TSTP Solution File: SWC010-1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWC010-1 : 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 : n008.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 20:53:24 EDT 2023
% Result : Unsatisfiable 5.39s 1.05s
% Output : Proof 5.57s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.11 % Problem : SWC010-1 : TPTP v8.1.2. Released v2.4.0.
% 0.00/0.12 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.11/0.33 % Computer : n008.cluster.edu
% 0.11/0.33 % Model : x86_64 x86_64
% 0.11/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.33 % Memory : 8042.1875MB
% 0.11/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.11/0.33 % CPULimit : 300
% 0.11/0.33 % WCLimit : 300
% 0.11/0.33 % DateTime : Mon Aug 28 16:24:47 EDT 2023
% 0.11/0.33 % CPUTime :
% 5.39/1.05 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 5.39/1.05
% 5.39/1.05 % SZS status Unsatisfiable
% 5.39/1.05
% 5.57/1.06 % SZS output start Proof
% 5.57/1.06 Take the following subset of the input axioms:
% 5.57/1.06 fof(clause110, axiom, ![U, V]: (~gt(U, V) | (~gt(V, U) | (~ssItem(U) | ~ssItem(V))))).
% 5.57/1.06 fof(clause111, axiom, ![U2, V2]: (U2!=V2 | (~lt(U2, V2) | (~ssItem(V2) | ~ssItem(U2))))).
% 5.57/1.06 fof(clause114, axiom, ![U2, V2]: (~lt(U2, V2) | (~lt(V2, U2) | (~ssItem(U2) | ~ssItem(V2))))).
% 5.57/1.06 fof(clause115, axiom, ![U2, V2]: (U2!=V2 | (~neq(U2, V2) | (~ssList(V2) | ~ssList(U2))))).
% 5.57/1.06 fof(clause117, axiom, ![U2, V2]: (U2!=V2 | (~neq(U2, V2) | (~ssItem(V2) | ~ssItem(U2))))).
% 5.57/1.06 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)))))))).
% 5.57/1.06 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))))))))))).
% 5.57/1.06 fof(clause63, axiom, ![U2]: (~lt(U2, U2) | ~ssItem(U2))).
% 5.57/1.06 fof(clause71, axiom, ![U2]: (~memberP(nil, U2) | ~ssItem(U2))).
% 5.57/1.06 fof(clause98, axiom, ![U2, V2]: (cons(U2, V2)!=nil | (~ssItem(U2) | ~ssList(V2)))).
% 5.57/1.06 fof(clause99, axiom, ![U2, V2]: (cons(U2, V2)!=V2 | (~ssItem(U2) | ~ssList(V2)))).
% 5.57/1.06 fof(co1_16, negated_conjecture, ![A, B, C]: (~ssItem(A) | (~ssList(B) | (~ssList(C) | (app(app(B, cons(A, nil)), C)!=sk2 | (app(B, C)!=sk1 | ~neq(sk4, nil))))))).
% 5.57/1.06 fof(co1_17, negated_conjecture, ssItem(sk5) | ~neq(sk4, nil)).
% 5.57/1.06 fof(co1_18, negated_conjecture, ssList(sk6) | ~neq(sk4, nil)).
% 5.57/1.06 fof(co1_19, negated_conjecture, ssList(sk7) | ~neq(sk4, nil)).
% 5.57/1.06 fof(co1_20, negated_conjecture, app(app(sk6, cons(sk5, nil)), sk7)=sk4 | ~neq(sk4, nil)).
% 5.57/1.06 fof(co1_21, negated_conjecture, app(sk6, sk7)=sk3 | ~neq(sk4, nil)).
% 5.57/1.06 fof(co1_5, negated_conjecture, sk2=sk4).
% 5.57/1.06 fof(co1_6, negated_conjecture, sk1=sk3).
% 5.57/1.06 fof(co1_7, negated_conjecture, neq(sk2, nil) | neq(sk2, nil)).
% 5.57/1.06
% 5.57/1.06 Now clausify the problem and encode Horn clauses using encoding 3 of
% 5.57/1.06 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 5.57/1.06 We repeatedly replace C & s=t => u=v by the two clauses:
% 5.57/1.06 fresh(y, y, x1...xn) = u
% 5.57/1.06 C => fresh(s, t, x1...xn) = v
% 5.57/1.06 where fresh is a fresh function symbol and x1..xn are the free
% 5.57/1.06 variables of u and v.
% 5.57/1.06 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 5.57/1.06 input problem has no model of domain size 1).
% 5.57/1.06
% 5.57/1.06 The encoding turns the above axioms into the following unit equations and goals:
% 5.57/1.06
% 5.57/1.06 Axiom 1 (co1_5): sk2 = sk4.
% 5.57/1.06 Axiom 2 (co1_6): sk1 = sk3.
% 5.57/1.06 Axiom 3 (co1_7): neq(sk2, nil) = true2.
% 5.57/1.06 Axiom 4 (co1_17): fresh20(X, X) = true2.
% 5.57/1.06 Axiom 5 (co1_18): fresh19(X, X) = true2.
% 5.57/1.06 Axiom 6 (co1_19): fresh18(X, X) = true2.
% 5.57/1.06 Axiom 7 (co1_20): fresh17(X, X) = sk4.
% 5.57/1.06 Axiom 8 (co1_21): fresh16(X, X) = sk3.
% 5.57/1.06 Axiom 9 (co1_17): fresh20(neq(sk4, nil), true2) = ssItem(sk5).
