TSTP Solution File: SET629^5 by Duper---1.0
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%------------------------------------------------------------------------------
% File : Duper---1.0
% Problem : SET629^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n018.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 14:47:10 EDT 2023
% Result : Theorem 3.64s 3.84s
% Output : Proof 3.64s
% 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 : SET629^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14 % Command : duper %s
% 0.14/0.36 % Computer : n018.cluster.edu
% 0.14/0.36 % Model : x86_64 x86_64
% 0.14/0.36 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.36 % Memory : 8042.1875MB
% 0.14/0.36 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.36 % CPULimit : 300
% 0.14/0.36 % WCLimit : 300
% 0.14/0.36 % DateTime : Sat Aug 26 12:35:00 EDT 2023
% 0.14/0.36 % CPUTime :
% 3.64/3.84 SZS status Theorem for theBenchmark.p
% 3.64/3.84 SZS output start Proof for theBenchmark.p
% 3.64/3.84 Clause #0 (by assumption #[]): Eq (Not (∀ (X Y : a → Prop), Not (Exists fun Xx => And (And (And (X Xx) (Y Xx)) (X Xx)) (Not (Y Xx))))) True
% 3.64/3.84 Clause #1 (by clausification #[0]): Eq (∀ (X Y : a → Prop), Not (Exists fun Xx => And (And (And (X Xx) (Y Xx)) (X Xx)) (Not (Y Xx)))) False
% 3.64/3.84 Clause #2 (by clausification #[1]): ∀ (a_1 : a → Prop),
% 3.64/3.84 Eq
% 3.64/3.84 (Not
% 3.64/3.84 (∀ (Y : a → Prop), Not (Exists fun Xx => And (And (And (skS.0 0 a_1 Xx) (Y Xx)) (skS.0 0 a_1 Xx)) (Not (Y Xx)))))
% 3.64/3.84 True
% 3.64/3.84 Clause #3 (by clausification #[2]): ∀ (a_1 : a → Prop),
% 3.64/3.84 Eq (∀ (Y : a → Prop), Not (Exists fun Xx => And (And (And (skS.0 0 a_1 Xx) (Y Xx)) (skS.0 0 a_1 Xx)) (Not (Y Xx))))
% 3.64/3.84 False
% 3.64/3.84 Clause #4 (by clausification #[3]): ∀ (a_1 a_2 : a → Prop),
% 3.64/3.84 Eq
% 3.64/3.84 (Not
% 3.64/3.84 (Not
% 3.64/3.84 (Exists fun Xx =>
% 3.64/3.84 And (And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (skS.0 0 a_1 Xx)) (Not (skS.0 1 a_1 a_2 Xx)))))
% 3.64/3.84 True
% 3.64/3.84 Clause #5 (by clausification #[4]): ∀ (a_1 a_2 : a → Prop),
% 3.64/3.84 Eq
% 3.64/3.84 (Not
% 3.64/3.84 (Exists fun Xx =>
% 3.64/3.84 And (And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (skS.0 0 a_1 Xx)) (Not (skS.0 1 a_1 a_2 Xx))))
% 3.64/3.84 False
% 3.64/3.84 Clause #6 (by clausification #[5]): ∀ (a_1 a_2 : a → Prop),
% 3.64/3.84 Eq
% 3.64/3.84 (Exists fun Xx => And (And (And (skS.0 0 a_1 Xx) (skS.0 1 a_1 a_2 Xx)) (skS.0 0 a_1 Xx)) (Not (skS.0 1 a_1 a_2 Xx)))
% 3.64/3.84 True
% 3.64/3.84 Clause #7 (by clausification #[6]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.64/3.84 Eq
% 3.64/3.84 (And
% 3.64/3.84 (And (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)))
% 3.64/3.84 (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)))
% 3.64/3.84 (Not (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))))
% 3.64/3.84 True
% 3.64/3.84 Clause #8 (by clausification #[7]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (Not (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))) True
% 3.64/3.84 Clause #9 (by clausification #[7]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.64/3.84 Eq
% 3.64/3.84 (And (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)))
% 3.64/3.84 (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)))
% 3.64/3.84 True
% 3.64/3.84 Clause #10 (by clausification #[8]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)) False
% 3.64/3.84 Clause #12 (by clausification #[9]): ∀ (a_1 a_2 : a → Prop) (a_3 : a),
% 3.64/3.84 Eq (And (skS.0 0 a_1 (skS.0 2 a_1 a_2 a_3)) (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3))) True
% 3.64/3.84 Clause #13 (by clausification #[12]): ∀ (a_1 a_2 : a → Prop) (a_3 : a), Eq (skS.0 1 a_1 a_2 (skS.0 2 a_1 a_2 a_3)) True
% 3.64/3.84 Clause #14 (by superposition #[13, 10]): Eq True False
% 3.64/3.84 Clause #15 (by clausification #[14]): False
% 3.64/3.84 SZS output end Proof for theBenchmark.p
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