%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : SYN406+1 : TPTP v9.2.0. Released v2.0.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n013.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 08:07:36 PM UTC 2025 % Result : Theorem 3.62s 3.80s % Output : Proof 3.62s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.06/0.12 % Problem : SYN406+1 : TPTP v9.2.0. Released v2.0.0. % 0.06/0.13 % Command : duper %s % 0.13/0.34 % Computer : n013.cluster.edu % 0.13/0.34 % Model : x86_64 x86_64 % 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz % 0.13/0.34 % Memory : 8042.1875MB % 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64 % 0.13/0.34 % CPULimit : 300 % 0.13/0.34 % WCLimit : 300 % 0.13/0.34 % DateTime : Thu Oct 2 15:28:23 EDT 2025 % 0.13/0.34 % CPUTime : % 3.62/3.80 SZS status Theorem for theBenchmark.p % 3.62/3.80 SZS output start Proof for theBenchmark.p % 3.62/3.80 Clause #0 (by assumption #[]): Eq (Not (And (∀ (X : Iota), f X → g X) (Exists fun Y => And (f Y) (h Y)) → Exists fun Z => And (g Z) (h Z))) True % 3.62/3.80 Clause #1 (by clausification #[0]): Eq (And (∀ (X : Iota), f X → g X) (Exists fun Y => And (f Y) (h Y)) → Exists fun Z => And (g Z) (h Z)) False % 3.62/3.80 Clause #2 (by clausification #[1]): Eq (And (∀ (X : Iota), f X → g X) (Exists fun Y => And (f Y) (h Y))) True % 3.62/3.80 Clause #3 (by clausification #[1]): Eq (Exists fun Z => And (g Z) (h Z)) False % 3.62/3.80 Clause #4 (by clausification #[2]): Eq (Exists fun Y => And (f Y) (h Y)) True % 3.62/3.80 Clause #5 (by clausification #[2]): Eq (∀ (X : Iota), f X → g X) True % 3.62/3.80 Clause #6 (by clausification #[4]): ∀ (a : Iota), Eq (And (f (skS.0 0 a)) (h (skS.0 0 a))) True % 3.62/3.80 Clause #7 (by clausification #[6]): ∀ (a : Iota), Eq (h (skS.0 0 a)) True % 3.62/3.80 Clause #8 (by clausification #[6]): ∀ (a : Iota), Eq (f (skS.0 0 a)) True % 3.62/3.80 Clause #9 (by clausification #[5]): ∀ (a : Iota), Eq (f a → g a) True % 3.62/3.80 Clause #10 (by clausification #[9]): ∀ (a : Iota), Or (Eq (f a) False) (Eq (g a) True) % 3.62/3.80 Clause #11 (by superposition #[8, 10]): ∀ (a : Iota), Or (Eq (g (skS.0 0 a)) True) (Eq False True) % 3.62/3.80 Clause #12 (by clausification #[3]): ∀ (a : Iota), Eq (And (g a) (h a)) False % 3.62/3.80 Clause #13 (by clausification #[12]): ∀ (a : Iota), Or (Eq (g a) False) (Eq (h a) False) % 3.62/3.80 Clause #14 (by clausification #[11]): ∀ (a : Iota), Eq (g (skS.0 0 a)) True % 3.62/3.80 Clause #15 (by superposition #[14, 13]): ∀ (a : Iota), Or (Eq True False) (Eq (h (skS.0 0 a)) False) % 3.62/3.80 Clause #16 (by clausification #[15]): ∀ (a : Iota), Eq (h (skS.0 0 a)) False % 3.62/3.80 Clause #17 (by superposition #[16, 7]): Eq False True % 3.62/3.80 Clause #18 (by clausification #[17]): False % 3.62/3.80 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------