%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SET590^5 : TPTP v9.2.0. Released v4.0.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n012.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 08:00:12 PM UTC 2025 % Result : Theorem 3.56s 3.71s % Output : Proof 3.56s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.12/0.12 % Problem : SET590^5 : TPTP v9.2.0. Released v4.0.0. % 0.12/0.14 % Command : duper %s % 0.13/0.35 % Computer : n012.cluster.edu % 0.13/0.35 % Model : x86_64 x86_64 % 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz % 0.13/0.35 % Memory : 8042.1875MB % 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64 % 0.13/0.35 % CPULimit : 300 % 0.13/0.35 % WCLimit : 300 % 0.13/0.35 % DateTime : Thu Oct 2 14:08:23 EDT 2025 % 0.13/0.35 % CPUTime : % 3.56/3.71 SZS status Theorem for theBenchmark.p % 3.56/3.71 SZS output start Proof for theBenchmark.p % 3.56/3.71 Clause #0 (by assumption #[]): Eq (Not (∀ (X Y : a → Prop) (Xx : a), And (X Xx) (Not (Y Xx)) → X Xx)) True % 3.56/3.71 Clause #1 (by clausification #[0]): Eq (∀ (X Y : a → Prop) (Xx : a), And (X Xx) (Not (Y Xx)) → X Xx) False % 3.56/3.71 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop), Eq (Not (∀ (Y : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (Not (Y Xx)) → skS.0 0 a_1 Xx)) True % 3.56/3.71 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop), Eq (∀ (Y : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (Not (Y Xx)) → skS.0 0 a_1 Xx) False % 3.56/3.71 Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop), Eq (Not (∀ (Xx : a), And (skS.0 0 a_1 Xx) (Not (skS.0 1 a_1 a_2 Xx)) → skS.0 0 a_1 Xx)) True % 3.56/3.71 Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop), Eq (∀ (Xx : a), And (skS.0 0 a_1 Xx) (Not (skS.0 1 a_1 a_2 Xx)) → skS.0 0 a_1 Xx) False % 3.56/3.71 Clause #6 (by clausification #[5]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), % 3.56/3.71 Eq % 3.56/3.71 (Not % 3.56/3.71 (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (Not (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))) → % 3.56/3.71 skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3))) % 3.56/3.71 True % 3.56/3.71 Clause #7 (by clausification #[6]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), % 3.56/3.71 Eq % 3.56/3.71 (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (Not (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))) → % 3.56/3.71 skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) % 3.56/3.71 False % 3.56/3.71 Clause #8 (by clausification #[7]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), % 3.56/3.71 Eq (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (Not (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)))) True % 3.56/3.71 Clause #9 (by clausification #[7]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) False % 3.56/3.71 Clause #11 (by clausification #[8]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) True % 3.56/3.71 Clause #13 (by superposition #[11, 9]): Eq True False % 3.56/3.71 Clause #14 (by clausification #[13]): False % 3.56/3.71 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------