Use a truth table to verify the first de morgan law (p ∧ q)’ ≡ p’ ∨ q’.

(p ∧ q)’ ≡ p’ ∨ q’

Step-by-step explanation:

First, p and q have just four (4) possibilities, p∧q is true (t) when p and q are both t.

p∧    q

ttt

tff

fft

fff

next step is getting the opposite

(p∧q)'

    f

    t

    t

    t

Then we get p' V q', V is true (t) when the first or the second is true.

p'V  q'

f  f  f

f  t  t

t  t  f

t  t  t

Let's compare them, ≡ is true if the first is equal to the second one.

(p∧q)'   ≡     (p' V q')

    f              f

    t              t

    t              t

    t              t

Both are true, so

(p ∧ q)’ ≡ p’ ∨ q’

irrational, cannot