Conjunction implies Disjunction

Theorem

$\vdash \paren {p \land q} \implies \paren {p \lor q}$

Proof by Truth Table

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.

$\begin{array}{|ccc|c|ccc|} \hline (p & \land & q) & \implies & (p & \lor & q) \\ \hline \F & \F & \F & \T & \F & \F & \F \\ \F & \F & \T & \T & \F & \T & \T \\ \T & \F & \F & \T & \T & \T & \F \\ \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$

$\blacksquare$

Proof 2

By the tableau method of natural deduction:

$\paren {p \land q} \implies \paren {p \lor q} $

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \land q$	Assumption	(None)
2		1	$p$	Rule of Simplification: $\land \EE_1$	1
3		1	$p \lor q$	Rule of Addition: $\lor \II_1$	2
4			$\paren {p \land q} \implies \paren {p \lor q}$	Rule of Implication: $\implies \II$	1 – 3	Assumption 1 has been discharged

$\blacksquare$