Prove for the following that one element of the pair is logically equivalent to the other one using logical equivalences:
a) ~d -> (a && b && c) = ~(a && ~d) && ((d || b) & (c || d))
b) (a->b) && (c->d) = (c NOR a) || (b && ~c) || (d && ~a) || (b && d)
c) (~a && ~b) <--> (c || d) = (a || b || c || d) && ~((a || b) && (c || d))
d) (a XOR b) -> (~c XOR d) = ((a NOR b) || (a && b)) || ((c NOR d) || (c && d))
a)
| a | b | c | d | (¬d → (a ∧ (b ∧ c))) |
|---|---|---|---|---|
| F | F | F | F | F |
| F | F | F | T | T |
| F | F | T | F | F |
| F | F | T | T | T |
| F | T | F | F | F |
| F | T | F | T | T |
| F | T | T | F | F |
| F | T | T | T | T |
| T | F | F | F | F |
| T | F | F | T | T |
| T | F | T | F | F |
| T | F | T | T | T |
| T | T | F | F | F |
| T | T | F | T | T |
| T | T | T | F | T |
| T | T | T | T | T |
| a | b | c | d | (¬(a ∧ ¬d) ∧ ((d ∨ b) ∧ (c ∨ d))) |
|---|---|---|---|---|
| F | F | F | F | F |
| F | F | F | T | T |
| F | F | T | F | F |
| F | F | T | T | T |
| F | T | F | F | F |
| F | T | F | T | T |
| F | T | T | F | T |
| F | T | T | T | T |
| T | F | F | F | F |
| T | F | F | T | T |
| T | F | T | F | F |
| T | F | T | T | T |
| T | T | F | F | F |
| T | T | F | T | T |
| T | T | T | F | F |
| T | T | T | T | T |
As last columns of both tables not same
~d -> (a && b && c) != ~(a && ~d) && ((d || b) & (c || d))
=====================================
b)
| a | b | c | d | ((a → b) ∧ (c → d)) |
|---|---|---|---|---|
| F | F | F | F | T |
| F | F | F | T | T |
| F | F | T | F | F |
| F | F | T | T | T |
| F | T | F | F | T |
| F | T | F | T | T |
| F | T | T | F | F |
| F | T | T | T | T |
| T | F | F | F | F |
| T | F | F | T | F |
| T | F | T | F | F |
| T | F | T | T | F |
| T | T | F | F | T |
| T | T | F | T | T |
| T | T | T | F | F |
| T | T | T | T | T |
| a | b | c | d | (¬(c ∨ a) ∨ ((b ∧ ¬c) ∨ ((d ∧ ¬a) ∨ (b ∧ d)))) |
|---|---|---|---|---|
| F | F | F | F | T |
| F | F | F | T | T |
| F | F | T | F | F |
| F | F | T | T | T |
| F | T | F | F | T |
| F | T | F | T | T |
| F | T | T | F | F |
| F | T | T | T | T |
| T | F | F | F | F |
| T | F | F | T | F |
| T | F | T | F | F |
| T | F | T | T | F |
| T | T | F | F | T |
| T | T | F | T | T |
| T | T | T | F | F |
| T | T | T | T | T |
As last columns of both tables are same
(a->b) && (c->d) = (c NOR a) || (b && ~c) || (d && ~a) || (b && d)
=====================================
c)
| a | b | c | d | ((¬a ∧ ¬b) ↔ (c ∨ d)) |
|---|---|---|---|---|
| F | F | F | F | F |
| F | F | F | T | T |
| F | F | T | F | T |
| F | F | T | T | T |
| F | T | F | F | T |
| F | T | F | T | F |
| F | T | T | F | F |
| F | T | T | T | F |
| T | F | F | F | T |
| T | F | F | T | F |
| T | F | T | F | F |
| T | F | T | T | F |
| T | T | F | F | T |
| T | T | F | T | F |
| T | T | T | F | F |
| T | T | T | T | F |
| a | b | c | d | ((a ∨ (b ∨ (c ∨ d))) ∧ ¬((a ∨ b) ∧ (c ∨ d))) |
|---|---|---|---|---|
| F | F | F | F | F |
| F | F | F | T | T |
| F | F | T | F | T |
| F | F | T | T | T |
| F | T | F | F | T |
| F | T | F | T | F |
| F | T | T | F | F |
| F | T | T | T | F |
| T | F | F | F | T |
| T | F | F | T | F |
| T | F | T | F | F |
| T | F | T | T | F |
| T | T | F | F | T |
| T | T | F | T | F |
| T | T | T | F | F |
| T | T | T | T | F |
As last columns of both tables are same
(~a && ~b) <--> (c || d) = (a || b || c || d) && ~((a || b) && (c || d))
=====================================
d)
| a | b | c | d | (a XOR b) -> (~c XOR d) |
|---|---|---|---|---|
| F | F | F | F | T |
| F | F | F | T | T |
| F | F | T | F | T |
| F | F | T | T | T |
| F | T | F | F | T |
| F | T | F | T | F |
| F | T | T | F | F |
| F | T | T | T | T |
| T | F | F | F | T |
| T | F | F | T | F |
| T | F | T | F | F |
| T | F | T | T | T |
| T | T | F | F | T |
| T | T | F | T | T |
| T | T | T | F | T |
| T | T | T | T | T |
| a | b | c | d | ((a NOR b) || (a && b)) || ((c NOR d) || (c && d)) |
|---|---|---|---|---|
| F | F | F | F | T |
| F | F | F | T | T |
| F | F | T | F | T |
| F | F | T | T | T |
| F | T | F | F | T |
| F | T | F | T | F |
| F | T | T | F | F |
| F | T | T | T | T |
| T | F | F | F | T |
| T | F | F | T | F |
| T | F | T | F | F |
| T | F | T | T | T |
| T | T | F | F | T |
| T | T | F | T | T |
| T | T | T | F | T |
| T | T | T | T | T |
As last columns of both tables are same
(a XOR b) -> (~c XOR d) = ((a NOR b) || (a && b)) || ((c NOR d) || (c && d))
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