%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : ITP209^2 : TPTP v9.2.0. Released v8.0.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n002.cluster.edu % Model : x86_64 x86_64 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz % Memory : 8042.1875MB % OS : Linux 3.10.0-693.el7.x86_64 % CPULimit : 300s % WCLimit : 300s % DateTime : Fri Oct 3 07:51:27 PM UTC 2025 % Result : Theorem 5.59s 5.82s % Output : Proof 5.59s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.07/0.12 % Problem : ITP209^2 : TPTP v9.2.0. Released v8.0.0. % 0.07/0.13 % Command : duper %s % 0.13/0.34 % Computer : n002.cluster.edu % 0.13/0.34 % Model : x86_64 x86_64 % 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz % 0.13/0.34 % Memory : 8042.1875MB % 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64 % 0.13/0.34 % CPULimit : 300 % 0.13/0.34 % WCLimit : 300 % 0.13/0.34 % DateTime : Thu Oct 2 23:33:23 EDT 2025 % 0.13/0.34 % CPUTime : % 5.59/5.82 SZS status Theorem for theBenchmark.p % 5.59/5.82 SZS output start Proof for theBenchmark.p % 5.59/5.82 Clause #1 (by assumption #[]): Eq (∀ (A2 B2 : a), Eq (f A2 B2) (f B2 A2)) True % 5.59/5.82 Clause #2 (by assumption #[]): Eq (∀ (A2 B2 C : a), Eq (f A2 (f B2 C)) (f B2 (f A2 C))) True % 5.59/5.82 Clause #12 (by assumption #[]): Eq (Not (Eq (f (f a2 b) c) (f (f a2 c) b))) True % 5.59/5.82 Clause #50 (by clausification #[1]): ∀ (a_1 : a), Eq (∀ (B2 : a), Eq (f a_1 B2) (f B2 a_1)) True % 5.59/5.82 Clause #51 (by clausification #[50]): ∀ (a_1 a_2 : a), Eq (Eq (f a_1 a_2) (f a_2 a_1)) True % 5.59/5.82 Clause #52 (by clausification #[51]): ∀ (a_1 a_2 : a), Eq (f a_1 a_2) (f a_2 a_1) % 5.59/5.82 Clause #130 (by clausification #[2]): ∀ (a_1 : a), Eq (∀ (B2 C : a), Eq (f a_1 (f B2 C)) (f B2 (f a_1 C))) True % 5.59/5.82 Clause #131 (by clausification #[130]): ∀ (a_1 a_2 : a), Eq (∀ (C : a), Eq (f a_1 (f a_2 C)) (f a_2 (f a_1 C))) True % 5.59/5.82 Clause #132 (by clausification #[131]): ∀ (a_1 a_2 a_3 : a), Eq (Eq (f a_1 (f a_2 a_3)) (f a_2 (f a_1 a_3))) True % 5.59/5.82 Clause #133 (by clausification #[132]): ∀ (a_1 a_2 a_3 : a), Eq (f a_1 (f a_2 a_3)) (f a_2 (f a_1 a_3)) % 5.59/5.82 Clause #139 (by superposition #[133, 52]): ∀ (a_1 a_2 a_3 : a), Eq (f a_1 (f a_2 a_3)) (f a_2 (f a_3 a_1)) % 5.59/5.82 Clause #177 (by superposition #[139, 52]): ∀ (a_1 a_2 a_3 : a), Eq (f (f a_1 a_2) a_3) (f a_2 (f a_3 a_1)) % 5.59/5.82 Clause #1572 (by clausification #[12]): Eq (Eq (f (f a2 b) c) (f (f a2 c) b)) False % 5.59/5.82 Clause #1573 (by clausification #[1572]): Ne (f (f a2 b) c) (f (f a2 c) b) % 5.59/5.82 Clause #1574 (by forward demodulation #[1573, 177]): Ne (f b (f c a2)) (f (f a2 c) b) % 5.59/5.82 Clause #1575 (by forward demodulation #[1574, 133]): Ne (f c (f b a2)) (f (f a2 c) b) % 5.59/5.82 Clause #1576 (by forward demodulation #[1575, 177]): Ne (f c (f b a2)) (f c (f b a2)) % 5.59/5.82 Clause #1577 (by eliminate resolved literals #[1576]): False % 5.59/5.82 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------