Obtain the simple logical expression of the following. Draw the corresponding switching circuit.
(p ∧ q ∧ ∼ p) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r)
Solution
Using the laws of logic, we have,
(p ∧ q ∧ ∼ p) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r)
≡ (p ∧∼ p ∧ q) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) .......(By Commutative Law)
≡ (F ∧ q) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) ...............(By Complement Law)
≡ F ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) ............(By Identity Law)
≡ (∼ p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) .........(By Identity Law)
≡ (∼ p ∨ p) ∧ (q ∧ r) ......(By Distributive Law)
≡ T ∧ (q ∧ r) ........(By Complement Law)
≡ q ∧ r .........(By Identity Law)
Hence, the simple logical expression of the given expression is q ∧ r.
Let q: the switch S2 is closed
r: the switch S3 is closed.
Then the corresponding switching circuit is:
