Disjunction Math Example 4

Follow the full solution, then compare it with the other examples linked below.

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Use a truth table to verify De Morgan's Law:

\neg(p \lor q) \equiv \neg p \land \neg q

1
Rows: $(T,T),(T,F),(F,T),(F,F)$ . Compute $p \lor q$ : $T,T,T,F$ . Then $\neg(p \lor q)$ : $F,F,F,T$ .
2
Compute $\neg p$ : $F,F,T,T$ ; $\neg q$ : $F,T,F,T$ . Then $\neg p \land \neg q$ : $F,F,F,T$ .
3
Both columns are $F,F,F,T$ . The law is verified.

Answer

\neg(p \lor q) \equiv \neg p \land \neg q

De Morgan's Laws connect negation with conjunction and disjunction. This one says: negating 'A or B' is the same as 'not A and not B'.

About Disjunction

A disjunction $P \vee Q$ is a compound statement that is true whenever at least one of its parts is true.

Let [formula]: '[formula] is even' and [formula]: '[formula] is odd'. Evaluate [formula] and explain

Find all real [formula] satisfying '[formula] or [formula]', and express as a union of intervals.

Evaluate: (a) 'F [formula] T', (b) 'F [formula] F', (c) 'T [formula] T'.