Write the dual of p ∧ ~ p ≡ F .

To find the dual of p ∧ ~ p ≡ F In dual ∧ → ∨ , ∨ → ∧ , F → T , T → F Now let us make a truth table: p ~ p p ∧ ~ p T F F F T F Now, let us make truth table of p ∨ ~ p : p ~ p p ∨ ~ p T F T F T T Which gives p ∨ ~ p ≡ T Therefore, the dual of p ∧ ~ p ≡ F is p ∨ ~ p ≡ T

Using truth table verify that ~ p ∨ q ≡ ~ p ∧ ~ q .

To verify that ~ p ∨ q ≡ ~ p ∧ ~ q p q p ∨ q ~ p ∨ q ~ p ~ q ~ p ∧ ~ q T T T F F F F T F T F F T F F T T F T F F F F F T T T T Hence, verified.

Express the following circuit in symbolic form:

Let p : the switch S 1 is closed. q : the switch S 2 is closed. r : the switch S 3 is closed. ~ p : the switch S 1 ' is closed or the switch S 1 is open. ~ q : the switch S 2 ' is closed or the switch S 2 is open. ~ r : the switch S 3 ' is closed or the switch S 3 is open. l : the lamp L is on. The symbolic form of the given circuit is: p ∨ ( q ∧ r ) = l l is generally dropped and it can be written as: p ∨ ( q ∧ r ) Input-Output table p q r q ∧ r p ∨ ( q ∧ r ) 1 1 1 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 0 1 0 1 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0

Write the dual of p ∧ ~ p ≡ F .

To find the dual of p ∧ ~ p ≡ F In dual ∧ → ∨ , ∨ → ∧ , F → T , T → F Now let us make a truth table: p ~ p p ∧ ~ p T F F F T F Now, let us make truth table of p ∨ ~ p : p ~ p p ∨ ~ p T F T F T T Which gives p ∨ ~ p ≡ T Therefore, the dual of p ∧ ~ p ≡ F is p ∨ ~ p ≡ T