Answer:
See below.
Step-by-step explanation:
To prove: A'UC' = A'UBUC' if and only if A ∩ B ∩ C = ∅
Proof:
First, we will show that A'UC' = A'UBUC' implies A ∩ B ∩ C = ∅.
Using De Morgan's Law on A'UC', we get:
A'UC' = (A∩C)' ∪ U ∪ (B∩C)'
Now, substituting A'UBUC' for A'UC', we get:
(A'UBUC') = (A∩B∩C)' ∪ U ∪ (A∩B∩C)'
Since U is the universal set, it contains all elements. Therefore, (A'UBUC') covers all elements that could be in A, B, and C.
Thus, (A'UBUC') = U, which implies that (A'UBUC') ∩ (A∩B∩C) = A∩B∩C.
Hence, we have (A'UC') ∩ (A∩B∩C) = A∩B∩C.
From here, we can deduce that (A'UC' ∩ A∩B∩C) = ∅.
Since A'UC' is the complement of A∩B∩C, we have:
(A'UC') ∩ (A∩B∩C) = ∅
(A'UC' ∩ A∩B∩C)' = U
(A'UC')' ∪ (A∩B∩C)' = U
(A∩(U∩C)') ∪ (A∩B∩C)' = U
(A ∩ C') ∪ (A∩B∩C)' = U
(A ∩ B ∩ C) = ∅
Therefore, A ∩ B ∩ C = ∅, which proves the first part of the theorem.
Now, we will prove that A ∩ B ∩ C = ∅ implies A'UC' = A'UBUC'.
Using De Morgan's Law, we get:
(A'UC')' = (A∩C)' ∪ (B∩C)'
(A'UC')' = (A'∪C) ∩ (B'∪C)
Using distributive law, we get:
(A'UC')' = (A'∩B') ∪ (A'∩C) ∪ (B'∩C) ∪ (C∩B)
Now, we will show that (A'∩B') ∪ (A'∩C) ∪ (B'∩C) = A'UBUC.
Let x be an arbitrary element of A'UBUC. Then, x ∈ A' or x ∈ B or x ∈ C.
If x ∈ A' or x ∈ B, then x ∈ A'∩B'. Therefore, x ∈ (A'∩B').
If x ∈ A' or x ∈ C, then x ∈ A'∩C. Therefore, x ∈ (A'∩C).
If x ∈ B' or x ∈ C, then x ∈ B'∩C. Therefore, x ∈ (B'∩C).
Hence, we have shown that any element in A'UBUC must belong to (A'∩B') ∪ (A'∩C) ∪ (B'∩C).
Conversely, let y be an arbitrary element of (A'∩B') ∪