Eduprobe
Question:
Which one of the following statements is not a tautology?
(p∧q)→p
p→(p∨q)
(p∧q)→(¬p)∨q
(p∨q)→(p∨(¬q))
Solution:
Correct option is D (p∨q)→(p∨(¬q))
Let's analyze each option using truth tables:
A. (p∧q)→p
| p | q | p∧q | (p∧q)→p |
|---|---|---|---|
| T | T | T | T |
| T | F | F | T |
| F | T | F | T |
| F | F | F | T |
| This is a tautology. |
B. p→(p∨q)
| p | q | p∨q | p→(p∨q) |
|---|---|---|---|
| T | T | T | T |
| T | F | T | T |
| F | T | T | T |
| F | F | F | T |
| This is a tautology. |
C. (p∧q)→(¬p)∨q
| p | q | p∧q | ¬p | (¬p)∨q | (p∧q)→((¬p)∨q) |
|---|---|---|---|---|---|
| T | T | T | F | T | T |
| T | F | F | F | F | T |
| F | T | F | T | T | T |
| F | F | F | T | T | T |
| This is a tautology. |
D. (p∨q)→(p∨(¬q))
| p | q | p∨q | ¬q | p∨(¬q) | (p∨q)→(p∨(¬q)) |
|---|---|---|---|---|---|
| T | T | T | F | T | T |
| T | F | T | T | T | T |
| F | T | T | F | F | F |
| F | F | F | T | T | T |
| This is not a tautology because it's false when p is F and q is T. |
Therefore, the correct option is D.