For any three sets A, B and C

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Prove or disprove the following proposition
For any three sets A, B and C (A?B)?C
A?(B?C)

First you need to decide what to doIf the
proposition looks like to be true, prove it for
arbitrary sets. But suppose you suspect that it
not always true It is sufficient to find a
single counterexample to disprove it!
To disprove that the proposition is true for any
sets its sufficient to find a single example of
three sets, such that the proposition is false.
2
?
(A?B)?C A?(B?C)
3
We are going to disprove by counterexample.
(A?B)?C A?(B?C)
B
A
3
4
6
A 1, 2, 3, 4 B 1, 3, 5, 6 C 1, 2,
5, 7
1
2
5
7
C
A?B 1, 3, (A?B)?C 1, 2, 3, 5, 7 B?C
1, 2, 3, 5, 6, 7, A?(B?C) 1, 2, 3
4
You can use a membership method here.
5
Theorem 5. Let A, B and C be arbitrary sets.
Prove that if A?B C?B and A?B C?B, then A
C.
What this theorem says?
Counterexample A 1, C 1, 2, B 1, 2
Counterexample A 1, C 1, 2, B 1
But A?B C?B and A?B C?B ? A C
6
A
C
A ? C
A?B C?B
7
A?B C?B
A
C
A ? C
8
Direct proof of p ? q Assume p A?B C?B and
A?B C?B to prove q A C
The equality of two sets can be proved as two
subset relations (double inclusion ) A?C ?
C?A ? AC So, there should be two parts of the
proof.
1). A?B C?B and A?B C?B ? A?C
x?A ?x?A?B ? x?C?B
2). A?B C?B and A?B C?B ? C?A
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Part 1 proof of A?C
Assume A?B C?B (1) and A?B C ?B (2). To
prove A?C we need to show that any element from A
belongs to C, ?x x?A ? x?C . So, take
arbitrary element x ?A (3) to show that x?C. By
the Theorem 1, (3) implies x ? A?B (4). By
assumption (1) we conclude from (4) that x ? C?B
(5), or by the definition of the set union that x
? C or x ? B (6). So, we have one of two cases
to consider. Case x ? C we are done. Case x ? B
we can imply x ? A?B (7) by the Theorem 2. By
assumption (2) it can be implied from (7) that x
? C?B (8). From (8) we can conclude that x ? C
making use of the Theorem 2. So, in both cases
we come to x ? C. QED.
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Consider an example of application of the Theorem
5.
Theorem 6. For any three sets A, B and C if
(A?B)?C A?(B?C) then C ? A.
Proof. We assume (A?B)?C A?(B?C), (1) to prove
C ? A, (2) (direct proof). (A?B)?C
(A?C)?(B?C), by distributive law A?(B?C),
by assumption (1). So we have (A?C)?(B?C)
A?(B?C), (3). By associative and commutative
laws we have (A?C)?(B?C) A?(B?C), (4). By
the Theorem 5, (3) and (4) imply that A?C A,
(5). By the Theorem 4a (C ? A ? A?C A), (5)
implies C ? A. QED
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We can use a membership table method to prove
that (A?B)?C A?(B?C) ? C ? A.
C ? A 1 1 1 1 0 1 0
1
12
B
A
C
(A?B)?C A?(B?C) ? C ? A
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Theorem. For any two sets A and B prove or
disprove Power(A)?Power(B)Power(A?B).
A
A a, c B b, c
A?B a, b, c
Power(A)?, a, c, a, c
Power(B)?, b, c, b, c
Power(A?B) ?, a, c, b, a, b, a, c,
b, c, a, b, c
To disprove that Power(A)?Power(B)Power(A?B) it
is sufficient to find a single counterexample of
sets A and B for which the proposition is false.
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But, we can prove another theorem
Theorem 7. For any two sets A and B prove or
disprove Power(A)?Power(B) ? Power(A?B).
x?Power(A)?Power(B)? x?Power(A)? x?Power(B)
? x?A?x?B? x? A?B ? x ?Power(A?B)
Proof. Take any element x?Power(A)?Power(B), (1),
or x?Power(A)?x?Power(B), (2) by the definition
of the ?. By the definition of the power set,
(2) means that x?A?x?B, (3). Since A ?A ?B and
B?A ?B by the Theorem 1, and by the Theorem 3,
(3) implies that in both cases x? A?B, (4). By
the definition of power sets, (4) means that x
?Power(A?B). QED (Quad Erat Demonstrandum,
which was to be proved).
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Theorem 8. For any two sets A and B (A?B)?(A?B)
?
Proof by contradiction. Assume (A?B)?(A?B) ? ?,
(1). i.e., that there exists at least one element
in the intersection, (A?B)?(A?B) ? ? ??x x?
(A?B)?(A?B) ??x x?(A?B)? x?(A?B) ??x (x?A ?
x ?B)?( x?A ? x?B) ??x x?A ? (x?B? x?B) ??x
x?A ? x? ? ??x x?A?? ??x x?? ?F
We come to the contradiction that proves the
original proposition.
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Theorem 9. For any sets A, B and C (A?B)?C iff
(A?C) ?B
Part 1 to prove (A?B)?C ? (A?C) ?B
Proof by contradiction. Assume (A?B)?C and (A?C)
?B and show that this results in contradiction.
/
p??q ? False
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(A?C) ? B ? ??x x?(A?C) ? x?B ???xx?(A?C)?
x?B ??x?x?(A?C)? x?B ??xx?(A?C)?
x?B ??x(x?A ? x?C)? x?B ??x(x?A ? x?B) ?
x?C ??xx?(A?B) ? x?C ??x ?x?(A?B) ? x
?C ???xx?(A?B)? x?C ???xx?(A?B) ?x?C ?
(A?B)?C ? ?p
?q
/
/
p??q ? p??p? False
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• Direct proof
• To prove an implication p? q we assume p and
• show that q can be implied.
• Indirect proof
• contrapositive ? q ? ?p
• To prove an implication p? q we assume ? q
  and
• show that ?p can be implied.
• contradiction p ? ? q
• To prove an implication p? q we assume that
  p?? q and
• show that this assumption results in
  contradiction
• (is always false).

