TSTP Solution File: SWC097-1 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SWC097-1 : TPTP v8.1.2. Released v2.4.0.
% 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 : n017.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 20:53:51 EDT 2023

% Result   : Unsatisfiable 3.56s 0.84s
% Output   : Proof 3.56s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12  % Problem  : SWC097-1 : TPTP v8.1.2. Released v2.4.0.
% 0.11/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.34  % Computer : n017.cluster.edu
% 0.12/0.34  % Model    : x86_64 x86_64
% 0.12/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34  % Memory   : 8042.1875MB
% 0.12/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34  % CPULimit : 300
% 0.12/0.34  % WCLimit  : 300
% 0.12/0.34  % DateTime : Mon Aug 28 15:48:27 EDT 2023
% 0.12/0.34  % CPUTime  : 
% 3.56/0.84  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 3.56/0.84  
% 3.56/0.84  % SZS status Unsatisfiable
% 3.56/0.84  
% 3.56/0.84  % SZS output start Proof
% 3.56/0.84  Take the following subset of the input axioms:
% 3.56/0.84    fof(clause110, axiom, ![U, V]: (~gt(U, V) | (~gt(V, U) | (~ssItem(U) | ~ssItem(V))))).
% 3.56/0.84    fof(clause111, axiom, ![U2, V2]: (U2!=V2 | (~lt(U2, V2) | (~ssItem(V2) | ~ssItem(U2))))).
% 3.56/0.84    fof(clause114, axiom, ![U2, V2]: (~lt(U2, V2) | (~lt(V2, U2) | (~ssItem(U2) | ~ssItem(V2))))).
% 3.56/0.84    fof(clause115, axiom, ![U2, V2]: (U2!=V2 | (~neq(U2, V2) | (~ssList(V2) | ~ssList(U2))))).
% 3.56/0.84    fof(clause117, axiom, ![U2, V2]: (U2!=V2 | (~neq(U2, V2) | (~ssItem(V2) | ~ssItem(U2))))).
% 3.56/0.84    fof(clause179, axiom, ![W, X, Y, U2, V2]: (app(app(U2, cons(V2, W)), cons(V2, X))!=Y | (~ssList(X) | (~ssList(W) | (~ssList(U2) | (~ssItem(V2) | (~duplicatefreeP(Y) | ~ssList(Y)))))))).
% 3.56/0.84    fof(clause185, axiom, ![Z, U2, V2, W2, X2, Y2]: (~leq(U2, V2) | (~leq(V2, U2) | (app(app(W2, cons(U2, X2)), cons(V2, Y2))!=Z | (~ssList(Y2) | (~ssList(X2) | (~ssList(W2) | (~ssItem(V2) | (~ssItem(U2) | (~cyclefreeP(Z) | ~ssList(Z))))))))))).
% 3.56/0.84    fof(clause63, axiom, ![U2]: (~lt(U2, U2) | ~ssItem(U2))).
% 3.56/0.84    fof(clause71, axiom, ![U2]: (~memberP(nil, U2) | ~ssItem(U2))).
% 3.56/0.84    fof(clause98, axiom, ![U2, V2]: (cons(U2, V2)!=nil | (~ssItem(U2) | ~ssList(V2)))).
% 3.56/0.84    fof(clause99, axiom, ![U2, V2]: (cons(U2, V2)!=V2 | (~ssItem(U2) | ~ssList(V2)))).
% 3.56/0.84    fof(co1_12, negated_conjecture, ![A]: (~ssItem(A) | (app(sk1, cons(A, nil))!=sk2 | ~neq(sk4, nil)))).
% 3.56/0.84    fof(co1_13, negated_conjecture, ssItem(sk5) | ~neq(sk4, nil)).
% 3.56/0.84    fof(co1_14, negated_conjecture, app(sk3, cons(sk5, nil))=sk4 | ~neq(sk4, nil)).
% 3.56/0.84    fof(co1_5, negated_conjecture, sk2=sk4).
% 3.56/0.84    fof(co1_6, negated_conjecture, sk1=sk3).
% 3.56/0.84    fof(co1_7, negated_conjecture, neq(sk2, nil) | neq(sk2, nil)).
