%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SET627^5 : TPTP v9.2.0. Released v4.0.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n027.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 08:00:17 PM UTC 2025 % Result : Theorem 5.15s 5.37s % Output : Proof 5.15s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.06/0.11 % Problem : SET627^5 : TPTP v9.2.0. Released v4.0.0. % 0.06/0.12 % Command : duper %s % 0.11/0.33 % Computer : n027.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 14:34:23 EDT 2025 % 0.11/0.33 % CPUTime : % 5.15/5.37 SZS status Theorem for theBenchmark.p % 5.15/5.37 SZS output start Proof for theBenchmark.p % 5.15/5.37 Clause #0 (by assumption #[]): Eq (Not (∀ (X : a → Prop), Not (Exists fun Xx => And (X Xx) False))) True % 5.15/5.37 Clause #1 (by clausification #[0]): Eq (∀ (X : a → Prop), Not (Exists fun Xx => And (X Xx) False)) False % 5.15/5.37 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop), Eq (Not (Not (Exists fun Xx => And (skS.0 0 a_1 Xx) False))) True % 5.15/5.37 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop), Eq (Not (Exists fun Xx => And (skS.0 0 a_1 Xx) False)) False % 5.15/5.37 Clause #4 (by clausification #[3]): ∀ (a_1 : a → Prop), Eq (Exists fun Xx => And (skS.0 0 a_1 Xx) False) True % 5.15/5.37 Clause #5 (by clausification #[4]): ∀ (a_1 : a → Prop) (a_2 : a), Eq (And (skS.0 0 a_1 (skS.0 1 a_1 a_2)) False) True % 5.15/5.37 Clause #6 (by clausification #[5]): Eq False True % 5.15/5.37 Clause #8 (by clausification #[6]): False % 5.15/5.37 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------