TSTP Solution File: SWC317-1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWC317-1 : 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 : n031.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 20:54:57 EDT 2023
% Result : Unsatisfiable 3.73s 0.85s
% Output : Proof 3.73s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : SWC317-1 : TPTP v8.1.2. Released v2.4.0.
% 0.13/0.13 % 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 : n031.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
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Mon Aug 28 17:14:55 EDT 2023
% 0.13/0.35 % CPUTime :
% 3.73/0.85 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 3.73/0.85
% 3.73/0.85 % SZS status Unsatisfiable
% 3.73/0.85
% 3.73/0.85 % SZS output start Proof
% 3.73/0.85 Take the following subset of the input axioms:
% 3.73/0.86 fof(clause110, axiom, ![U, V]: (~gt(U, V) | (~gt(V, U) | (~ssItem(U) | ~ssItem(V))))).
% 3.73/0.86 fof(clause111, axiom, ![U2, V2]: (U2!=V2 | (~lt(U2, V2) | (~ssItem(V2) | ~ssItem(U2))))).
% 3.73/0.86 fof(clause114, axiom, ![U2, V2]: (~lt(U2, V2) | (~lt(V2, U2) | (~ssItem(U2) | ~ssItem(V2))))).
% 3.73/0.86 fof(clause115, axiom, ![U2, V2]: (U2!=V2 | (~neq(U2, V2) | (~ssList(V2) | ~ssList(U2))))).
% 3.73/0.86 fof(clause117, axiom, ![U2, V2]: (U2!=V2 | (~neq(U2, V2) | (~ssItem(V2) | ~ssItem(U2))))).
% 3.73/0.86 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)))))))).
% 3.73/0.86 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))))))))))).
% 3.73/0.86 fof(clause63, axiom, ![U2]: (~lt(U2, U2) | ~ssItem(U2))).
% 3.73/0.86 fof(clause71, axiom, ![U2]: (~memberP(nil, U2) | ~ssItem(U2))).
% 3.73/0.86 fof(clause98, axiom, ![U2, V2]: (cons(U2, V2)!=nil | (~ssItem(U2) | ~ssList(V2)))).
% 3.73/0.86 fof(clause99, axiom, ![U2, V2]: (cons(U2, V2)!=V2 | (~ssItem(U2) | ~ssList(V2)))).
% 3.73/0.86 fof(co1_14, negated_conjecture, ![A, B]: (~ssItem(A) | (~ssList(B) | (app(cons(A, nil), B)!=sk1 | (app(B, cons(A, nil))!=sk2 | ~neq(sk4, nil)))))).
% 3.73/0.86 fof(co1_15, negated_conjecture, ssItem(sk5) | ~neq(sk4, nil)).
% 3.73/0.86 fof(co1_16, negated_conjecture, ssList(sk6) | ~neq(sk4, nil)).
% 3.73/0.86 fof(co1_17, negated_conjecture, app(cons(sk5, nil), sk6)=sk3 | ~neq(sk4, nil)).
% 3.73/0.86 fof(co1_18, negated_conjecture, app(sk6, cons(sk5, nil))=sk4 | ~neq(sk4, nil)).
% 3.73/0.86 fof(co1_5, negated_conjecture, sk2=sk4).
% 3.73/0.86 fof(co1_6, negated_conjecture, sk1=sk3).
% 3.73/0.86 fof(co1_7, negated_conjecture, neq(sk2, nil) | neq(sk2, nil)).
% 3.73/0.86
% 3.73/0.86 Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.73/0.86 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.73/0.86 We repeatedly replace C & s=t => u=v by the two clauses:
% 3.73/0.86 fresh(y, y, x1...xn) = u
% 3.73/0.86 C => fresh(s, t, x1...xn) = v
% 3.73/0.86 where fresh is a fresh function symbol and x1..xn are the free
% 3.73/0.86 variables of u and v.
% 3.73/0.86 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.73/0.86 input problem has no model of domain size 1).
% 3.73/0.86
% 3.73/0.86 The encoding turns the above axioms into the following unit equations and goals:
% 3.73/0.86
% 3.73/0.86 Axiom 1 (co1_5): sk2 = sk4.
% 3.73/0.86 Axiom 2 (co1_6): sk1 = sk3.
% 3.73/0.86 Axiom 3 (co1_7): neq(sk2, nil) = true2.
% 3.73/0.86 Axiom 4 (co1_15): fresh19(X, X) = true2.
% 3.73/0.86 Axiom 5 (co1_16): fresh18(X, X) = true2.
% 3.73/0.86 Axiom 6 (co1_17): fresh17(X, X) = sk3.
% 3.73/0.86 Axiom 7 (co1_18): fresh16(X, X) = sk4.
