%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SET585^5 : TPTP v9.2.0. Released v4.0.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n010.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 08:00:11 PM UTC 2025 % Result : Theorem 5.56s 5.76s % Output : Proof 5.56s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.06/0.11 % Problem : SET585^5 : TPTP v9.2.0. Released v4.0.0. % 0.06/0.12 % Command : duper %s % 0.11/0.33 % Computer : n010.cluster.edu % 0.11/0.33 % Model : x86_64 x86_64 % 0.11/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz % 0.11/0.33 % Memory : 8042.1875MB % 0.11/0.33 % OS : Linux 3.10.0-693.el7.x86_64 % 0.11/0.33 % CPULimit : 300 % 0.11/0.33 % WCLimit : 300 % 0.11/0.33 % DateTime : Thu Oct 2 14:57:38 EDT 2025 % 0.11/0.33 % CPUTime : % 5.56/5.76 SZS status Theorem for theBenchmark.p % 5.56/5.76 SZS output start Proof for theBenchmark.p % 5.56/5.76 Clause #0 (by assumption #[]): Eq (Not (∀ (X Y Z : a → Prop) (Xx : a), And (X Xx) (Y Xx) → Or (X Xx) (Z Xx))) True % 5.56/5.76 Clause #1 (by clausification #[0]): Eq (∀ (X Y Z : a → Prop) (Xx : a), And (X Xx) (Y Xx) → Or (X Xx) (Z Xx)) False % 5.56/5.76 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop), % 5.56/5.76 Eq (Not (∀ (Y Z : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (Y Xx) → Or (skS.0 0 a_1 Xx) (Z Xx))) True % 5.56/5.76 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop), Eq (∀ (Y Z : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (Y Xx) → Or (skS.0 0 a_1 Xx) (Z Xx)) False % 5.56/5.76 Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop), % 5.56/5.76 Eq (Not (∀ (Z : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → Or (skS.0 0 a_1 Xx) (Z Xx))) True % 5.56/5.76 Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop), % 5.56/5.76 Eq (∀ (Z : a → Prop) (Xx : a), And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → Or (skS.0 0 a_1 Xx) (Z Xx)) False % 5.56/5.76 Clause #6 (by clausification #[5]): ∀ (a_1 a_2 a_3 : a → Prop), % 5.56/5.76 Eq (Not (∀ (Xx : a), And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → Or (skS.0 0 a_1 Xx) (skS.0 2 a_1 a_2 a_3 Xx))) True % 5.56/5.76 Clause #7 (by clausification #[6]): ∀ (a_1 a_2 a_3 : a → Prop), % 5.56/5.76 Eq (∀ (Xx : a), And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx) → Or (skS.0 0 a_1 Xx) (skS.0 2 a_1 a_2 a_3 Xx)) False % 5.56/5.76 Clause #8 (by clausification #[7]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), % 5.56/5.76 Eq % 5.56/5.76 (Not % 5.56/5.76 (And (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3 a_4)) → % 5.56/5.76 Or (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_3 a_4)))) % 5.56/5.76 True % 5.56/5.76 Clause #9 (by clausification #[8]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), % 5.56/5.76 Eq % 5.56/5.76 (And (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3 a_4)) → % 5.56/5.76 Or (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_3 a_4))) % 5.56/5.76 False % 5.56/5.76 Clause #10 (by clausification #[9]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), % 5.56/5.76 Eq (And (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 1 a_1 a_2 (skS.0 3 a_1 a_2 a_3 a_4))) True % 5.56/5.76 Clause #11 (by clausification #[9]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), % 5.56/5.76 Eq (Or (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) (skS.0 2 a_1 a_2 a_3 (skS.0 3 a_1 a_2 a_3 a_4))) False % 5.56/5.76 Clause #13 (by clausification #[10]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Eq (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) True % 5.56/5.76 Clause #15 (by clausification #[11]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Eq (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3 a_4)) False % 5.56/5.76 Clause #16 (by superposition #[15, 13]): Eq False True % 5.56/5.76 Clause #17 (by clausification #[16]): False % 5.56/5.76 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------