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Proof for 6: By the definition of the equality of sets, we need to prove that ∀x [x ∈ A ∪ (B ∩ C)] if and only if x ∈ (A ∪ B) ∩ (A ∪ C).
For that, considering the Universal Generalization rule, we need to show that for an arbitrary element in the universe x, x ∈ A ∪ (B ∩ C) if and only if x ∈ (A ∪ B) ∩ (A ∪ C).
Here the only if part is going to be proven. The if part can be proven similarly.
x ∈ A ∪ (B ∩ C) ⇔ x ∈ A ∨ x ∈ (B ∩ C) by the definition of ∪
⇔ x ∈ A ∨ (x ∈ B ⋀ x ∈ C) by the definition of ∩
⇔ (x ∈ A ∨ x ∈ B) ⋀ (x ∈ A ∨ x ∈ C) by the distribution from the equivalences of propositional logic
⇔ x ∈ (A ∪ B) ⋀ x ∈ (A ∪ C) by the definition of ∪.
⇔ x ∈ (A ∪ B) ∩ (A ∪ C) by the definition of ∩.
Proof for 8: (a) If A⊆B then A ∪ B = B.
Let x be an arbitrary element in the universe.
Then x ∈ A ∪ B ⇔ x ∈ A ∨ x ∈ B.
Since A⊆B, x ∈ A → x ∈ B
Also x ∈ B → x ∈ B
Hence x ∈ A ∪ B → x ∈ B.
Hence A ∪ B ⊆ B
Since B ⊆ A ∪ B (use "addition" rule), A ∪ B = B follows.
(b) Similarly for A ∩ B = A.
Alternative proof:
These can also be proven using 8, 14, and 15. For example, (b) can be proven as follows:
First by 15, A ∩B ⊆A.
Then since A ⊆A, and A ⊆B, by 7 A ∩A ⊆A ∩B.
Since A ∩A = A by 3, A ⊆A ∩B.
Proof for 9: Let x be an arbitrary element in the universe.
Then [x ∈ A ∪ (B - A)] ⇔ [x ∈ A ∨ (x ∈ B ⋀ x ∉ A)]
⇔ [(x ∈ A ∨ x ∈ B) ⋀ (x ∈ A ∨ x ∉ A)]
⇔ [(x ∈ A ∨ x ∈ B) ⋀ True]
⇔ [x ∈ A ∨ x ∈ B]
Hence A ∪ (B - A) = A ∪ B.
Alternative proof
This can also proven using set properties as follows.
A ∪( B - A ) = A ∪( B ∩ ) by the definition of ( B - A ) .
= ( A ∪B ) ∩( A ∪ ) by the distribution.
= (A ∪B ) ∩U
= ( A ∪B ) by 1.
Proof for 10: Suppose A ∩ (B - A) ≠ ∅.
Then there is an element x that is in A ∩ (B - A), i.e.
x ∈ A ∩ (B - A) ⇔ x ∈ A ⋀ x ∈ B – A
⇔ x ∈ A ⋀ (x ∈ B ⋀ x ∉ A)
⇔ (x ∈ A ⋀ x ∉ A) ⋀ x ∈ B
⇔ False
Hence A ∩ (B - A) ≠ ∅ does not hold.
Hence A ∩ (B - A) = ∅.
This can also be proven in the similar manner to 9 above.
Proof for 11: Let x be an arbitrary element in the universe.
Then x ∈ A - (B ∪ C) ⇔ x ∈ A ⋀ x ∉ B ∪ C
⇔ x ∈ A ⋀ ¬(x ∈ B ⋁ x ∈C)
⇔ x ∈ A ⋀ (x ∉ B ⋀ x ∉C)
⇔ (x ∈ A ⋀ x ∉ B) ⋀ x ∉C
⇔ (x ∈ A ⋀ x ∉ B) ⋀ (x ∈ A ⋀ x ∉C)
⇔ x ∈ A - B ⋀ x ∈ A - C
⇔ x ∈ (A - B) ∩ (A - C)
Hence A - (B ∪ C) = (A – B) ∩ (A - C)
Proof for 12:
(a) A ∪ B = U ⋀ A ∩ B = ∅ ⇒ B = ?
(b) Try to prove B ⊆ and ⊆ B.
Let x be an arbitrary element in the universe.
Then if x ∈ B, then x ∉A since A ∩ B = ∅. Hence x ∈ .
Hence B ⊆ .
If x ∈ , then x ∉A. Since x ∈ A ∨ x ∈ B (from A ∪ B = U ), x ∈ B
must hold. Hence ⊆ B.
Hence B = . (c) B = ⇒ A ∪ B = U ⋀ A ∩ B = ∅ ?
Since B = , A ∪ B = A ∪ (U-A) = A ∪ U = U since A ⊆ U
Also A ∩ B = A ∩ (U - A) = ∅ by 10 above.
Proof for 13: Since ∪ = ∪ , ∪ = U
Also since ∩ = ∩ , ∩ = ∅.
Hence satisfies the conditions for the complement of .
Hence .
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