%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SEU936^5 : TPTP v9.2.0. Released v4.0.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n019.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 08:02:39 PM UTC 2025 % Result : Theorem 3.66s 3.81s % Output : Proof 3.66s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.11/0.12 % Problem : SEU936^5 : TPTP v9.2.0. Released v4.0.0. % 0.11/0.13 % Command : duper %s % 0.13/0.34 % Computer : n019.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 : Fri Oct 3 05:30:38 EDT 2025 % 0.13/0.34 % CPUTime : % 3.66/3.81 SZS status Theorem for theBenchmark.p % 3.66/3.81 SZS output start Proof for theBenchmark.p % 3.66/3.81 Clause #0 (by assumption #[]): Eq % 3.66/3.81 (Not % 3.66/3.81 (∀ (F : a → b) (G : b → c), % 3.66/3.81 (∀ (Xx Xy : a), Eq (G (F Xx)) (G (F Xy)) → Eq Xx Xy) → ∀ (Xx Xy : a), Eq (F Xx) (F Xy) → Eq Xx Xy)) % 3.66/3.81 True % 3.66/3.81 Clause #1 (by clausification #[0]): Eq % 3.66/3.81 (∀ (F : a → b) (G : b → c), % 3.66/3.81 (∀ (Xx Xy : a), Eq (G (F Xx)) (G (F Xy)) → Eq Xx Xy) → ∀ (Xx Xy : a), Eq (F Xx) (F Xy) → Eq Xx Xy) % 3.66/3.81 False % 3.66/3.81 Clause #2 (by clausification #[1]): ∀ (a_1 : a → b), % 3.66/3.81 Eq % 3.66/3.81 (Not % 3.66/3.81 (∀ (G : b → c), % 3.66/3.81 (∀ (Xx Xy : a), Eq (G (skS.0 0 a_1 Xx)) (G (skS.0 0 a_1 Xy)) → Eq Xx Xy) → % 3.66/3.81 ∀ (Xx Xy : a), Eq (skS.0 0 a_1 Xx) (skS.0 0 a_1 Xy) → Eq Xx Xy)) % 3.66/3.81 True % 3.66/3.81 Clause #3 (by clausification #[2]): ∀ (a_1 : a → b), % 3.66/3.81 Eq % 3.66/3.81 (∀ (G : b → c), % 3.66/3.81 (∀ (Xx Xy : a), Eq (G (skS.0 0 a_1 Xx)) (G (skS.0 0 a_1 Xy)) → Eq Xx Xy) → % 3.66/3.81 ∀ (Xx Xy : a), Eq (skS.0 0 a_1 Xx) (skS.0 0 a_1 Xy) → Eq Xx Xy) % 3.66/3.81 False % 3.66/3.81 Clause #4 (by clausification #[3]): ∀ (a_1 : a → b) (a_2 : b → c), % 3.66/3.81 Eq % 3.66/3.81 (Not % 3.66/3.81 ((∀ (Xx Xy : a), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xx)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xy)) → Eq Xx Xy) → % 3.66/3.81 ∀ (Xx Xy : a), Eq (skS.0 0 a_1 Xx) (skS.0 0 a_1 Xy) → Eq Xx Xy)) % 3.66/3.81 True % 3.66/3.81 Clause #5 (by clausification #[4]): ∀ (a_1 : a → b) (a_2 : b → c), % 3.66/3.81 Eq % 3.66/3.81 ((∀ (Xx Xy : a), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xx)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xy)) → Eq Xx Xy) → % 3.66/3.81 ∀ (Xx Xy : a), Eq (skS.0 0 a_1 Xx) (skS.0 0 a_1 Xy) → Eq Xx Xy) % 3.66/3.81 False % 3.66/3.81 Clause #6 (by clausification #[5]): ∀ (a_1 : a → b) (a_2 : b → c), % 3.66/3.81 Eq (∀ (Xx Xy : a), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xx)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xy)) → Eq Xx Xy) True % 3.66/3.81 Clause #7 (by clausification #[5]): ∀ (a_1 : a → b), Eq (∀ (Xx Xy : a), Eq (skS.0 0 a_1 Xx) (skS.0 0 a_1 Xy) → Eq Xx Xy) False % 3.66/3.81 Clause #8 (by clausification #[6]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 : a), % 3.66/3.81 Eq (∀ (Xy : a), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_3)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 