TSTP Solution File: SEU931^5 by Duper---1.0
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% File : Duper---1.0
% Problem : SEU931^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n028.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 : Thu Aug 31 16:44:20 EDT 2023
% Result : Theorem 3.90s 4.11s
% Output : Proof 3.90s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.13/0.13 % Problem : SEU931^5 : TPTP v8.1.2. Released v4.0.0.
% 0.13/0.14 % Command : duper %s
% 0.14/0.35 % Computer : n028.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 : Wed Aug 23 14:29:33 EDT 2023
% 0.14/0.35 % CPUTime :
% 3.90/4.11 SZS status Theorem for theBenchmark.p
% 3.90/4.11 SZS output start Proof for theBenchmark.p
% 3.90/4.11 Clause #0 (by assumption #[]): Eq
% 3.90/4.11 (Not
% 3.90/4.11 ((Eq (fun Xx => f (g Xx)) fun Xx => g (f Xx)) →
% 3.90/4.11 (Exists fun Xx => And (Eq (g Xx) Xx) (∀ (Xz : Iota), Eq (g Xz) Xz → Eq Xz Xx)) → Exists fun Xy => Eq (f Xy) Xy))
% 3.90/4.11 True
% 3.90/4.11 Clause #1 (by clausification #[0]): Eq
% 3.90/4.11 ((Eq (fun Xx => f (g Xx)) fun Xx => g (f Xx)) →
% 3.90/4.11 (Exists fun Xx => And (Eq (g Xx) Xx) (∀ (Xz : Iota), Eq (g Xz) Xz → Eq Xz Xx)) → Exists fun Xy => Eq (f Xy) Xy)
% 3.90/4.11 False
% 3.90/4.11 Clause #2 (by clausification #[1]): Eq (Eq (fun Xx => f (g Xx)) fun Xx => g (f Xx)) True
% 3.90/4.11 Clause #3 (by clausification #[1]): Eq ((Exists fun Xx => And (Eq (g Xx) Xx) (∀ (Xz : Iota), Eq (g Xz) Xz → Eq Xz Xx)) → Exists fun Xy => Eq (f Xy) Xy)
% 3.90/4.11 False
% 3.90/4.11 Clause #4 (by clausification #[2]): Eq (fun Xx => f (g Xx)) fun Xx => g (f Xx)
% 3.90/4.11 Clause #5 (by argument congruence #[4]): ∀ (a : Iota), Eq ((fun Xx => f (g Xx)) a) ((fun Xx => g (f Xx)) a)
% 3.90/4.11 Clause #6 (by betaEtaReduce #[5]): ∀ (a : Iota), Eq (f (g a)) (g (f a))
% 3.90/4.11 Clause #7 (by clausification #[3]): Eq (Exists fun Xx => And (Eq (g Xx) Xx) (∀ (Xz : Iota), Eq (g Xz) Xz → Eq Xz Xx)) True
% 3.90/4.11 Clause #8 (by clausification #[3]): Eq (Exists fun Xy => Eq (f Xy) Xy) False
% 3.90/4.11 Clause #9 (by clausification #[7]): ∀ (a : Iota), Eq (And (Eq (g (skS.0 0 a)) (skS.0 0 a)) (∀ (Xz : Iota), Eq (g Xz) Xz → Eq Xz (skS.0 0 a))) True
% 3.90/4.11 Clause #10 (by clausification #[9]): ∀ (a : Iota), Eq (∀ (Xz : Iota), Eq (g Xz) Xz → Eq Xz (skS.0 0 a)) True
% 3.90/4.11 Clause #11 (by clausification #[9]): ∀ (a : Iota), Eq (Eq (g (skS.0 0 a)) (skS.0 0 a)) True
% 3.90/4.11 Clause #12 (by clausification #[10]): ∀ (a a_1 : Iota), Eq (Eq (g a) a → Eq a (skS.0 0 a_1)) True
% 3.90/4.11 Clause #13 (by clausification #[12]): ∀ (a a_1 : Iota), Or (Eq (Eq (g a) a) False) (Eq (Eq a (skS.0 0 a_1)) True)
% 3.90/4.11 Clause #14 (by clausification #[13]): ∀ (a a_1 : Iota), Or (Eq (Eq a (skS.0 0 a_1)) True) (Ne (g a) a)
% 3.90/4.11 Clause #15 (by clausification #[14]): ∀ (a a_1 : Iota), Or (Ne (g a) a) (Eq a (skS.0 0 a_1))
% 3.90/4.11 Clause #16 (by clausification #[8]): ∀ (a : Iota), Eq (Eq (f a) a) False
% 3.90/4.11 Clause #17 (by clausification #[16]): ∀ (a : Iota), Ne (f a) a
% 3.90/4.11 Clause #20 (by clausification #[11]): ∀ (a : Iota), Eq (g (skS.0 0 a)) (skS.0 0 a)
% 3.90/4.11 Clause #21 (by superposition #[20, 6]): ∀ (a : Iota), Eq (f (skS.0 0 a)) (g (f (skS.0 0 a)))
% 3.90/4.11 Clause #22 (by superposition #[20, 15]): ∀ (a a_1 : Iota), Or (Ne (skS.0 0 a) (skS.0 0 a)) (Eq (skS.0 0 a) (skS.0 0 a_1))
% 3.90/4.11 Clause #33 (by superposition #[21, 15]): ∀ (a a_1 : Iota), Or (Ne (f (skS.0 0 a)) (f (skS.0 0 a))) (Eq (f (skS.0 0 a)) (skS.0 0 a_1))
% 3.90/4.11 Clause #34 (by eliminate resolved literals #[22]): ∀ (a a_1 : Iota), Eq (skS.0 0 a) (skS.0 0 a_1)
% 3.90/4.11 Clause #112 (by eliminate resolved literals #[33]): ∀ (a a_1 : Iota), Eq (f (skS.0 0 a)) (skS.0 0 a_1)
% 3.90/4.11 Clause #114 (by superposition #[112, 17]): ∀ (a a_1 : Iota), Ne (skS.0 0 a) (skS.0 0 a_1)
% 3.90/4.11 Clause #128 (by forward contextual literal cutting #[114, 34]): False
% 3.90/4.11 SZS output end Proof for theBenchmark.p
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