{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|Instance 2 of the Hilbert-style systems}} {{TableauLine  | n = 1  | f = \paren {q \land r} \implies q  | rlnk = Rule of Simplification/Sequent Form/Formulation 2/Form 1  | rtxt = Rule of Simplification  | c = $q / p, r / q$ }} {{TableauLine  | n = 2  | f = \paren {\paren {q \land r} \implies q} \implies \paren {\paren {p \lor \paren {q \land r} } \implies \paren {p \lor q} }  | rlnk = Definition:Hilbert Proof System/Instance 2  | rtxt = Axiom $A 4$  | c = $\paren {q \land r} \, / \, q, q \, / \, r$ }} {{TableauLine  | n = 3  | f = \paren {p \lor \paren {q \land r} } \implies \paren {p \lor q}  | rlnk = Definition:Hilbert Proof System/Instance 2  | rtxt = Rule $RST \, 3$  | dep = 1, 2 }} {{TableauLine  | n = 4  | f = \paren {q \land r} \implies \paren {r \land q}  | rlnk = Rule of Commutation/Conjunction/Formulation 2  | rtxt = Rule of Commutation  | c = $q / p, r / q$ }} {{TableauLine  | n = 5  | f = \paren {r \land q} \implies r  | rlnk = Rule of Simplification/Sequent Form/Formulation 2/Form 1  | rtxt = Rule of Simplification  | c = $r / p$ }} {{TableauLine  | n = 6  | f = \paren {q \land r} \implies r  | rlnk = Hypothetical Syllogism/Formulation 1/Proof 3  | rtxt = Hypothetical Syllogism  | dep = 4, 5 }} {{TableauLine  | n = 7  | f = \paren {\paren {q \land r} \implies r} \implies \paren {\paren {p \lor \paren {q \land r} } \implies \paren {p \lor r} }  | rlnk = Definition:Hilbert Proof System/Instance 2  | rtxt = Axiom $A4$  | c = $\paren {q \land r} \, / \, q$ }} {{TableauLine  | n = 8  | f = \paren {p \lor \paren {q \land r} } \implies \paren {p \lor r}  | rlnk = Definition:Hilbert Proof System/Instance 2  | rtxt = Rule $RST \, 3$  | dep = 6, 7 }} {{TableauLine  | n = 9  | f = \paren {p \lor q} \implies \paren {\paren {p \lor r} \implies \paren {\paren {p \lor q} \land \paren {p \lor r} } }  | rlnk = Rule of Conjunction/Sequent Form/Formulation 2  | rtxt = Rule of Conjunction  | c = $\paren {p \lor q} \, / \, p, \paren {p \lor r} \, / \, q$ }} {{TableauLine  | n = 10  | f = \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor r} \implies \paren {\paren {p \lor q} \land \paren {p \lor r} } }  | rlnk = Hypothetical Syllogism/Formulation 1/Proof 3  | rtxt = Hypothetical Syllogism  | dep = 3, 9 }} {{TableauLine  | n = 11  | f = \paren {\paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor r} \implies \paren {p \lor q} \land \paren {p \lor r} } } \\ \implies \paren {\paren {p \lor r} \implies \paren {\paren {p \lor \paren {q \land r} } \implies \paren {p \lor q} \land \paren {p \lor r} } }  | rlnk = Principle of Commutation/Forward Implication/Formulation 2  | rtxt = Principle of Commutation  | c = $\paren {p \lor \paren {q \land r} } \, / \, p, \paren {p \lor r} \, / \, q, \paren {\paren {p \lor q} \land \paren {p \lor r} } \, / \, r$ }} {{TableauLine  | n = 12  | f = \paren {p \lor r} \implies \paren {\paren {p \lor \paren {q \land r} } \implies \paren {p \lor q} \land \paren {p \lor r} }  | rlnk = Definition:Hilbert Proof System/Instance 2  | rtxt = Rule $RST \, 3$  | dep = 10, 11 }} {{TableauLine  | n = 13  | f = \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor \paren {q \land r} } \implies \paren {p \lor q} \land \paren {p \lor r} }  | rlnk = Hypothetical Syllogism/Formulation 1/Proof 3  | rtxt = Hypothetical Syllogism  | dep = 8, 12 }} {{TableauLine  | n = 14  | f = \paren {p \lor \paren {q \land r} } \implies \paren {p \lor q} \land \paren {p \lor r} }} {{EndTableau}} {{Qed}}