TSTP Solution File: GRP710-10 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : GRP710-10 : TPTP v8.1.2. Released v8.1.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof

% Computer : n021.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 01:19:50 EDT 2023

% Result   : Unsatisfiable 0.20s 0.39s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12  % Problem  : GRP710-10 : TPTP v8.1.2. Released v8.1.0.
% 0.07/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.34  % Computer : n021.cluster.edu
% 0.14/0.34  % Model    : x86_64 x86_64
% 0.14/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34  % Memory   : 8042.1875MB
% 0.14/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34  % CPULimit : 300
% 0.14/0.34  % WCLimit  : 300
% 0.14/0.34  % DateTime : Tue Aug 29 02:29:43 EDT 2023
% 0.14/0.34  % CPUTime  : 
% 0.20/0.39  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.20/0.39  
% 0.20/0.39  % SZS status Unsatisfiable
% 0.20/0.39  
% 0.20/0.39  % SZS output start Proof
% 0.20/0.39  Axiom 1 (f01): mult(X, unit) = X.
% 0.20/0.39  Axiom 2 (f02): mult(unit, X) = X.
% 0.20/0.39  Axiom 3 (f04): mult(X, i(X)) = unit.
% 0.20/0.39  Axiom 4 (f05): mult(i(X), X) = unit.
% 0.20/0.39  Axiom 5 (f03): mult(X, mult(Y, mult(Y, Z))) = mult(mult(mult(X, Y), Y), Z).
% 0.20/0.39  
% 0.20/0.39  Lemma 6: mult(i(X), mult(X, mult(X, Y))) = mult(X, Y).
% 0.20/0.39  Proof:
% 0.20/0.39    mult(i(X), mult(X, mult(X, Y)))
% 0.20/0.39  = { by axiom 5 (f03) }
% 0.20/0.39    mult(mult(mult(i(X), X), X), Y)
% 0.20/0.39  = { by axiom 4 (f05) }
% 0.20/0.39    mult(mult(unit, X), Y)
% 0.20/0.39  = { by axiom 2 (f02) }
% 0.20/0.39    mult(X, Y)
% 0.20/0.39  
% 0.20/0.39  Lemma 7: mult(X, mult(i(X), mult(i(X), Y))) = mult(i(X), Y).
% 0.20/0.39  Proof:
% 0.20/0.39    mult(X, mult(i(X), mult(i(X), Y)))
% 0.20/0.39  = { by axiom 5 (f03) }
% 0.20/0.39    mult(mult(mult(X, i(X)), i(X)), Y)
% 0.20/0.39  = { by axiom 3 (f04) }
% 0.20/0.39    mult(mult(unit, i(X)), Y)
% 0.20/0.39  = { by axiom 2 (f02) }
% 0.20/0.39    mult(i(X), Y)
% 0.20/0.39  
% 0.20/0.39  Lemma 8: mult(i(mult(X, X)), mult(X, mult(X, Y))) = Y.
% 0.20/0.39  Proof:
% 0.20/0.39    mult(i(mult(X, X)), mult(X, mult(X, Y)))
% 0.20/0.39  = { by axiom 5 (f03) }
% 0.20/0.39    mult(mult(mult(i(mult(X, X)), X), X), Y)
% 0.20/0.39  = { by axiom 1 (f01) R->L }
% 0.20/0.39    mult(mult(mult(mult(i(mult(X, X)), X), X), unit), Y)
% 0.20/0.39  = { by axiom 5 (f03) R->L }
% 0.20/0.39    mult(mult(i(mult(X, X)), mult(X, mult(X, unit))), Y)
% 0.20/0.39  = { by axiom 1 (f01) }
% 0.20/0.39    mult(mult(i(mult(X, X)), mult(X, X)), Y)
% 0.20/0.39  = { by axiom 4 (f05) }
% 0.20/0.39    mult(unit, Y)
% 0.20/0.39  = { by axiom 2 (f02) }
% 0.20/0.39    Y
% 0.20/0.39  
% 0.20/0.39  Goal 1 (goal): mult(x0, X) = x1.
% 0.20/0.39  The goal is true when:
% 0.20/0.39    X = mult(i(x0), mult(i(x0), mult(x0, x1)))
% 0.20/0.39  
% 0.20/0.39  Proof:
% 0.20/0.39    mult(x0, mult(i(x0), mult(i(x0), mult(x0, x1))))
% 0.20/0.39  = { by lemma 7 }
% 0.20/0.39    mult(i(x0), mult(x0, x1))
% 0.20/0.39  = { by lemma 8 R->L }
% 0.20/0.39    mult(i(mult(x0, x0)), mult(x0, mult(x0, mult(i(x0), mult(x0, x1)))))
% 0.20/0.39  = { by lemma 6 R->L }
% 0.20/0.39    mult(i(mult(x0, x0)), mult(x0, mult(x0, mult(i(x0), mult(i(x0), mult(x0, mult(x0, x1)))))))
% 0.20/0.39  = { by lemma 7 }
% 0.20/0.39    mult(i(mult(x0, x0)), mult(x0, mult(i(x0), mult(x0, mult(x0, x1)))))
% 0.20/0.39  = { by lemma 6 }
% 0.20/0.39    mult(i(mult(x0, x0)), mult(x0, mult(x0, x1)))
% 0.20/0.39  = { by lemma 8 }
% 0.20/0.39    x1
% 0.20/0.39  % SZS output end Proof
% 0.20/0.39  
% 0.20/0.39  RESULT: Unsatisfiable (the axioms are contradictory).
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