%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SYN368+1 : TPTP v9.2.0. Released v2.0.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n010.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 08:07:31 PM UTC 2025 % Result : Theorem 5.45s 5.69s % Output : Proof 5.45s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.10/0.11 % Problem : SYN368+1 : TPTP v9.2.0. Released v2.0.0. % 0.10/0.12 % Command : duper %s % 0.11/0.33 % Computer : n010.cluster.edu % 0.11/0.33 % Model : x86_64 x86_64 % 0.11/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz % 0.11/0.33 % Memory : 8042.1875MB % 0.11/0.33 % OS : Linux 3.10.0-693.el7.x86_64 % 0.11/0.33 % CPULimit : 300 % 0.11/0.33 % WCLimit : 300 % 0.11/0.33 % DateTime : Thu Oct 2 15:45:53 EDT 2025 % 0.11/0.33 % CPUTime : % 5.45/5.69 SZS status Theorem for theBenchmark.p % 5.45/5.69 SZS output start Proof for theBenchmark.p % 5.45/5.69 Clause #0 (by assumption #[]): Eq (Not (Exists fun Y => ∀ (X : Iota), big_p Y → big_p X)) True % 5.45/5.69 Clause #1 (by clausification #[0]): Eq (Exists fun Y => ∀ (X : Iota), big_p Y → big_p X) False % 5.45/5.69 Clause #2 (by clausification #[1]): ∀ (a : Iota), Eq (∀ (X : Iota), big_p a → big_p X) False % 5.45/5.69 Clause #3 (by clausification #[2]): ∀ (a a_1 : Iota), Eq (Not (big_p a → big_p (skS.0 0 a a_1))) True % 5.45/5.69 Clause #4 (by clausification #[3]): ∀ (a a_1 : Iota), Eq (big_p a → big_p (skS.0 0 a a_1)) False % 5.45/5.69 Clause #5 (by clausification #[4]): ∀ (a : Iota), Eq (big_p a) True % 5.45/5.69 Clause #6 (by clausification #[4]): ∀ (a a_1 : Iota), Eq (big_p (skS.0 0 a a_1)) False % 5.45/5.69 Clause #7 (by superposition #[6, 5]): Eq False True % 5.45/5.69 Clause #8 (by clausification #[7]): False % 5.45/5.69 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------