% 5.57/1.06 Axiom 10 (co1_18): fresh19(neq(sk4, nil), true2) = ssList(sk6).
% 5.57/1.06 Axiom 11 (co1_19): fresh18(neq(sk4, nil), true2) = ssList(sk7).
% 5.57/1.06 Axiom 12 (co1_21): fresh16(neq(sk4, nil), true2) = app(sk6, sk7).
% 5.57/1.06 Axiom 13 (co1_20): fresh17(neq(sk4, nil), true2) = app(app(sk6, cons(sk5, nil)), sk7).
% 5.57/1.06
% 5.57/1.06 Lemma 14: neq(sk4, nil) = true2.
% 5.57/1.06 Proof:
% 5.57/1.06 neq(sk4, nil)
% 5.57/1.06 = { by axiom 1 (co1_5) R->L }
% 5.57/1.07 neq(sk2, nil)
% 5.57/1.07 = { by axiom 3 (co1_7) }
% 5.57/1.07 true2
% 5.57/1.07
% 5.57/1.07 Goal 1 (co1_16): tuple6(app(X, Y), app(app(X, cons(Z, nil)), Y), ssList(X), ssList(Y), ssItem(Z), neq(sk4, nil)) = tuple6(sk1, sk2, true2, true2, true2, true2).
% 5.57/1.07 The goal is true when:
% 5.57/1.07 X = sk6
% 5.57/1.07 Y = sk7
% 5.57/1.07 Z = sk5
% 5.57/1.07
% 5.57/1.07 Proof:
% 5.57/1.07 tuple6(app(sk6, sk7), app(app(sk6, cons(sk5, nil)), sk7), ssList(sk6), ssList(sk7), ssItem(sk5), neq(sk4, nil))
% 5.57/1.07 = { by lemma 14 }
% 5.57/1.07 tuple6(app(sk6, sk7), app(app(sk6, cons(sk5, nil)), sk7), ssList(sk6), ssList(sk7), ssItem(sk5), true2)
% 5.57/1.07 = { by axiom 13 (co1_20) R->L }
% 5.57/1.07 tuple6(app(sk6, sk7), fresh17(neq(sk4, nil), true2), ssList(sk6), ssList(sk7), ssItem(sk5), true2)
% 5.57/1.07 = { by lemma 14 }
% 5.57/1.07 tuple6(app(sk6, sk7), fresh17(true2, true2), ssList(sk6), ssList(sk7), ssItem(sk5), true2)
% 5.57/1.07 = { by axiom 7 (co1_20) }
% 5.57/1.07 tuple6(app(sk6, sk7), sk4, ssList(sk6), ssList(sk7), ssItem(sk5), true2)
% 5.57/1.07 = { by axiom 12 (co1_21) R->L }
% 5.57/1.07 tuple6(fresh16(neq(sk4, nil), true2), sk4, ssList(sk6), ssList(sk7), ssItem(sk5), true2)
% 5.57/1.07 = { by lemma 14 }
% 5.57/1.07 tuple6(fresh16(true2, true2), sk4, ssList(sk6), ssList(sk7), ssItem(sk5), true2)
% 5.57/1.07 = { by axiom 8 (co1_21) }
% 5.57/1.07 tuple6(sk3, sk4, ssList(sk6), ssList(sk7), ssItem(sk5), true2)
% 5.57/1.07 = { by axiom 2 (co1_6) R->L }
% 5.57/1.07 tuple6(sk1, sk4, ssList(sk6), ssList(sk7), ssItem(sk5), true2)
% 5.57/1.07 = { by axiom 10 (co1_18) R->L }
% 5.57/1.07 tuple6(sk1, sk4, fresh19(neq(sk4, nil), true2), ssList(sk7), ssItem(sk5), true2)
% 5.57/1.07 = { by lemma 14 }
% 5.57/1.07 tuple6(sk1, sk4, fresh19(true2, true2), ssList(sk7), ssItem(sk5), true2)
% 5.57/1.07 = { by axiom 5 (co1_18) }
% 5.57/1.07 tuple6(sk1, sk4, true2, ssList(sk7), ssItem(sk5), true2)
% 5.57/1.07 = { by axiom 11 (co1_19) R->L }
% 5.57/1.07 tuple6(sk1, sk4, true2, fresh18(neq(sk4, nil), true2), ssItem(sk5), true2)
% 5.57/1.07 = { by lemma 14 }
% 5.57/1.07 tuple6(sk1, sk4, true2, fresh18(true2, true2), ssItem(sk5), true2)
% 5.57/1.07 = { by axiom 6 (co1_19) }
% 5.57/1.07 tuple6(sk1, sk4, true2, true2, ssItem(sk5), true2)
% 5.57/1.07 = { by axiom 9 (co1_17) R->L }
% 5.57/1.07 tuple6(sk1, sk4, true2, true2, fresh20(neq(sk4, nil), true2), true2)
% 5.57/1.07 = { by lemma 14 }
% 5.57/1.07 tuple6(sk1, sk4, true2, true2, fresh20(true2, true2), true2)
% 5.57/1.07 = { by axiom 4 (co1_17) }
% 5.57/1.07 tuple6(sk1, sk4, true2, true2, true2, true2)
% 5.57/1.07 = { by axiom 1 (co1_5) R->L }
% 5.57/1.07 tuple6(sk1, sk2, true2, true2, true2, true2)
% 5.57/1.07 % SZS output end Proof
% 5.57/1.07
% 5.57/1.07 RESULT: Unsatisfiable (the axioms are contradictory).
%------------------------------------------------------------------------------