A
• (A?A) - (A?A)
• (A) - (A)
• Ø

• Definition of ?
• Idempotent Laws
• Definition of -

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 AB 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?BA?C ? BC
• False! A Ø, Ba, Cb
• A?BA?C ? BC
• False! Aa, B Ø, Ca

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.