TSTP Solution File: SEV241^5 by Duper---1.0
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% File : Duper---1.0
% Problem : SEV241^5 : 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 19:24:34 EDT 2023
% Result : Theorem 3.50s 3.88s
% Output : Proof 3.50s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13 % Problem : SEV241^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.15 % Command : duper %s
% 0.15/0.36 % Computer : n028.cluster.edu
% 0.15/0.36 % Model : x86_64 x86_64
% 0.15/0.36 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.36 % Memory : 8042.1875MB
% 0.15/0.36 % OS : Linux 3.10.0-693.el7.x86_64
% 0.15/0.36 % CPULimit : 300
% 0.15/0.36 % WCLimit : 300
% 0.15/0.36 % DateTime : Thu Aug 24 03:23:04 EDT 2023
% 0.15/0.36 % CPUTime :
% 3.50/3.88 SZS status Theorem for theBenchmark.p
% 3.50/3.88 SZS output start Proof for theBenchmark.p
% 3.50/3.88 Clause #0 (by assumption #[]): Eq (Not (∀ (Xx : a), And (cU Xx) (cW Xx) → ∀ (S : a → Prop), Or (Eq S cU) (Eq S cW) → S Xx)) True
% 3.50/3.88 Clause #1 (by clausification #[0]): Eq (∀ (Xx : a), And (cU Xx) (cW Xx) → ∀ (S : a → Prop), Or (Eq S cU) (Eq S cW) → S Xx) False
% 3.50/3.88 Clause #2 (by clausification #[1]): ∀ (a_1 : a),
% 3.50/3.88 Eq (Not (And (cU (skS.0 0 a_1)) (cW (skS.0 0 a_1)) → ∀ (S : a → Prop), Or (Eq S cU) (Eq S cW) → S (skS.0 0 a_1))) True
% 3.50/3.88 Clause #3 (by clausification #[2]): ∀ (a_1 : a),
% 3.50/3.88 Eq (And (cU (skS.0 0 a_1)) (cW (skS.0 0 a_1)) → ∀ (S : a → Prop), Or (Eq S cU) (Eq S cW) → S (skS.0 0 a_1)) False
% 3.50/3.88 Clause #4 (by clausification #[3]): ∀ (a_1 : a), Eq (And (cU (skS.0 0 a_1)) (cW (skS.0 0 a_1))) True
% 3.50/3.88 Clause #5 (by clausification #[3]): ∀ (a_1 : a), Eq (∀ (S : a → Prop), Or (Eq S cU) (Eq S cW) → S (skS.0 0 a_1)) False
% 3.50/3.88 Clause #6 (by clausification #[4]): ∀ (a_1 : a), Eq (cW (skS.0 0 a_1)) True
% 3.50/3.88 Clause #7 (by clausification #[4]): ∀ (a_1 : a), Eq (cU (skS.0 0 a_1)) True
% 3.50/3.88 Clause #8 (by clausification #[5]): ∀ (a_1 : a) (a_2 : a → Prop),
% 3.50/3.88 Eq (Not (Or (Eq (skS.0 1 a_1 a_2) cU) (Eq (skS.0 1 a_1 a_2) cW) → skS.0 1 a_1 a_2 (skS.0 0 a_1))) True
% 3.50/3.88 Clause #9 (by clausification #[8]): ∀ (a_1 : a) (a_2 : a → Prop),
% 3.50/3.88 Eq (Or (Eq (skS.0 1 a_1 a_2) cU) (Eq (skS.0 1 a_1 a_2) cW) → skS.0 1 a_1 a_2 (skS.0 0 a_1)) False
% 3.50/3.88 Clause #10 (by clausification #[9]): ∀ (a_1 : a) (a_2 : a → Prop), Eq (Or (Eq (skS.0 1 a_1 a_2) cU) (Eq (skS.0 1 a_1 a_2) cW)) True
% 3.50/3.88 Clause #11 (by clausification #[9]): ∀ (a_1 : a) (a_2 : a → Prop), Eq (skS.0 1 a_1 a_2 (skS.0 0 a_1)) False
% 3.50/3.88 Clause #12 (by clausification #[10]): ∀ (a_1 : a) (a_2 : a → Prop), Or (Eq (Eq (skS.0 1 a_1 a_2) cU) True) (Eq (Eq (skS.0 1 a_1 a_2) cW) True)
% 3.50/3.88 Clause #13 (by clausification #[12]): ∀ (a_1 : a) (a_2 : a → Prop), Or (Eq (Eq (skS.0 1 a_1 a_2) cW) True) (Eq (skS.0 1 a_1 a_2) cU)
% 3.50/3.88 Clause #14 (by clausification #[13]): ∀ (a_1 : a) (a_2 : a → Prop), Or (Eq (skS.0 1 a_1 a_2) cU) (Eq (skS.0 1 a_1 a_2) cW)
% 3.50/3.88 Clause #16 (by argument congruence #[14]): ∀ (a_1 : a) (a_2 : a → Prop) (a_3 : a), Or (Eq (skS.0 1 a_1 a_2) cU) (Eq (skS.0 1 a_1 a_2 a_3) (cW a_3))
% 3.50/3.88 Clause #19 (by superposition #[16, 11]): ∀ (a_1 : a) (a_2 : a → Prop), Or (Eq (skS.0 1 a_1 fun x => a_2 x) cU) (Eq (cW (skS.0 0 a_1)) False)
% 3.50/3.88 Clause #21 (by betaEtaReduce #[19]): ∀ (a_1 : a) (a_2 : a → Prop), Or (Eq (skS.0 1 a_1 a_2) cU) (Eq (cW (skS.0 0 a_1)) False)
% 3.50/3.88 Clause #22 (by forward demodulation #[21, 6]): ∀ (a_1 : a) (a_2 : a → Prop), Or (Eq (skS.0 1 a_1 a_2) cU) (Eq True False)
% 3.50/3.88 Clause #23 (by clausification #[22]): ∀ (a_1 : a) (a_2 : a → Prop), Eq (skS.0 1 a_1 a_2) cU
% 3.50/3.88 Clause #25 (by argument congruence #[23]): ∀ (a_1 : a) (a_2 : a → Prop) (a_3 : a), Eq (skS.0 1 a_1 a_2 a_3) (cU a_3)
% 3.50/3.88 Clause #28 (by superposition #[25, 11]): ∀ (a_1 : a), Eq (cU (skS.0 0 a_1)) False
% 3.50/3.88 Clause #29 (by superposition #[28, 7]): Eq False True
% 3.50/3.88 Clause #30 (by clausification #[29]): False
% 3.50/3.88 SZS output end Proof for theBenchmark.p
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