{{BeginTableau|\vdash \left({p \lor \left({q \lor r}\right)}\right) \implies \left({\left({p \lor q}\right) \lor r}\right)|Instance 2 of the Hilbert-style systems}} {{TableauLine  |n = 1  |f = \paren{ \paren{ p \lor q } \lor r } \implies \paren{ p \lor \paren{ q \lor r } }  |rlnk = Rule of Association/Disjunction/Formulation 2/Reverse Implication  |rtxt = Rule of Association: Reverse Implication }} {{TableauLine  |n = 2  |f = \paren{ q \lor r } \implies \paren{ r \lor q }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A3$  |c = $q \,/\, p, r \,/\, q$ }} {{TableauLine  |n = 3  |f = \paren{ \paren{ q \lor r } \implies \paren{ r \lor q } } \implies \paren{ \paren{ p \lor \paren{ q \lor r } } \implies \paren{ p \lor \paren{ r \lor q } } }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A4$  |c = $\paren{ q \lor r } \,/\, q, \paren{ r \lor q } \,/\, r$ }} {{TableauLine  |n = 4  |f = \paren{ p \lor \paren{ q \lor r } } \implies \paren{ p \lor \paren{ r \lor q } }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 3$  |dep  = 2,3 }} {{TableauLine  |n = 5  |f = \paren{ \paren{ p \lor q } \lor r } \implies \paren{ p \lor \paren{ r \lor q } }  |rlnk = Hypothetical Syllogism/Formulation 1/Proof 3  |rtxt = Hypothetical Syllogism  |dep  = 1,4 }} {{TableauLine  |n = 6  |f = \paren{ p \lor \paren{ r \lor q } } \implies \paren{ \paren{ r \lor q } \lor p }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A3$  |c = $\paren{ r \lor q } \,/\, q$ }} {{TableauLine  |n = 7  |f = \paren{ \paren{ p \lor q } \lor r } \implies \paren{ \paren{ r \lor q } \lor p }  |rlnk = Hypothetical Syllogism/Formulation 1/Proof 3  |rtxt = Hypothetical Syllogism  |dep  = 5,6 }} {{TableauLine  |n = 8  |f = \paren{ r \lor \paren{ p \lor q } } \implies \paren{ \paren{ p \lor q } \lor r }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A3$  |c = $r \,/\, p, \paren{ p \lor q } \,/\, q$ }} {{TableauLine  |n = 9  |f = \paren{ r \lor \paren{ p \lor q } } \implies \paren{ \paren{ r \lor q } \lor p }  |rlnk = Hypothetical Syllogism/Formulation 1/Proof 3  |rtxt = Hypothetical Syllogism  |dep  = 7,8 }} {{TableauLine  |n = 10  |f = \paren{ q \lor p } \implies \paren{ p \lor q }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A3$  |c = $p/q, q/p$ }} {{TableauLine  |n = 11  |f = \paren{ \paren{ q \lor p } \implies \paren{ p \lor q } } \implies \paren{ \paren{ r \lor \paren{ q \lor p } } \implies \paren{ r \lor \paren{ p \lor q } } }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A4$  |c = $r \,/\, p, \paren{ q \lor p } \,/\, q, \paren{ p \lor q } \,/\, r$ }} {{TableauLine  |n = 12  |f = \paren{ r \lor \paren{ q \lor p } } \implies \paren{ r \lor \paren{ p \lor q } }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 3$  |dep  = 10,11 }} {{TableauLine  |n = 13  |f = \paren{ r \lor \paren{ q \lor p } } \implies \paren{ \paren{ r \lor q } \lor p }  |rlnk = Hypothetical Syllogism/Formulation 1/Proof 3  |rtxt = Hypothetical Syllogism  |dep  = 9,12 }} {{TableauLine  |n = 14  |f = \paren{ p \lor \paren{ q \lor r } } \implies \paren{ \paren{ p \lor q } \lor r }  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 1$  |dep = 13  |c = $r/p, p/r$ }} {{EndTableau}} {{qed}}