TSTP Solution File: COL080-2 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : COL080-2 : TPTP v8.1.2. Released v1.2.0.
% 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 : n012.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 : Wed Aug 30 18:32:02 EDT 2023

% Result   : Unsatisfiable 0.19s 0.49s
% Output   : Proof 0.19s
% 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  : COL080-2 : TPTP v8.1.2. Released v1.2.0.
% 0.00/0.14  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35  % Computer : n012.cluster.edu
% 0.13/0.35  % Model    : x86_64 x86_64
% 0.13/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35  % Memory   : 8042.1875MB
% 0.13/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35  % CPULimit : 300
% 0.13/0.35  % WCLimit  : 300
% 0.13/0.35  % DateTime : Sun Aug 27 04:11:25 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 0.19/0.49  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.19/0.49  
% 0.19/0.49  % SZS status Unsatisfiable
% 0.19/0.49  
% 0.19/0.50  % SZS output start Proof
% 0.19/0.50  Take the following subset of the input axioms:
% 0.19/0.50    fof(abstraction, axiom, ![X, Y, Z]: apply(apply(apply(abstraction, X), Y), Z)=apply(apply(X, k(Z)), apply(Y, Z))).
% 0.19/0.50    fof(extensionality2, axiom, ![X2, Y2]: (X2=Y2 | apply(X2, n(X2, Y2))!=apply(Y2, n(X2, Y2)))).
% 0.19/0.50    fof(k_definition, axiom, ![X2, Y2]: apply(k(X2), Y2)=X2).
% 0.19/0.50    fof(prove_TRC2b, negated_conjecture, k(b)!=apply(abstraction, apply(abstraction, k(b)))).
% 0.19/0.50  
% 0.19/0.50  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.50  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.50  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.50    fresh(y, y, x1...xn) = u
% 0.19/0.50    C => fresh(s, t, x1...xn) = v
% 0.19/0.50  where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.50  variables of u and v.
% 0.19/0.50  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.50  input problem has no model of domain size 1).
% 0.19/0.50  
% 0.19/0.50  The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.50  
% 0.19/0.50  Axiom 1 (k_definition): apply(k(X), Y) = X.
% 0.19/0.50  Axiom 2 (extensionality2): fresh(X, X, Y, Z) = Z.
% 0.19/0.50  Axiom 3 (abstraction): apply(apply(apply(abstraction, X), Y), Z) = apply(apply(X, k(Z)), apply(Y, Z)).
% 0.19/0.50  Axiom 4 (extensionality2): fresh(apply(X, n(X, Y)), apply(Y, n(X, Y)), X, Y) = X.
% 0.19/0.50  
% 0.19/0.50  Goal 1 (prove_TRC2b): k(b) = apply(abstraction, apply(abstraction, k(b))).
% 0.19/0.50  Proof:
% 0.19/0.50    k(b)
% 0.19/0.50  = { by axiom 2 (extensionality2) R->L }
% 0.19/0.50    fresh(b, b, apply(abstraction, apply(abstraction, k(b))), k(b))
% 0.19/0.50  = { by axiom 4 (extensionality2) R->L }
% 0.19/0.50    fresh(fresh(apply(b, n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))), apply(apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))), n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))), b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b)))), b, apply(abstraction, apply(abstraction, k(b))), k(b))
% 0.19/0.50  = { by axiom 3 (abstraction) }
% 0.19/0.50    fresh(fresh(apply(b, n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))), apply(apply(apply(abstraction, k(b)), k(n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b)))))), apply(n(apply(abstraction, apply(abstraction, k(b))), k(b)), n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b)))))), b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b)))), b, apply(abstraction, apply(abstraction, k(b))), k(b))
% 0.19/0.50  = { by axiom 3 (abstraction) }
% 0.19/0.50    fresh(fresh(apply(b, n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))), apply(apply(k(b), k(apply(n(apply(abstraction, apply(abstraction, k(b))), k(b)), n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))))), apply(k(n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))), apply(n(apply(abstraction, apply(abstraction, k(b))), k(b)), n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))))), b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b)))), b, apply(abstraction, apply(abstraction, k(b))), k(b))
% 0.19/0.50  = { by axiom 1 (k_definition) }
% 0.19/0.50    fresh(fresh(apply(b, n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))), apply(b, apply(k(n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))), apply(n(apply(abstraction, apply(abstraction, k(b))), k(b)), n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))))), b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b)))), b, apply(abstraction, apply(abstraction, k(b))), k(b))
% 0.19/0.50  = { by axiom 1 (k_definition) }
% 0.19/0.50    fresh(fresh(apply(b, n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))), apply(b, n(b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))))), b, apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b)))), b, apply(abstraction, apply(abstraction, k(b))), k(b))
% 0.19/0.50  = { by axiom 2 (extensionality2) }
% 0.19/0.50    fresh(apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))), b, apply(abstraction, apply(abstraction, k(b))), k(b))
% 0.19/0.50  = { by axiom 1 (k_definition) R->L }
% 0.19/0.50    fresh(apply(apply(abstraction, apply(abstraction, k(b))), n(apply(abstraction, apply(abstraction, k(b))), k(b))), apply(k(b), n(apply(abstraction, apply(abstraction, k(b))), k(b))), apply(abstraction, apply(abstraction, k(b))), k(b))
% 0.19/0.50  = { by axiom 4 (extensionality2) }
% 0.19/0.50    apply(abstraction, apply(abstraction, k(b)))
% 0.19/0.50  % SZS output end Proof
% 0.19/0.50  
% 0.19/0.50  RESULT: Unsatisfiable (the axioms are contradictory).
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