Fsc Part 1 Mathematics (Complete Solution)
Q5
Prove that p∨(∼p∧∼q)∨(p∧q)=p∨(∼p∧∼q)
Solution:
| p | q | ∼p | ∼q | p∧q | ∼p∧∼q | p∨(∼p∧∼q) | p∨(∼p∧∼q)∨(p∧q) |
|---|---|---|---|---|---|---|---|
| T | T | F | F | T | F | T | T |
| T | F | F | T | F | F | T | T |
| F | T | T | F | F | F | F | F |
| F | F | T | T | F | T | T | T |
The last two columns of the above table show that the statements p∨(∼p∧∼q)∨(p∧q) and p∨(∼p∧∼q) are equivalent to each other. So p∨(∼p∧∼q)∨(p∧q)=p∨(∼p∧∼q) hence proved..