TSTP Solution File: ROB021-1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : ROB021-1 : TPTP v8.1.2. Released v1.0.0.
% 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 : n020.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 14:09:10 EDT 2023

% Result   : Unsatisfiable 0.19s 0.39s
% 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
% 0.00/0.12  % Problem  : ROB021-1 : TPTP v8.1.2. Released v1.0.0.
% 0.00/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34  % Computer : n020.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34  % CPULimit : 300
% 0.13/0.34  % WCLimit  : 300
% 0.13/0.34  % DateTime : Mon Aug 28 06:51:44 EDT 2023
% 0.13/0.34  % CPUTime  : 
% 0.19/0.39  Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.19/0.39  
% 0.19/0.39  % SZS status Unsatisfiable
% 0.19/0.39  
% 0.19/0.39  % SZS output start Proof
% 0.19/0.39  Take the following subset of the input axioms:
% 0.19/0.39    fof(commutativity_of_add, axiom, ![X, Y]: add(X, Y)=add(Y, X)).
% 0.19/0.39    fof(negative_equality_implies_positive_equality, hypothesis, ![X2, Y2]: (negate(X2)!=negate(Y2) | X2=Y2)).
% 0.19/0.39    fof(prove_huntingtons_axiom, negated_conjecture, add(negate(add(a, negate(b))), negate(add(negate(a), negate(b))))!=b).
% 0.19/0.39    fof(robbins_axiom, axiom, ![X2, Y2]: negate(add(negate(add(X2, Y2)), negate(add(X2, negate(Y2)))))=X2).
% 0.19/0.39  
% 0.19/0.39  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.39  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.39  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.39    fresh(y, y, x1...xn) = u
% 0.19/0.39    C => fresh(s, t, x1...xn) = v
% 0.19/0.39  where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.39  variables of u and v.
% 0.19/0.39  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.39  input problem has no model of domain size 1).
% 0.19/0.39  
% 0.19/0.39  The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.39  
% 0.19/0.39  Axiom 1 (commutativity_of_add): add(X, Y) = add(Y, X).
% 0.19/0.39  Axiom 2 (negative_equality_implies_positive_equality): fresh(X, X, Y, Z) = Z.
% 0.19/0.39  Axiom 3 (negative_equality_implies_positive_equality): fresh(negate(X), negate(Y), X, Y) = X.
% 0.19/0.39  Axiom 4 (robbins_axiom): negate(add(negate(add(X, Y)), negate(add(X, negate(Y))))) = X.
% 0.19/0.39  
% 0.19/0.39  Goal 1 (prove_huntingtons_axiom): add(negate(add(a, negate(b))), negate(add(negate(a), negate(b)))) = b.
% 0.19/0.39  Proof:
% 0.19/0.39    add(negate(add(a, negate(b))), negate(add(negate(a), negate(b))))
% 0.19/0.39  = { by axiom 1 (commutativity_of_add) R->L }
% 0.19/0.39    add(negate(add(a, negate(b))), negate(add(negate(b), negate(a))))
% 0.19/0.39  = { by axiom 1 (commutativity_of_add) R->L }
% 0.19/0.39    add(negate(add(negate(b), a)), negate(add(negate(b), negate(a))))
% 0.19/0.39  = { by axiom 2 (negative_equality_implies_positive_equality) R->L }
% 0.19/0.39    fresh(negate(b), negate(b), b, add(negate(add(negate(b), a)), negate(add(negate(b), negate(a)))))
% 0.19/0.39  = { by axiom 4 (robbins_axiom) R->L }
% 0.19/0.39    fresh(negate(b), negate(add(negate(add(negate(b), a)), negate(add(negate(b), negate(a))))), b, add(negate(add(negate(b), a)), negate(add(negate(b), negate(a)))))
% 0.19/0.39  = { by axiom 3 (negative_equality_implies_positive_equality) }
% 0.19/0.39    b
% 0.19/0.39  % SZS output end Proof
% 0.19/0.39  
% 0.19/0.39  RESULT: Unsatisfiable (the axioms are contradictory).
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