%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SEU661^2 : TPTP v9.2.0. Released v3.7.0. % 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 08:02:09 PM UTC 2025 % Result : Theorem 3.91s 4.08s % Output : Proof 3.91s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.06/0.12 % Problem : SEU661^2 : TPTP v9.2.0. Released v3.7.0. % 0.06/0.13 % Command : duper %s % 0.13/0.35 % Computer : n022.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 : Fri Oct 3 05:48:53 EDT 2025 % 0.13/0.35 % CPUTime : % 3.91/4.08 SZS status Theorem for theBenchmark.p % 3.91/4.08 SZS output start Proof for theBenchmark.p % 3.91/4.08 Clause #0 (by assumption #[]): Eq (Not (∀ (A B Xx : Iota), in Xx A → ∀ (Xy : Iota), in Xy B → Eq (kpair Xx Xy) (kpair Xx Xy))) True % 3.91/4.08 Clause #1 (by clausification #[0]): Eq (∀ (A B Xx : Iota), in Xx A → ∀ (Xy : Iota), in Xy B → Eq (kpair Xx Xy) (kpair Xx Xy)) False % 3.91/4.08 Clause #2 (by clausification #[1]): ∀ (a : Iota), % 3.91/4.08 Eq (Not (∀ (B Xx : Iota), in Xx (skS.0 0 a) → ∀ (Xy : Iota), in Xy B → Eq (kpair Xx Xy) (kpair Xx Xy))) True % 3.91/4.08 Clause #3 (by clausification #[2]): ∀ (a : Iota), Eq (∀ (B Xx : Iota), in Xx (skS.0 0 a) → ∀ (Xy : Iota), in Xy B → Eq (kpair Xx Xy) (kpair Xx Xy)) False % 3.91/4.08 Clause #4 (by clausification #[3]): ∀ (a a_1 : Iota), % 3.91/4.08 Eq (Not (∀ (Xx : Iota), in Xx (skS.0 0 a) → ∀ (Xy : Iota), in Xy (skS.0 1 a a_1) → Eq (kpair Xx Xy) (kpair Xx Xy))) % 3.91/4.08 True % 3.91/4.08 Clause #5 (by clausification #[4]): ∀ (a a_1 : Iota), % 3.91/4.08 Eq (∀ (Xx : Iota), in Xx (skS.0 0 a) → ∀ (Xy : Iota), in Xy (skS.0 1 a a_1) → Eq (kpair Xx Xy) (kpair Xx Xy)) False % 3.91/4.08 Clause #6 (by clausification #[5]): ∀ (a a_1 a_2 : Iota), % 3.91/4.08 Eq % 3.91/4.08 (Not % 3.91/4.08 (in (skS.0 2 a a_1 a_2) (skS.0 0 a) → % 3.91/4.08 ∀ (Xy : Iota), in Xy (skS.0 1 a a_1) → Eq (kpair (skS.0 2 a a_1 a_2) Xy) (kpair (skS.0 2 a a_1 a_2) Xy))) % 3.91/4.08 True % 3.91/4.08 Clause #7 (by clausification #[6]): ∀ (a a_1 a_2 : Iota), % 3.91/4.08 Eq % 3.91/4.08 (in (skS.0 2 a a_1 a_2) (skS.0 0 a) → % 3.91/4.08 ∀ (Xy : Iota), in Xy (skS.0 1 a a_1) → Eq (kpair (skS.0 2 a a_1 a_2) Xy) (kpair (skS.0 2 a a_1 a_2) Xy)) % 3.91/4.08 False % 3.91/4.08 Clause #9 (by clausification #[7]): ∀ (a a_1 a_2 : Iota), % 3.91/4.08 Eq (∀ (Xy : Iota), in Xy (skS.0 1 a a_1) → Eq (kpair (skS.0 2 a a_1 a_2) Xy) (kpair (skS.0 2 a a_1 a_2) Xy)) False % 3.91/4.08 Clause #10 (by clausification #[9]): ∀ (a a_1 a_2 a_3 : Iota), % 3.91/4.08 Eq % 3.91/4.08 (Not % 3.91/4.08 (in (skS.0 3 a a_1 a_2 a_3) (skS.0 1 a a_1) → % 3.91/4.08 Eq (kpair (skS.0 2 a a_1 a_2) (skS.0 3 a a_1 a_2 a_3)) (kpair (skS.0 2 a a_1 a_2) (skS.0 3 a a_1 a_2 a_3)))) % 3.91/4.08 True % 3.91/4.08 Clause #11 (by clausification #[10]): ∀ (a a_1 a_2 a_3 : Iota), % 3.91/4.08 Eq % 3.91/4.08 (in (skS.0 3 a a_1 a_2 a_3) (skS.0 1 a a_1) → % 3.91/4.08 Eq (kpair (skS.0 2 a a_1 a_2) (skS.0 3 a a_1 a_2 a_3)) (kpair (skS.0 2 a a_1 a_2) (skS.0 3 a a_1 a_2 a_3))) % 3.91/4.08 False % 3.91/4.08 Clause #13 (by clausification #[11]): ∀ (a a_1 a_2 a_3 : Iota), % 3.91/4.08 Eq (Eq (kpair (skS.0 2 a a_1 a_2) (skS.0 3 a a_1 a_2 a_3)) (kpair (skS.0 2 a a_1 a_2) (skS.0 3 a a_1 a_2 a_3))) False % 3.91/4.08 Clause #14 (by clausification #[13]): ∀ (a a_1 a_2 a_3 : Iota), % 3.91/4.08 Ne (kpair (skS.0 2 a a_1 a_2) (skS.0 3 a a_1 a_2 a_3)) (kpair (skS.0 2 a a_1 a_2) (skS.0 3 a a_1 a_2 a_3)) % 3.91/4.08 Clause #15 (by eliminate resolved literals #[14]): False % 3.91/4.08 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------