Xy)) → Eq a_3 Xy) True % 3.66/3.81 Clause #9 (by clausification #[8]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 a_4 : a), % 3.66/3.81 Eq (Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_3)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_4)) → Eq a_3 a_4) True % 3.66/3.81 Clause #10 (by clausification #[9]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 a_4 : a), % 3.66/3.81 Or (Eq (Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_3)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_4))) False) (Eq (Eq a_3 a_4) True) % 3.66/3.81 Clause #11 (by clausification #[10]): ∀ (a_1 a_2 : a) (a_3 : a → b) (a_4 : b → c), % 3.66/3.81 Or (Eq (Eq a_1 a_2) True) (Ne (skS.0 1 a_3 a_4 (skS.0 0 a_3 a_1)) (skS.0 1 a_3 a_4 (skS.0 0 a_3 a_2))) % 3.66/3.81 Clause #12 (by clausification #[11]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 a_4 : a), % 3.66/3.81 Or (Ne (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_3)) (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_4))) (Eq a_3 a_4) % 3.66/3.81 Clause #14 (by clausification #[7]): ∀ (a_1 : a → b) (a_2 : a), % 3.66/3.81 Eq (Not (∀ (Xy : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 Xy) → Eq (skS.0 2 a_1 a_2) Xy)) True % 3.66/3.81 Clause #15 (by clausification #[14]): ∀ (a_1 : a → b) (a_2 : a), % 3.66/3.81 Eq (∀ (Xy : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 Xy) → Eq (skS.0 2 a_1 a_2) Xy) False % 3.66/3.81 Clause #16 (by clausification #[15]): ∀ (a_1 : a → b) (a_2 a_3 : a), % 3.66/3.81 Eq % 3.66/3.81 (Not % 3.66/3.81 (Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3)) → % 3.66/3.81 Eq (skS.0 2 a_1 a_2) (skS.0 3 a_1 a_2 a_3))) % 3.66/3.81 True % 3.66/3.81 Clause #17 (by clausification #[16]): ∀ (a_1 : a → b) (a_2 a_3 : a), % 3.66/3.81 Eq % 3.66/3.81 (Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3)) → % 3.66/3.81 Eq (skS.0 2 a_1 a_2) (skS.0 3 a_1 a_2 a_3)) % 3.66/3.81 False % 3.66/3.81 Clause #18 (by clausification #[17]): ∀ (a_1 : a → b) (a_2 a_3 : a), Eq (Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3))) True % 3.66/3.81 Clause #19 (by clausification #[17]): ∀ (a_1 : a → b) (a_2 a_3 : a), Eq (Eq (skS.0 2 a_1 a_2) (skS.0 3 a_1 a_2 a_3)) False % 3.66/3.81 Clause #20 (by clausification #[18]): ∀ (a_1 : a → b) (a_2 a_3 : a), Eq (skS.0 0 a_1 (skS.0 2 a_1 a_2)) (skS.0 0 a_1 (skS.0 3 a_1 a_2 a_3)) % 3.66/3.81 Clause #21 (by superposition #[20, 12]): ∀ (a_1 : a → b) (a_2 : b → c) (a_3 a_4 a_5 : a), % 3.66/3.81 Or (Ne (skS.0 1 a_1 a_2 (skS.0 0 a_1 (skS.0 2 a_1 a_3))) (skS.0 1 a_1 a_2 (skS.0 0 a_1 a_4))) % 3.66/3.81 (Eq (skS.0 3 a_1 a_3 a_5) a_4) % 3.66/3.81 Clause #23 (by clausification #[19]): ∀ (a_1 : a → b) (a_2 a_3 : a), Ne (skS.0 2 a_1 a_2) (skS.0 3 a_1 a_2 a_3) % 3.66/3.81 Clause #24 (by equality resolution #[21]): ∀ (a_1 : a → b) (a_2 a_3 : a), Eq (skS.0 3 a_1 a_2 a_3) (skS.0 2 a_1 a_2) % 3.66/3.81 Clause #26 (by forward contextual literal cutting #[24, 23]): False % 3.66/3.81 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------