% 3.56/0.84  
% 3.56/0.84  Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.56/0.84  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.56/0.84  We repeatedly replace C & s=t => u=v by the two clauses:
% 3.56/0.84    fresh(y, y, x1...xn) = u
% 3.56/0.84    C => fresh(s, t, x1...xn) = v
% 3.56/0.84  where fresh is a fresh function symbol and x1..xn are the free
% 3.56/0.84  variables of u and v.
% 3.56/0.84  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.56/0.84  input problem has no model of domain size 1).
% 3.56/0.84  
% 3.56/0.84  The encoding turns the above axioms into the following unit equations and goals:
% 3.56/0.84  
% 3.56/0.84  Axiom 1 (co1_5): sk2 = sk4.
% 3.56/0.84  Axiom 2 (co1_6): sk1 = sk3.
% 3.56/0.84  Axiom 3 (co1_7): neq(sk2, nil) = true2.
% 3.56/0.84  Axiom 4 (co1_13): fresh18(X, X) = true2.
% 3.56/0.84  Axiom 5 (co1_14): fresh17(X, X) = sk4.
% 3.56/0.84  Axiom 6 (co1_13): fresh18(neq(sk4, nil), true2) = ssItem(sk5).
% 3.56/0.84  Axiom 7 (co1_14): fresh17(neq(sk4, nil), true2) = app(sk3, cons(sk5, nil)).
% 3.56/0.84  
% 3.56/0.84  Goal 1 (co1_12): tuple5(app(sk1, cons(X, nil)), ssItem(X), neq(sk4, nil)) = tuple5(sk2, true2, true2).
% 3.56/0.84  The goal is true when:
% 3.56/0.84    X = sk5
% 3.56/0.84  
% 3.56/0.84  Proof:
% 3.56/0.84    tuple5(app(sk1, cons(sk5, nil)), ssItem(sk5), neq(sk4, nil))
% 3.56/0.84  = { by axiom 1 (co1_5) R->L }
% 3.56/0.84    tuple5(app(sk1, cons(sk5, nil)), ssItem(sk5), neq(sk2, nil))
% 3.56/0.84  = { by axiom 3 (co1_7) }
% 3.56/0.84    tuple5(app(sk1, cons(sk5, nil)), ssItem(sk5), true2)
% 3.56/0.84  = { by axiom 6 (co1_13) R->L }
% 3.56/0.84    tuple5(app(sk1, cons(sk5, nil)), fresh18(neq(sk4, nil), true2), true2)
% 3.56/0.84  = { by axiom 1 (co1_5) R->L }
% 3.56/0.84    tuple5(app(sk1, cons(sk5, nil)), fresh18(neq(sk2, nil), true2), true2)
% 3.56/0.84  = { by axiom 3 (co1_7) }
% 3.56/0.84    tuple5(app(sk1, cons(sk5, nil)), fresh18(true2, true2), true2)
% 3.56/0.84  = { by axiom 4 (co1_13) }
% 3.56/0.84    tuple5(app(sk1, cons(sk5, nil)), true2, true2)
% 3.56/0.84  = { by axiom 2 (co1_6) }
% 3.56/0.84    tuple5(app(sk3, cons(sk5, nil)), true2, true2)
% 3.56/0.84  = { by axiom 7 (co1_14) R->L }
% 3.56/0.84    tuple5(fresh17(neq(sk4, nil), true2), true2, true2)
% 3.56/0.84  = { by axiom 1 (co1_5) R->L }
% 3.56/0.84    tuple5(fresh17(neq(sk2, nil), true2), true2, true2)
% 3.56/0.84  = { by axiom 3 (co1_7) }
% 3.56/0.84    tuple5(fresh17(true2, true2), true2, true2)
% 3.56/0.84  = { by axiom 5 (co1_14) }
% 3.56/0.84    tuple5(sk4, true2, true2)
% 3.56/0.84  = { by axiom 1 (co1_5) R->L }
% 3.56/0.84    tuple5(sk2, true2, true2)
% 3.56/0.84  % SZS output end Proof
% 3.56/0.84  
% 3.56/0.84  RESULT: Unsatisfiable (the axioms are contradictory).
%------------------------------------------------------------------------------