Computer Science
From premises A ⇒ B and B ⇒ A, conclude B' + A.B.
Boolean Algebra
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Answer
We need to prove:
(A ⇒ B) ^ (B ⇒ A) ⇒ (B' + A.B) = 1
By using truth table:
| A | B | B' | A⇒B | B⇒A | A⇒B ^ B⇒A | A.B | B'+A.B | A⇒B ^ B⇒A ⇒ B'+A.B |
|---|---|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
All the elements of the column (A ⇒ B) ^ (B ⇒ A) ⇒ (B' + A.B) are 1, i.e. a tautology. Hence, B' + A.B is the conclusion of premises A ⇒ B and B ⇒ A.
By using algebraic method:
(A ⇒ B) ^ (B ⇒ A) ⇒ (B' + A.B)
((A ⇒ B) . (B ⇒ A))' + (B' + A.B) [a⇒b=a'+b]
((A' + B) . (B' + A))' + (B' + A.B)
(A' + B)' + (A + B')' + (B' + A.B) [Demorgan's Law]
(A'' . B') + (A' + B'') + (B' + A.B) [Demorgan's Law]
A.B' + A'.B + (B' + A.B)
A.B' + A'.B + [(A + B').(B + B')] [Distributive Law]
A.B' + A'.B + [(A + B'). 1] [B+B'=1]
A.B' + A'.B + A + B' [(A+B').1=A+B']
A.B' + B' + (A'.B + A)
A.B' + B' + [(A + A').(A + B)] [Distributive Law]
A.B' + B' + [1.(A + B)] [A+A'=1]
A.B' + B' + A + B
(A.B' + B) + B' + A
[(A + B).(B + B')] + B' + A [Distributive Law]
[(A + B).1] + B' + A [B+B'=1]
A + B + B' + A [B+B'=1]
A + 1 + A [1+A=1]
A + 1
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