Use a truth table to verify the first De Morgan law (p ∧ q)’ ≡ p’ ∨ q’.
1 answer:
Answer:
(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
t t t
t f f
f f t
f f f
next step is getting the opposite
(p∧q)'
<em>f</em>
<em> t</em>
<em> t</em>
<em> t</em>
Then we get p' V q', V is true (t) when the first or the second is true.
p' V q'
f <em>f</em> f
f <em>t</em> t
t <em>t</em> f
t <em>t</em> t
Let's compare them, ≡ is true if the first is equal to the second one.
(p∧q)' ≡ (p' V q')
<em>f f </em>
<em> t t</em>
<em> t t</em>
<em> t t</em>
Both are true, so
(p ∧ q)’ ≡ p’ ∨ q’
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