%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : NUM694^1 : TPTP v9.2.0. Released v3.7.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n023.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 07:57:06 PM UTC 2025 % Result : Theorem 4.67s 4.90s % Output : Proof 4.67s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.00/0.09 % Problem : NUM694^1 : TPTP v9.2.0. Released v3.7.0. % 0.00/0.09 % Command : duper %s % 0.09/0.29 % Computer : n023.cluster.edu % 0.09/0.29 % Model : x86_64 x86_64 % 0.09/0.29 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz % 0.09/0.29 % Memory : 8042.1875MB % 0.09/0.29 % OS : Linux 3.10.0-693.el7.x86_64 % 0.09/0.29 % CPULimit : 300 % 0.09/0.29 % WCLimit : 300 % 0.09/0.29 % DateTime : Fri Oct 3 08:02:08 EDT 2025 % 0.09/0.29 % CPUTime : % 4.67/4.90 SZS status Theorem for theBenchmark.p % 4.67/4.90 SZS output start Proof for theBenchmark.p % 4.67/4.90 Clause #0 (by assumption #[]): Eq (∀ (Xx Xy : nat), moreis Xx Xy → lessis Xy Xx) True % 4.67/4.90 Clause #1 (by assumption #[]): Eq (∀ (Xx : nat), moreis Xx n_1) True % 4.67/4.90 Clause #2 (by assumption #[]): Eq (Not (lessis n_1 x)) True % 4.67/4.90 Clause #3 (by clausification #[2]): Eq (lessis n_1 x) False % 4.67/4.90 Clause #4 (by clausification #[1]): ∀ (a : nat), Eq (moreis a n_1) True % 4.67/4.90 Clause #5 (by clausification #[0]): ∀ (a : nat), Eq (∀ (Xy : nat), moreis a Xy → lessis Xy a) True % 4.67/4.90 Clause #6 (by clausification #[5]): ∀ (a a_1 : nat), Eq (moreis a a_1 → lessis a_1 a) True % 4.67/4.90 Clause #7 (by clausification #[6]): ∀ (a a_1 : nat), Or (Eq (moreis a a_1) False) (Eq (lessis a_1 a) True) % 4.67/4.90 Clause #8 (by superposition #[7, 4]): ∀ (a : nat), Or (Eq (lessis n_1 a) True) (Eq False True) % 4.67/4.90 Clause #9 (by clausification #[8]): ∀ (a : nat), Eq (lessis n_1 a) True % 4.67/4.90 Clause #10 (by superposition #[9, 3]): Eq True False % 4.67/4.90 Clause #11 (by clausification #[10]): False % 4.67/4.90 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------