Prove the demorgans second law (anb)'=(a'ub'). Sho that au(bnc)=(aub)n(auc).
To prove De Morgan's second law, which states (A ∩ B)' = A' ∪ B', we need to show that the left-hand side of the equation, (A ∩ B)', is equivalent to the right-hand side, A' ∪ B'.
Proof:
1. Let's start by considering an arbitrary element x.
2. Assume that x belongs to (A ∩ B)'. This means that x is not in the intersection of A and B. It could either be not in A, not in B, or not in both.
3. In the case where x is not in A, it must belong to A'. Similarly, in the case where x is not in B, it must belong to B'.
4. Therefore, x must belong to A' ∪ B', which means (A ∩ B)' is a subset of A' ∪ B'.
Next, we need to show that A' ∪ B' is a subset of (A ∩ B)'.
Proof:
1. Let's start by considering an arbitrary element y.
2. Assume that y belongs to A' ∪ B'. This means that y either belongs to A' or B', or both.
3. In the case where y belongs to A', it is not in A. Therefore, it cannot simultaneously be in A and B, which implies y is not in (A ∩ B).
4. Similarly, in the case where y belongs to B', it is not in B. Therefore, y is also not in (A ∩ B).
5. In either case, y does not belong to (A ∩ B), and hence, it must be in (A ∩ B)'.
6. Therefore, A' ∪ B' is a subset of (A ∩ B)'.
Since we have shown that both (A ∩ B)' is a subset of A' ∪ B' and A' ∪ B' is a subset of (A ∩ B)', we can conclude that they are equal. Hence, the De Morgan's second law is proven.
Now let's move on to the second part of your question:
To show that Au(B ∩ C) = (A ∪ B) ∩ (A ∪ C), we need to prove that the left-hand side, Au(B ∩ C), is equal to the right-hand side, (A ∪ B) ∩ (A ∪ C).
Proof:
1. Let's start by considering an arbitrary element z.
2. Assume that z belongs to Au(B ∩ C). This means that z either belongs to A or to (B ∩ C), or to both.
3. In the case where z belongs to A, it must belong to A ∪ B and A ∪ C, which means it belongs to (A ∪ B) ∩ (A ∪ C).
4. In the case where z belongs to (B ∩ C), it must belong to A ∪ B and A ∪ C as well, hence it also belongs to (A ∪ B) ∩ (A ∪ C).
5. In both cases, z belongs to (A ∪ B) ∩ (A ∪ C), so Au(B ∩ C) is a subset of (A ∪ B) ∩ (A ∪ C).
Next, we need to show that (A ∪ B) ∩ (A ∪ C) is a subset of Au(B ∩ C).
Proof:
1. Let's start by considering an arbitrary element w.
2. Assume that w belongs to (A ∪ B) ∩ (A ∪ C). This means that w belongs to both A ∪ B and A ∪ C.
3. If w belongs to A, it also belongs to A ∪ (B ∩ C), as it is a subset of A. Therefore, w belongs to Au(B ∩ C).
4. If w belongs to B or C (or both), it also belongs to the intersection of B and C, which is (B ∩ C). Thus, w also belongs to Au(B ∩ C).
5. In either case, w belongs to Au(B ∩ C), and hence, (A ∪ B) ∩ (A ∪ C) is a subset of Au(B ∩ C).
Since we have shown that both Au(B ∩ C) is a subset of (A ∪ B) ∩ (A ∪ C) and (A ∪ B) ∩ (A ∪ C) is a subset of Au(B ∩ C), we can conclude that they are equal. Hence, the equality Au(B ∩ C) = (A ∪ B) ∩ (A ∪ C) is proven.