TSTP Solution File: SEV163^5 by Duper---1.0
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Duper---1.0
% Problem : SEV163^5 : TPTP v8.1.2. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : duper %s
% Computer : n025.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 19:24:23 EDT 2023
% Result : Theorem 3.65s 3.81s
% Output : Proof 3.65s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SEV163^5 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.14 % Command : duper %s
% 0.14/0.35 % Computer : n025.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 : Thu Aug 24 04:01:21 EDT 2023
% 0.14/0.35 % CPUTime :
% 3.65/3.81 SZS status Theorem for theBenchmark.p
% 3.65/3.81 SZS output start Proof for theBenchmark.p
% 3.65/3.81 Clause #0 (by assumption #[]): Eq
% 3.65/3.81 (Not
% 3.65/3.81 (∀ (Xp : (a → a → a) → a),
% 3.65/3.81 (Eq Xp fun Xg => Xg (Xp fun Xx Xy => Xx) (Xp fun Xx Xy => Xy)) →
% 3.65/3.81 Eq (fun Xg => Xg (Xp fun Xx Xy => Xx) (Xp fun Xx Xy => Xy)) Xp))
% 3.65/3.81 True
% 3.65/3.81 Clause #1 (by clausification #[0]): Eq
% 3.65/3.81 (∀ (Xp : (a → a → a) → a),
% 3.65/3.81 (Eq Xp fun Xg => Xg (Xp fun Xx Xy => Xx) (Xp fun Xx Xy => Xy)) →
% 3.65/3.81 Eq (fun Xg => Xg (Xp fun Xx Xy => Xx) (Xp fun Xx Xy => Xy)) Xp)
% 3.65/3.81 False
% 3.65/3.81 Clause #2 (by clausification #[1]): ∀ (a_1 : (a → a → a) → a),
% 3.65/3.81 Eq
% 3.65/3.81 (Not
% 3.65/3.81 ((Eq (skS.0 0 a_1) fun Xg => Xg (skS.0 0 a_1 fun Xx Xy => Xx) (skS.0 0 a_1 fun Xx Xy => Xy)) →
% 3.65/3.81 Eq (fun Xg => Xg (skS.0 0 a_1 fun Xx Xy => Xx) (skS.0 0 a_1 fun Xx Xy => Xy)) (skS.0 0 a_1)))
% 3.65/3.81 True
% 3.65/3.81 Clause #3 (by clausification #[2]): ∀ (a_1 : (a → a → a) → a),
% 3.65/3.81 Eq
% 3.65/3.81 ((Eq (skS.0 0 a_1) fun Xg => Xg (skS.0 0 a_1 fun Xx Xy => Xx) (skS.0 0 a_1 fun Xx Xy => Xy)) →
% 3.65/3.81 Eq (fun Xg => Xg (skS.0 0 a_1 fun Xx Xy => Xx) (skS.0 0 a_1 fun Xx Xy => Xy)) (skS.0 0 a_1))
% 3.65/3.81 False
% 3.65/3.81 Clause #4 (by clausification #[3]): ∀ (a_1 : (a → a → a) → a),
% 3.65/3.81 Eq (Eq (skS.0 0 a_1) fun Xg => Xg (skS.0 0 a_1 fun Xx Xy => Xx) (skS.0 0 a_1 fun Xx Xy => Xy)) True
% 3.65/3.81 Clause #5 (by clausification #[3]): ∀ (a_1 : (a → a → a) → a),
% 3.65/3.81 Eq (Eq (fun Xg => Xg (skS.0 0 a_1 fun Xx Xy => Xx) (skS.0 0 a_1 fun Xx Xy => Xy)) (skS.0 0 a_1)) False
% 3.65/3.81 Clause #6 (by clausification #[4]): ∀ (a_1 : (a → a → a) → a), Eq (skS.0 0 a_1) fun Xg => Xg (skS.0 0 a_1 fun Xx Xy => Xx) (skS.0 0 a_1 fun Xx Xy => Xy)
% 3.65/3.81 Clause #18 (by clausification #[5]): ∀ (a_1 : (a → a → a) → a), Ne (fun Xg => Xg (skS.0 0 a_1 fun Xx Xy => Xx) (skS.0 0 a_1 fun Xx Xy => Xy)) (skS.0 0 a_1)
% 3.65/3.81 Clause #19 (by forward positive simplify reflect #[18, 6]): False
% 3.65/3.81 SZS output end Proof for theBenchmark.p
%------------------------------------------------------------------------------