Question

Complete the truth table given below to verify the first de morgan law ¬(p ∧ q) ≡ ¬p ∨ ¬q.

Answer 1

`| p | q | ¬(p ∧ q) | ¬p ∨ ¬q ||---|---|----------|--------|`

`| T | T | F | F || T | F | T | T || F | T | T | T || F | F | T | T |`

Step-by-step explanation:

The truth table demonstrates that
`¬(p ∧ q)` is equivalent to
`¬p ∨ ¬q f`or all possible truth values of p and q. According to the first De Morgan law, the negation of a conjunction is equivalent to the disjunction of the negations of the individual propositions. In the truth table:

1. When both p and q are true
`(T), ¬(p ∧ q) is false (F), and ¬p ∨ ¬q`is also false (F).

2. When p is true (T) and q is false
`(F), ¬(p ∧ q) is true (T), and ¬p ∨ ¬q`is also true (T).

3. When p is false (F) and q is true (T),
`¬(p ∧ q) is true (T), and ¬p ∨ ¬q` is also true (T).

4. When both p and q are false
`(F), ¬(p ∧ q) is true (T), and ¬p ∨ ¬q` is also true (T).

Thus, the truth values match for both expressions, confirming the validity of the first De Morgan law
`¬(p ∧ q) ≡ ¬p ∨ ¬q.`