Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Forward Implication/Proof 1

From ProofWiki
Jump to navigation Jump to search


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


This proof is derived in the context of the following proof system: instance 1 of a Gentzen proof system.

By the tableau method:

$\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }$
Line Pool Formula Rule Depends upon Notes
1 $\neg p, p, q$ Axiom
2 $\neg p, p \lor q$ $\beta$-Rule: $\beta \lor$ 1
3 $\neg p, p, r$ Axiom
4 $\neg p, p \lor r$ $\beta$-Rule: $\beta \lor$ 3
5 $\neg p, \paren {p \lor q} \land \paren {p \lor r}$ $\alpha$-Rule: $\alpha \land$ 2, 4
6 $\neg q, \neg r, p, q$ Axiom
7 $\neg q, \neg r, p \lor q$ $\beta$-Rule: $\beta \lor$ 6
8 $\neg q, \neg r, p, r$ Axiom
9 $\neg q, \neg r, p \lor r$ $\beta$-Rule: $\beta \lor$ 8
10 $\neg q, \neg r, \paren {p \lor q} \land \paren {p \lor r}$ $\alpha$-Rule: $\alpha \land$ 7, 9
11 $\neg \paren {q \land r}, \paren {p \lor q} \land \paren {p \lor r}$ $\beta$-Rule: $\beta \land$ 10
12 $\neg \paren {p \lor \paren {q \land r} }, \paren {p \lor q} \land \paren {p \lor r}$ $\alpha$-Rule: $\alpha \land$ 5, 11
13 $\paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }$ $\beta$-Rule: $\beta \implies$ 12