%------------------------------------------------------------------------------ % File : Duper---1.0 % Problem : PUZ140^1 : TPTP v9.2.0. Released v6.1.0. % Transfm : none % Format : tptp:raw % Command : duper %s % Computer : n024.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 07:58:25 PM UTC 2025 % Result : Theorem 5.24s 5.43s % Output : Proof 5.24s % Verified : % SZS Type : - % Comments : %------------------------------------------------------------------------------ %----WARNING: Could not form TPTP format derivation %------------------------------------------------------------------------------ %----ORIGINAL SYSTEM OUTPUT % 0.03/0.11 % Problem : PUZ140^1 : TPTP v9.2.0. Released v6.1.0. % 0.12/0.13 % Command : duper %s % 0.13/0.34 % Computer : n024.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 20:58:08 EDT 2025 % 0.13/0.34 % CPUTime : % 5.24/5.43 SZS status Theorem for theBenchmark.p % 5.24/5.43 SZS output start Proof for theBenchmark.p % 5.24/5.43 Clause #0 (by assumption #[]): Eq (∀ (B : beverage), hot (heat B)) True % 5.24/5.43 Clause #3 (by assumption #[]): Eq (Not (Exists fun Mixture => ∀ (S : syrup), Exists fun B => And (Eq (Mixture coffee S) B) (hot B))) True % 5.24/5.43 Clause #4 (by clausification #[0]): ∀ (a : beverage), Eq (hot (heat a)) True % 5.24/5.43 Clause #12 (by clausification #[3]): Eq (Exists fun Mixture => ∀ (S : syrup), Exists fun B => And (Eq (Mixture coffee S) B) (hot B)) False % 5.24/5.43 Clause #13 (by clausification #[12]): ∀ (a : beverage → syrup → beverage), Eq (∀ (S : syrup), Exists fun B => And (Eq (a coffee S) B) (hot B)) False % 5.24/5.43 Clause #14 (by clausification #[13]): ∀ (a : beverage → syrup → beverage) (a_1 : syrup), % 5.24/5.43 Eq (Not (Exists fun B => And (Eq (a coffee (skS.0 0 a a_1)) B) (hot B))) True % 5.24/5.43 Clause #15 (by clausification #[14]): ∀ (a : beverage → syrup → beverage) (a_1 : syrup), % 5.24/5.43 Eq (Exists fun B => And (Eq (a coffee (skS.0 0 a a_1)) B) (hot B)) False % 5.24/5.43 Clause #16 (by clausification #[15]): ∀ (a : beverage → syrup → beverage) (a_1 : syrup) (a_2 : beverage), % 5.24/5.43 Eq (And (Eq (a coffee (skS.0 0 a a_1)) a_2) (hot a_2)) False % 5.24/5.43 Clause #17 (by clausification #[16]): ∀ (a : beverage → syrup → beverage) (a_1 : syrup) (a_2 : beverage), % 5.24/5.43 Or (Eq (Eq (a coffee (skS.0 0 a a_1)) a_2) False) (Eq (hot a_2) False) % 5.24/5.43 Clause #18 (by clausification #[17]): ∀ (a : beverage) (a_1 : beverage → syrup → beverage) (a_2 : syrup), % 5.24/5.43 Or (Eq (hot a) False) (Ne (a_1 coffee (skS.0 0 a_1 a_2)) a) % 5.24/5.43 Clause #19 (by superposition #[18, 4]): ∀ (a : beverage → syrup → beverage) (a_1 : syrup) (a_2 : beverage), % 5.24/5.43 Or (Ne (a coffee (skS.0 0 a a_1)) (heat a_2)) (Eq False True) % 5.24/5.43 Clause #63 (by clausification #[19]): ∀ (a : beverage → syrup → beverage) (a_1 : syrup) (a_2 : beverage), Ne (a coffee (skS.0 0 a a_1)) (heat a_2) % 5.24/5.43 Clause #64 (by equality resolution #[63]): False % 5.24/5.43 SZS output end Proof for theBenchmark.p %------------------------------------------------------------------------------