%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SET626^5 : TPTP v9.2.0. Released v4.0.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n022.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 08:00:17 PM UTC 2025 % Result : Theorem 3.55s 3.72s % Output : Proof 3.55s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.11/0.12 % Problem : SET626^5 : TPTP v9.2.0. Released v4.0.0. % 0.11/0.13 % Command : duper %s % 0.13/0.34 % Computer : n022.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 13:51:08 EDT 2025 % 0.13/0.34 % CPUTime : % 3.55/3.72 SZS status Theorem for theBenchmark.p % 3.55/3.72 SZS output start Proof for theBenchmark.p % 3.55/3.72 Clause #0 (by assumption #[]): Eq (Not (∀ (X Y : a → Prop), (Exists fun Xx => And (And (X Xx) (Y Xx)) (cZ Xx)) → Exists fun Xx => And (X Xx) (Y Xx))) % 3.55/3.72 True % 3.55/3.72 Clause #1 (by clausification #[0]): Eq (∀ (X Y : a → Prop), (Exists fun Xx => And (And (X Xx) (Y Xx)) (cZ Xx)) → Exists fun Xx => And (X Xx) (Y Xx)) False % 3.55/3.72 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop), % 3.55/3.72 Eq % 3.55/3.72 (Not % 3.55/3.72 (∀ (Y : a → Prop), % 3.55/3.72 (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (cZ Xx)) → Exists fun Xx => And (skS.0 0 a_1 Xx) (Y Xx))) % 3.55/3.72 True % 3.55/3.72 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop), % 3.55/3.72 Eq % 3.55/3.72 (∀ (Y : a → Prop), % 3.55/3.72 (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (Y Xx)) (cZ Xx)) → Exists fun Xx => And (skS.0 0 a_1 Xx) (Y Xx)) % 3.55/3.72 False % 3.55/3.72 Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop), % 3.55/3.72 Eq % 3.55/3.72 (Not % 3.55/3.72 ((Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (cZ Xx)) → % 3.55/3.72 Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx))) % 3.55/3.72 True % 3.55/3.72 Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop), % 3.55/3.72 Eq % 3.55/3.72 ((Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (cZ Xx)) → % 3.55/3.72 Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) % 3.55/3.72 False % 3.55/3.72 Clause #6 (by clausification #[5]): ∀ (a_1 a_2 : a → Prop), Eq (Exists fun Xx => And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (cZ Xx)) True % 3.55/3.72 Clause #7 (by clausification #[5]): ∀ (a_1 a_2 : a → Prop), Eq (Exists fun Xx => And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) False % 3.55/3.72 Clause #8 (by clausification #[6]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), % 3.55/3.72 Eq (And (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))) (cZ (skS.0 2 a_1 a_2 a_3))) % 3.55/3.72 True % 3.55/3.72 Clause #10 (by clausification #[8]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), % 3.55/3.72 Eq (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))) True % 3.55/3.72 Clause #11 (by clausification #[7]): ∀ (a_1 : a → Prop) (a_2 : a) (a_3 : a → Prop), Eq (And (skS.0 0 a_1 a_2) (skS.0 1 a_1 a_3 a_2)) False % 3.55/3.72 Clause #12 (by clausification #[11]): ∀ (a_1 : a → Prop) (a_2 : a) (a_3 : a → Prop), Or (Eq (skS.0 0 a_1 a_2) False) (Eq (skS.0 1 a_1 a_3 a_2) False) % 3.55/3.72 Clause #13 (by clausification #[10]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)) True % 3.55/3.72 Clause #14 (by clausification #[10]): ∀ (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 % 3.55/3.72 Clause #15 (by superposition #[14, 12]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Or (Eq True False) (Eq (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_3 a_4)) False) % 3.55/3.72 Clause #16 (by clausification #[15]): ∀ (a_1 a_2 a_3 : a → Prop) (a_4 : a), Eq (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_3 a_4)) False % 3.55/3.72 Clause #17 (by superposition #[16, 13]): Eq False True % 3.55/3.72 Clause #18 (by clausification #[17]): False % 3.55/3.72 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------