snick

snack
Supplement:
Worked Set Proofs
Based on work by Meghan Allen
Proofs with Sets #1 version 1
Prove that for all sets A and B:
A  B = A  B
Proof #1:
A  B =
=
=
=
=
Def’n of
{x  U | x  (A  B)}
{x  U | ~(x  A  x  B)}Def’n of 
{x  U | x  A  x  B} De Morgan’s
{x  U | x  A  x  B} Def’n of
Def’n of 
A  B
Proofs with Sets #1 version 2
Remember that for any two sets C and D,
C = D iff C is a subset of D and D is a
subset of C.
C = D  [C  D  D  C]
Proofs with Sets #1 version 2
a) Prove that: A  B  A  B
Pick an arbitrary x  A  B,
Then x  A  B. Def’n of
~(x  A  x  B)
Def’n of 
xA  x B
De Morgan’s
Def’n of
xA  x B
Def’n of 
x  (A  B)
Proofs with Sets #1 version 2
b) Prove that: A  B  A  B
Pick an arbitrary x  A  B
Then,
xA  x B
xA  x B
~(x  A  x  B)
x  A  B
xA  B
Proofs with Sets #1 version 2
conclusion
We have shown that A  B  A  B and
A  B A  B .
A  B = A  B.
Therefore,
Proof with Sets #2
Prove that, for all sets A, B and C:
(A  B) – C = (A – C)  (B – C)
This time we’ll use the Set Identities
LHS:
(A  B) – C = (A  B)  C
= C  (A  B)
= (C  A)  (C  B)
Set Difference Law
= (A  C)  (B  C)
= (A - C)  (B – C)
Commutative Law
Commutative Law
Distributive Law
Set Difference Law
Proofs with Sets #3
Prove or disprove, for all sets A and B,
A–B=B–A
We can disprove this with a
counterexample.
Let A = {1,2} and B = {2,3}
Then, A-B = {1} and B-A = {3}
A–B≠B–A
Proofs with Sets #4
Prove that for all sets A and B, if A  B then
A  B = 
Proof by contradiction.
Assume A  B and A  B  . Then, there exists an
element x  A  B. By the definition of intersection,
that means x  A and x  B. Since x  B, x  B by
the definition of complement. But, A  B, so since x 
A, x  B by definition of subset. Thus, x  B and x 
B which is a contradiction. Hence the assumption
is false and therefore for all sets A and B, if A  B
then A  B = 
Proofs with Sets #5
Prove that for all sets A and B, if A  B then
P(A)  P(B)
WLOG, let A and B be sets.
Proof by antecedent assumption:
Assume A  B. Now, consider X  P(A).
X  A by the definition of power set. But,
because A  B, X  B by the transitive
property of subsets* and thus, by the definition of
power set, X  P(B). This proves that for all X,
if X  P(A) then X  P(B) and so P(A) 
P(B).
*Epp -Theorem 5.2.1
Proofs with Sets #6 version 1
Prove that for all sets S,   S.
We proceed by contradiction.
Assume that there is a set S such that the
empty set is not a subset of S. Then, by
definition of subset, there must be some
element x of the empty set that is not an
element of S. However, the empty set has
no elements. This is a contradiction.
QED
Proofs with Sets #6 version 2
Prove that for all sets S,   S.
Note that   S  x  U, x    x  S
 x  U, x    x  S.
We proceed by proving this last statement.
We know nothing is an element of the empty
set: x  U, x  .
By generalization, it follows that:
x  U, x    x  S.
QED