Which of the following equations are equivalent? 1 [((ab +bc + cd)’ + (a + b…

We check each expression to see whether it simplifies to the constant 1 (tautology).

1. Expression 1: [( ((ab + bc + cd)' + (a + b + c) )' c )' + c

   Let X = ((ab + bc + cd)' + (a + b + c))'. Then expression 1 is (X·c)' + c. Using De Morgan, (X·c)' = X' + c'. Therefore (X·c)' + c = X' + c' + c = X' + 1 = 1. So expression 1 is identically 1.

2. Expression 2: (ab' + bc') + (b'c + a'b)' + b + c

   If b = 1 or c = 1 then b + c = 1 and the whole expression is 1. If b = 0 and c = 0 then b' = c' = 1 and ab' + bc' = a, while b'c + a'b = 0 so (b'c + a'b)' = 1. Then the expression equals a + 1 + 0 + 0 = 1. Thus in every case the expression equals 1, so expression 2 is identically 1.

3. Expression 3: [(a + b)(b + c)(c + d)]' + [ab + bc + cd]'

   This expression is not a tautology. For example, take a = b = c = d = 1. Then (a + b) = (b + c) = (c + d) = 1, so (a + b)(b + c)(c + d) = 1 and its complement is 0. Also ab + bc + cd = 1, so its complement is 0. The sum is 0 + 0 = 0. Therefore expression 3 can be 0 and is not equivalent to 1.

4. Expression 4: ab' + b + a'

   Use the identity b + ab' = a + b. Then ab' + b + a' = (b + ab') + a' = (a + b) + a' = a + a' + b = 1 + b = 1. So expression 4 is identically 1.

Conclusion: Expressions 1, 2, and 4 each simplify to the constant 1, while expression 3 does not. Therefore the equivalent set is expressions 1, 2 and 4.