TSTP Solution File: SYN268-1 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN268-1 : TPTP v8.1.2. Released v1.1.0.
% 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 : n004.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 : Fri Sep  1 03:33:51 EDT 2023

% Result   : Unsatisfiable 26.17s 3.94s
% Output   : Proof 26.17s
% 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  : SYN268-1 : TPTP v8.1.2. Released v1.1.0.
% 0.12/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35  % Computer : n004.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 : Sat Aug 26 16:55:53 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 26.17/3.94  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 26.17/3.94  
% 26.17/3.94  % SZS status Unsatisfiable
% 26.17/3.94  
% 26.17/3.95  % SZS output start Proof
% 26.17/3.95  Take the following subset of the input axioms:
% 26.17/3.95    fof(axiom_19, axiom, ![X, Y]: m0(X, d, Y)).
% 26.17/3.95    fof(prove_this, negated_conjecture, ![X2, Y2]: ~p4(a, X2, Y2)).
% 26.17/3.95    fof(rule_107, axiom, ![A2]: (q1(e, A2, A2) | (~m0(A2, d, A2) | ~m0(e, d, A2)))).
% 26.17/3.95    fof(rule_129, axiom, ![J, A]: (k2(J, J) | ~q1(A, J, J))).
% 26.17/3.95    fof(rule_194, axiom, ![G, F]: (k3(F, F, G) | ~k2(G, F))).
% 26.17/3.95    fof(rule_287, axiom, ![C, B]: (p4(B, C, B) | ~k3(B, B, C))).
% 26.17/3.95  
% 26.17/3.95  Now clausify the problem and encode Horn clauses using encoding 3 of
% 26.17/3.95  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 26.17/3.95  We repeatedly replace C & s=t => u=v by the two clauses:
% 26.17/3.95    fresh(y, y, x1...xn) = u
% 26.17/3.95    C => fresh(s, t, x1...xn) = v
% 26.17/3.95  where fresh is a fresh function symbol and x1..xn are the free
% 26.17/3.95  variables of u and v.
% 26.17/3.95  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 26.17/3.95  input problem has no model of domain size 1).
% 26.17/3.95  
% 26.17/3.95  The encoding turns the above axioms into the following unit equations and goals:
% 26.17/3.95  
% 26.17/3.95  Axiom 1 (axiom_19): m0(X, d, Y) = true2.
% 26.17/3.95  Axiom 2 (rule_107): fresh301(X, X, Y) = q1(e, Y, Y).
% 26.17/3.95  Axiom 3 (rule_107): fresh300(X, X, Y) = true2.
% 26.17/3.95  Axiom 4 (rule_129): fresh270(X, X, Y) = true2.
% 26.17/3.95  Axiom 5 (rule_194): fresh184(X, X, Y, Z) = true2.
% 26.17/3.95  Axiom 6 (rule_287): fresh62(X, X, Y, Z) = true2.
% 26.17/3.95  Axiom 7 (rule_107): fresh301(m0(e, d, X), true2, X) = fresh300(m0(X, d, X), true2, X).
% 26.17/3.95  Axiom 8 (rule_129): fresh270(q1(X, Y, Y), true2, Y) = k2(Y, Y).
% 26.17/3.95  Axiom 9 (rule_194): fresh184(k2(X, Y), true2, Y, X) = k3(Y, Y, X).
% 26.17/3.95  Axiom 10 (rule_287): fresh62(k3(X, X, Y), true2, X, Y) = p4(X, Y, X).
% 26.17/3.95  
% 26.17/3.95  Goal 1 (prove_this): p4(a, X, Y) = true2.
% 26.17/3.95  The goal is true when:
% 26.17/3.95    X = a
% 26.17/3.95    Y = a
% 26.17/3.95  
% 26.17/3.95  Proof:
% 26.17/3.95    p4(a, a, a)
% 26.17/3.95  = { by axiom 10 (rule_287) R->L }
% 26.17/3.95    fresh62(k3(a, a, a), true2, a, a)
% 26.17/3.95  = { by axiom 9 (rule_194) R->L }
% 26.17/3.95    fresh62(fresh184(k2(a, a), true2, a, a), true2, a, a)
% 26.17/3.95  = { by axiom 8 (rule_129) R->L }
% 26.17/3.95    fresh62(fresh184(fresh270(q1(e, a, a), true2, a), true2, a, a), true2, a, a)
% 26.17/3.95  = { by axiom 2 (rule_107) R->L }
% 26.17/3.95    fresh62(fresh184(fresh270(fresh301(true2, true2, a), true2, a), true2, a, a), true2, a, a)
% 26.17/3.95  = { by axiom 1 (axiom_19) R->L }
% 26.17/3.95    fresh62(fresh184(fresh270(fresh301(m0(e, d, a), true2, a), true2, a), true2, a, a), true2, a, a)
% 26.17/3.95  = { by axiom 7 (rule_107) }
% 26.17/3.95    fresh62(fresh184(fresh270(fresh300(m0(a, d, a), true2, a), true2, a), true2, a, a), true2, a, a)
% 26.17/3.95  = { by axiom 1 (axiom_19) }
% 26.17/3.95    fresh62(fresh184(fresh270(fresh300(true2, true2, a), true2, a), true2, a, a), true2, a, a)
% 26.17/3.95  = { by axiom 3 (rule_107) }
% 26.17/3.95    fresh62(fresh184(fresh270(true2, true2, a), true2, a, a), true2, a, a)
% 26.17/3.95  = { by axiom 4 (rule_129) }
% 26.17/3.95    fresh62(fresh184(true2, true2, a, a), true2, a, a)
% 26.17/3.95  = { by axiom 5 (rule_194) }
% 26.17/3.95    fresh62(true2, true2, a, a)
% 26.17/3.95  = { by axiom 6 (rule_287) }
% 26.17/3.95    true2
% 26.17/3.95  % SZS output end Proof
% 26.17/3.95  
% 26.17/3.95  RESULT: Unsatisfiable (the axioms are contradictory).
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