Simplify the following so that the new circuit has a minimum number of switches. Also, draw the simplified circuit.

#### Solution

Let p: the switch S_{1 }is closed

q: the switch S_{2} is closed

r: the switch S_{3} is closed

s: the switch S_{4} is closed

t: the switch S_{5} 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

∼s: the switch S_{4}′ is closed or the switch S_{4} is open

∼t: the switch S_{5}′ is closed or the switch S_{5} is open.

Then the given circuit in symbolic form is:

[(p ∧ q) ∨ ∼r ∨ ∼s ∨ ∼t] ∧ [(p ∧ q) ∨ (r ∧ s ∧t )]

Using the laws of logic, we have,

[(p ∧ q) ∨ ∼r ∨ ∼s ∨ ∼t] ∧ [(p ∧ q) ∨ (r ∧ s ∧t )]

≡ [(p ∧ q) ∨ ∼ (r ∧ s ∧ t)] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)] .........(By De Morgan’s Law)

≡ (p ∧ q) ∨ [∼ (r ∧ s ∧ t) ∧ (r ∧ s ∧ t)] ......(By Distributive Law)

≡ (p ∧ q) ∨ F ...........(By Complement Law)

≡ p ∧ q .......(By Identity Law)

Hence, the alternative simplified circuit is: