%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SEU920^5 : TPTP v9.2.0. Released v4.0.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n011.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 08:02:38 PM UTC 2025 % Result : Theorem 4.67s 4.85s % Output : Proof 4.67s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.08/0.13 % Problem : SEU920^5 : TPTP v9.2.0. Released v4.0.0. % 0.08/0.14 % Command : duper %s % 0.15/0.36 % Computer : n011.cluster.edu % 0.15/0.36 % Model : x86_64 x86_64 % 0.15/0.36 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz % 0.15/0.36 % Memory : 8042.1875MB % 0.15/0.36 % OS : Linux 3.10.0-693.el7.x86_64 % 0.15/0.36 % CPULimit : 300 % 0.15/0.36 % WCLimit : 300 % 0.15/0.36 % DateTime : Fri Oct 3 06:01:53 EDT 2025 % 0.15/0.36 % CPUTime : % 4.67/4.85 SZS status Theorem for theBenchmark.p % 4.67/4.85 SZS output start Proof for theBenchmark.p % 4.67/4.85 Clause #0 (by assumption #[]): Eq % 4.67/4.85 (Not (∀ (F : a → b) (G : b → c), (∀ (Y : c), Exists fun X => Eq (G (F X)) Y) → ∀ (Y : c), Exists fun X => Eq (G X) Y)) % 4.67/4.85 True % 4.67/4.85 Clause #1 (by clausification #[0]): Eq (∀ (F : a → b) (G : b → c), (∀ (Y : c), Exists fun X => Eq (G (F X)) Y) → ∀ (Y : c), Exists fun X => Eq (G X) Y) % 4.67/4.85 False % 4.67/4.85 Clause #2 (by clausification #[1]): ∀ (a_1 : a → b), % 4.67/4.85 Eq % 4.67/4.85 (Not (∀ (G : b → c), (∀ (Y : c), Exists fun X => Eq (G (skS.0 0 a_1 X)) Y) → ∀ (Y : c), Exists fun X => Eq (G X) Y)) % 4.67/4.85 True % 4.67/4.85 Clause #3 (by clausification #[2]): ∀ (a_1 : a → b), % 4.67/4.85 Eq (∀ (G : b → c), (∀ (Y : c), Exists fun X => Eq (G (skS.0 0 a_1 X)) Y) → ∀ (Y : c), Exists fun X => Eq (G X) Y) % 4.67/4.85 False % 4.67/4.85 Clause #4 (by clausification #[3]): ∀ (a_1 : a → b) (a_2 : b → c), % 4.67/4.85 Eq % 4.67/4.85 (Not % 4.67/4.85 ((∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 X)) Y) → % 4.67/4.85 ∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 X) Y)) % 4.67/4.85 True % 4.67/4.85 Clause #5 (by clausification #[4]): ∀ (a_1 : a → b) (a_2 : b → c), % 4.67/4.85 Eq % 4.67/4.85 ((∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 X)) Y) → % 4.67/4.85 ∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 X) Y) % 4.67/4.85 False % 4.67/4.85 Clause #6 (by clausification #[5]): ∀ (a_1 : a → b) (a_2 : b → c), Eq (∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 X)) Y) True % 4.67/4.85 Clause #7 (by clausification #[5]): ∀ (a_1 : a → b) (a_2 : b → c), Eq (∀ (Y : c), Exists fun X => Eq (skS.0 1 a_1 a_2 X) Y) False % 4.67/4.85 Clause #8 (by clausification #[6]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : c), Eq (Exists fun X => Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 X)) a_3) True % 4.67/4.85 Clause #9 (by clausification #[8]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : c) (a_4 : a), % 4.67/4.85 Eq (Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3 a_4))) a_3) True % 4.67/4.85 Clause #10 (by clausification #[9]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : c) (a_4 : a), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3 a_4))) a_3 % 4.67/4.85 Clause #11 (by clausification #[7]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : c), Eq (Not (Exists fun X => Eq (skS.0 1 a_1 a_2 X) (skS.0 3 a_1 a_2 a_3))) True % 4.67/4.85 Clause #12 (by clausification #[11]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : c), Eq (Exists fun X => Eq (skS.0 1 a_1 a_2 X) (skS.0 3 a_1 a_2 a_3)) False % 4.67/4.85 Clause #13 (by clausification #[12]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : b) (a_4 : c), Eq (Eq (skS.0 1 a_1 a_2 a_3) (skS.0 3 a_1 a_2 a_4)) False % 4.67/4.85 Clause #14 (by clausification #[13]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : b) (a_4 : c), Ne (skS.0 1 a_1 a_2 a_3) (skS.0 3 a_1 a_2 a_4) % 4.67/4.85 Clause #15 (by superposition #[14, 10]): ∀ (a_1 : c) (a_2 : a → b) (a_3 : b → c) (a_4 : c), Ne a_1 (skS.0 3 a_2 a_3 a_4) % 4.67/4.85 Clause #16 (by destructive equality resolution #[15]): False % 4.67/4.85 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------