Biconditional iff Disjunction implies Conjunction/Formulation 1
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Theorem
- $p \iff q \dashv \vdash \paren {p \lor q} \implies \paren {p \land q}$
This can be expressed as two separate theorems:
Forward Implication
- $p \iff q \vdash \paren {p \lor q} \implies \paren {p \land q}$
Reverse Implication
- $\paren {p \lor q} \implies \paren {p \land q} \vdash p \iff q$
Proof
We apply the Method of Truth Tables.
As can be seen by inspection, in all cases the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||ccccccc|} \hline p & \iff & q & (p & \lor & q) & \implies & (p & \land & q) \\ \hline F & T & F & F & F & F & T & F & F & F \\ F & F & T & F & T & T & F & F & F & T \\ T & F & F & T & T & F & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$
$\blacksquare$