Exercise 1.6 | Q 6.4 | Page 16
Using the truth table, verify
~(p ∨ q) ∨ (~ p ∧ q) ≡ ~ p
1 | 2 | 3 | 4 | 5 | 6 | 7 |
p | q | ~p | (p∨q) | ~(p∨q) | ~p∧q | ~(p∨q)∨(~p∧q) |
T | T | F | T | F | F | F |
T | F | F | T | F | F | F |
F | T | T | T | F | T | T |
F | F | T | F | T | F | T |
In the above truth table, the entries in columns 3 and 7 are identical.
∴ ~(p ∨ q) ∨ (~ p ∧ q) ≡ ~ p
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilyFd11AyMYXvGoklYFqXUpeZxvjXuaEfG3q-0Z2YX2L3yofOZQEaIc3nwRq5YHet2OkSryE__4n03_qEt9dyZdxrG2UdzG1XwvOzSSXGIzj__pXB23NfByYMn9ZZkhDZt9TyXycc6ewo/s16000-rw/MATHEMATICAL+LOGIC.001.jpeg)