Exercise 1.9 | Q 1.2 | Page 22
Without using truth table, show that
p ∧ [(~ p ∨ q) ∨ ~ q] ≡ p
L.H.S.
≡ p ∧ [(~ p ∨ q) ∨ ~ q]
≡ p ∧ [(~ p ∨ (q ∨ ~ q)] .....[Associative law]
≡ p ∧ (~ p ∨ T) .....[Complement law]
≡ p ∧ T .....[Identity law]
≡ p .....[Identity law]
≡ R.H.S.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2EBtdKJJdYc7-Q_gvZ8sYbB_L0lM99mDMyrYDFr90eRtfNg1-qQTn5DaJNXVrJeV5LeMZJlAibZ-i-Q04sdhnPaHxMM3nB2in8Kqxd1fgK4s0uFwBOYOuexDDfexLBm-J7uk7qnm93RY/s16000-rw/hacker-2883632_640.jpg)