Factor Principles/Conjunction on Right/Formulation 1/Proof 1

From ProofWiki
Jump to navigation Jump to search


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


By the tableau method of natural deduction:

$p \implies q \vdash \paren {p \land r} \implies \paren {q \land r} $
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Premise (None)
2 $r \implies r$ Law of Identity (None) This is a theorem so depends on nothing
3 1 $\paren {p \implies q} \land \paren {r \implies r}$ Rule of Conjunction: $\land \II$ 1, 2
4 1 $\paren {p \land r} \implies \paren {q \land r}$ Sequent Introduction 3 Praeclarum Theorema