% 3.73/0.86 Axiom 8 (co1_15): fresh19(neq(sk4, nil), true2) = ssItem(sk5).
% 3.73/0.86 Axiom 9 (co1_16): fresh18(neq(sk4, nil), true2) = ssList(sk6).
% 3.73/0.86 Axiom 10 (co1_17): fresh17(neq(sk4, nil), true2) = app(cons(sk5, nil), sk6).
% 3.73/0.86 Axiom 11 (co1_18): fresh16(neq(sk4, nil), true2) = app(sk6, cons(sk5, nil)).
% 3.73/0.86
% 3.73/0.86 Lemma 12: neq(sk4, nil) = true2.
% 3.73/0.86 Proof:
% 3.73/0.86 neq(sk4, nil)
% 3.73/0.86 = { by axiom 1 (co1_5) R->L }
% 3.73/0.86 neq(sk2, nil)
% 3.73/0.86 = { by axiom 3 (co1_7) }
% 3.73/0.86 true2
% 3.73/0.86
% 3.73/0.86 Goal 1 (co1_14): tuple6(app(X, cons(Y, nil)), app(cons(Y, nil), X), ssList(X), ssItem(Y), neq(sk4, nil)) = tuple6(sk2, sk1, true2, true2, true2).
% 3.73/0.86 The goal is true when:
% 3.73/0.86 X = sk6
% 3.73/0.86 Y = sk5
% 3.73/0.86
% 3.73/0.86 Proof:
% 3.73/0.86 tuple6(app(sk6, cons(sk5, nil)), app(cons(sk5, nil), sk6), ssList(sk6), ssItem(sk5), neq(sk4, nil))
% 3.73/0.86 = { by lemma 12 }
% 3.73/0.86 tuple6(app(sk6, cons(sk5, nil)), app(cons(sk5, nil), sk6), ssList(sk6), ssItem(sk5), true2)
% 3.73/0.86 = { by axiom 10 (co1_17) R->L }
% 3.73/0.86 tuple6(app(sk6, cons(sk5, nil)), fresh17(neq(sk4, nil), true2), ssList(sk6), ssItem(sk5), true2)
% 3.73/0.86 = { by lemma 12 }
% 3.73/0.86 tuple6(app(sk6, cons(sk5, nil)), fresh17(true2, true2), ssList(sk6), ssItem(sk5), true2)
% 3.73/0.86 = { by axiom 6 (co1_17) }
% 3.73/0.86 tuple6(app(sk6, cons(sk5, nil)), sk3, ssList(sk6), ssItem(sk5), true2)
% 3.73/0.86 = { by axiom 2 (co1_6) R->L }
% 3.73/0.86 tuple6(app(sk6, cons(sk5, nil)), sk1, ssList(sk6), ssItem(sk5), true2)
% 3.73/0.86 = { by axiom 11 (co1_18) R->L }
% 3.73/0.86 tuple6(fresh16(neq(sk4, nil), true2), sk1, ssList(sk6), ssItem(sk5), true2)
% 3.73/0.86 = { by lemma 12 }
% 3.73/0.86 tuple6(fresh16(true2, true2), sk1, ssList(sk6), ssItem(sk5), true2)
% 3.73/0.86 = { by axiom 7 (co1_18) }
% 3.73/0.86 tuple6(sk4, sk1, ssList(sk6), ssItem(sk5), true2)
% 3.73/0.86 = { by axiom 9 (co1_16) R->L }
% 3.73/0.86 tuple6(sk4, sk1, fresh18(neq(sk4, nil), true2), ssItem(sk5), true2)
% 3.73/0.86 = { by lemma 12 }
% 3.73/0.86 tuple6(sk4, sk1, fresh18(true2, true2), ssItem(sk5), true2)
% 3.73/0.86 = { by axiom 5 (co1_16) }
% 3.73/0.86 tuple6(sk4, sk1, true2, ssItem(sk5), true2)
% 3.73/0.86 = { by axiom 8 (co1_15) R->L }
% 3.73/0.86 tuple6(sk4, sk1, true2, fresh19(neq(sk4, nil), true2), true2)
% 3.73/0.86 = { by lemma 12 }
% 3.73/0.86 tuple6(sk4, sk1, true2, fresh19(true2, true2), true2)
% 3.73/0.86 = { by axiom 4 (co1_15) }
% 3.73/0.86 tuple6(sk4, sk1, true2, true2, true2)
% 3.73/0.86 = { by axiom 1 (co1_5) R->L }
% 3.73/0.86 tuple6(sk2, sk1, true2, true2, true2)
% 3.73/0.86 % SZS output end Proof
% 3.73/0.86
% 3.73/0.86 RESULT: Unsatisfiable (the axioms are contradictory).
%------------------------------------------------------------------------------