%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : NUM550+1 : TPTP v9.2.0. Released v4.0.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n014.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 07:56:42 PM UTC 2025 % Result : Theorem 5.88s 6.08s % Output : Proof 5.88s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.08/0.12 % Problem : NUM550+1 : TPTP v9.2.0. Released v4.0.0. % 0.08/0.14 % Command : duper %s % 0.14/0.35 % Computer : n014.cluster.edu % 0.14/0.35 % Model : x86_64 x86_64 % 0.14/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz % 0.14/0.35 % Memory : 8042.1875MB % 0.14/0.35 % OS : Linux 3.10.0-693.el7.x86_64 % 0.14/0.35 % CPULimit : 300 % 0.14/0.35 % WCLimit : 300 % 0.14/0.35 % DateTime : Fri Oct 3 08:38:38 EDT 2025 % 0.14/0.35 % CPUTime : % 5.88/6.08 SZS status Theorem for theBenchmark.p % 5.88/6.08 SZS output start Proof for theBenchmark.p % 5.88/6.08 Clause #23 (by assumption #[]): Eq (aElementOf0 sz00 szNzAzT0) True % 5.88/6.08 Clause #51 (by assumption #[]): Eq (Eq (slbdtrb0 sz00) slcrc0) True % 5.88/6.08 Clause #55 (by assumption #[]): Eq (∀ (W0 : Iota), aElementOf0 W0 szNzAzT0 → Eq (sbrdtbr0 (slbdtrb0 W0)) W0) True % 5.88/6.08 Clause #61 (by assumption #[]): Eq (And (And (aSet0 xS) (aSet0 xT)) (Ne xk sz00)) True % 5.88/6.08 Clause #65 (by assumption #[]): Eq (And (And (aSet0 xQ) (isFinite0 xQ)) (Eq (sbrdtbr0 xQ) xk)) True % 5.88/6.08 Clause #66 (by assumption #[]): Eq (Not (Ne xQ slcrc0)) True % 5.88/6.08 Clause #81 (by clausification #[66]): Eq (Ne xQ slcrc0) False % 5.88/6.08 Clause #82 (by clausification #[81]): Eq xQ slcrc0 % 5.88/6.08 Clause #90 (by clausification #[51]): Eq (slbdtrb0 sz00) slcrc0 % 5.88/6.08 Clause #243 (by clausification #[55]): ∀ (a : Iota), Eq (aElementOf0 a szNzAzT0 → Eq (sbrdtbr0 (slbdtrb0 a)) a) True % 5.88/6.08 Clause #244 (by clausification #[243]): ∀ (a : Iota), Or (Eq (aElementOf0 a szNzAzT0) False) (Eq (Eq (sbrdtbr0 (slbdtrb0 a)) a) True) % 5.88/6.08 Clause #245 (by clausification #[244]): ∀ (a : Iota), Or (Eq (aElementOf0 a szNzAzT0) False) (Eq (sbrdtbr0 (slbdtrb0 a)) a) % 5.88/6.08 Clause #246 (by superposition #[245, 23]): Or (Eq (sbrdtbr0 (slbdtrb0 sz00)) sz00) (Eq False True) % 5.88/6.08 Clause #266 (by clausification #[246]): Eq (sbrdtbr0 (slbdtrb0 sz00)) sz00 % 5.88/6.08 Clause #267 (by forward demodulation #[266, 90]): Eq (sbrdtbr0 slcrc0) sz00 % 5.88/6.08 Clause #355 (by clausification #[61]): Eq (Ne xk sz00) True % 5.88/6.08 Clause #357 (by clausification #[355]): Ne xk sz00 % 5.88/6.08 Clause #908 (by clausification #[65]): Eq (Eq (sbrdtbr0 xQ) xk) True % 5.88/6.08 Clause #910 (by clausification #[908]): Eq (sbrdtbr0 xQ) xk % 5.88/6.08 Clause #911 (by forward demodulation #[910, 82]): Eq (sbrdtbr0 slcrc0) xk % 5.88/6.08 Clause #912 (by superposition #[911, 267]): Eq xk sz00 % 5.88/6.08 Clause #913 (by forward contextual literal cutting #[912, 357]): False % 5.88/6.08 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------