You can
put this solution on YOUR website!
We'll do each case separately then we'll
put the four cases into a truth table:
(P → Q) ↔ (星 ∨ Q)
Let "T" mean "true" and let "F" be "false:
Case 1: P is true and Q is true:
(P → Q) ↔ (星 ∨ Q)
(T → T) ↔ (曷 ∨ T)
T ↔ ( F ∨ T)
T ↔ T
T
Case 2: P is true and Q is false:
(P → Q) ↔ (星 ∨ Q)
(T → F) ↔ (曷 ∨ F)
F ↔ ( F ∨ F)
F ↔ F
T
Case 3: P is false and Q is true:
(P → Q) ↔ (星 ∨ Q)
(F → T) ↔ (政 ∨ T)
T ↔ ( T ∨ T)
T ↔ T
T
Case 4: P is false and Q is false:
(P → Q) ↔ (星 ∨ Q)
(F → F) ↔ (政 ∨ F)
T ↔ ( T ∨ F)
T ↔ T
T
It is true in all four cases, so it is a tautology.
Or you can do it as a truth table which is equivalent to
the four cases above:
| P | Q | 星 | P → Q | 星 ∨ Q | (P → Q) ↔ (星 ∨ Q |
case 1: | T | T | F | T | T | T |
case 2: | T | F | F | F | F | T |
case 3: | F | T | T | T | T | T |
case 4: | F | F | T | T | T | T |
Since the last column came out all trues, the original statement
is a tautology.
Edwin