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Theorem. If A ?B, then A?B ? A?B.
A ?B ? A?B ? A?B
A?B A?B ? A B
We can prove the equivalent contrapositive
proposition
Proof. To prove A?B A?B ? A B assume A?B
A?B and prove 1) A ?B and 2) B ? A
1). x?A ? x? A ? B ? x? A ? B ? x?B
2) x?B ? x? A ? B ? x? A ? B ? x?A
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Cartesian Product of two sets
Definition. Let A and B denote two sets. Define
the Cartesian product of A of B as A?B(a, b)
a?A and b?B. The notation (a, b) denotes
an ordered pair (a 2-tuple). The order is
significant, that is two ordered pairs are equal,
i.e. (a, b) (c, d) iff ac and bd.

• In general A?B ? B ?A.

• A?B B?A only in two cases

1) AB
2) either A or B is an empty set.
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Product Rule. The cardinality of the Cartesian
product of two sets A?BAB. This is the
number of possible pairs formed from the
elements of two sets, when the first element of
a pair is taken from set A and the second element
is taken from set B.
More generally the Cartesian product of n sets
A1, A2,, An, is defined as a set of ordered
n-tuples A1?A2 ? ?An(a1, a2, , an) ai?Ai
for 1? i ? n
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Theorem. Let X and Y be two sets. Suppose A?X
and B?Y. Then A?B ? X?Y.
Proof. Assume A?X and B?Y. To prove A?B ? X?Y
take any element z? A?B. By the definition of
Cartesian product we can imply that z(a, b)
where a?A and b?B. From assumption A?X and B?Y
we can imply that a?X and b?Y, i. e. z (a, b)?
X?Y. In this way we proved that A?B ? X?Y.
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Let A A1?A2 and B B1?B2 . Consider A?B.
A?B(a, b) a ?A and b ?B (a, b)
(a?A1? a?A2) ? (b?B1? b?B2) (a, b)
(a?A1 ? b?B1) ? (a?A1 ? b?B2)? ? (a?A2 ? b?B1)
? (a?A2 ? b?B2) A1?B1 ? A1?B2 ?
A2?B1 ? A2?B2