Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Using truth table prove that p ↔ q = (p ∧ q) ∨ (~p ∧ ~q).
Advertisement Remove all ads
Solution
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| p | q | p ↔ q | ~p | ~q | p ∧ q | ~p ∧ ~q | (p ∧ q) ∨(~p∧~q) |
| T | T | T | F | F | T | F | T |
| T | F | F | F | T | F | F | F |
| F | T | F | T | F | F | F | F |
| F | F | T | T | T | F | T | T |
The entries in columns 3 and 8 are identical.
p ↔ q = (p ∧ q) ∨ (~p ∧ ~q).
Notes
[1 mark each for column 3 and column 8]
Concept: Logical Connective, Simple and Compound Statements
Is there an error in this question or solution?
