TSTP Solution File: SWC254-1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWC254-1 : 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 : n022.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 20:54:38 EDT 2023
% Result : Unsatisfiable 5.94s 1.15s
% Output : Proof 5.94s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SWC254-1 : TPTP v8.1.2. Released v2.4.0.
% 0.07/0.13 % 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 : n022.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 17:29:54 EDT 2023
% 0.13/0.35 % CPUTime :
% 5.94/1.15 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 5.94/1.15
% 5.94/1.15 % SZS status Unsatisfiable
% 5.94/1.15
% 5.94/1.15 % SZS output start Proof
% 5.94/1.15 Take the following subset of the input axioms:
% 5.94/1.15 fof(clause116, axiom, ![U, V]: (cons(U, nil)!=V | (~ssItem(U) | (~ssList(V) | singletonP(V))))).
% 5.94/1.15 fof(co1_1, negated_conjecture, ssList(sk1)).
% 5.94/1.15 fof(co1_14, negated_conjecture, ssItem(sk5) | ~neq(sk4, nil)).
% 5.94/1.15 fof(co1_16, negated_conjecture, cons(sk5, nil)=sk3 | ~neq(sk4, nil)).
% 5.94/1.15 fof(co1_18, negated_conjecture, ~singletonP(sk1) | ~neq(sk4, nil)).
% 5.94/1.15 fof(co1_5, negated_conjecture, sk2=sk4).
% 5.94/1.15 fof(co1_6, negated_conjecture, sk1=sk3).
% 5.94/1.15 fof(co1_7, negated_conjecture, neq(sk2, nil) | neq(sk2, nil)).
% 5.94/1.15
% 5.94/1.15 Now clausify the problem and encode Horn clauses using encoding 3 of
% 5.94/1.15 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 5.94/1.15 We repeatedly replace C & s=t => u=v by the two clauses:
% 5.94/1.15 fresh(y, y, x1...xn) = u
% 5.94/1.15 C => fresh(s, t, x1...xn) = v
% 5.94/1.15 where fresh is a fresh function symbol and x1..xn are the free
% 5.94/1.15 variables of u and v.
% 5.94/1.15 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 5.94/1.15 input problem has no model of domain size 1).
% 5.94/1.15
% 5.94/1.15 The encoding turns the above axioms into the following unit equations and goals:
% 5.94/1.15
% 5.94/1.15 Axiom 1 (co1_5): sk2 = sk4.
% 5.94/1.15 Axiom 2 (co1_6): sk1 = sk3.
% 5.94/1.15 Axiom 3 (co1_1): ssList(sk1) = true2.
% 5.94/1.15 Axiom 4 (co1_7): neq(sk2, nil) = true2.
% 5.94/1.15 Axiom 5 (co1_14): fresh18(X, X) = true2.
% 5.94/1.15 Axiom 6 (co1_16): fresh16(X, X) = sk3.
% 5.94/1.15 Axiom 7 (clause116): fresh259(X, X, Y) = true2.
% 5.94/1.15 Axiom 8 (co1_14): fresh18(neq(sk4, nil), true2) = ssItem(sk5).
% 5.94/1.15 Axiom 9 (co1_16): fresh16(neq(sk4, nil), true2) = cons(sk5, nil).
% 5.94/1.15 Axiom 10 (clause116): fresh258(X, X, Y, Z) = fresh259(cons(Y, nil), Z, Z).
% 5.94/1.15 Axiom 11 (clause116): fresh78(X, X, Y, Z) = singletonP(Z).
% 5.94/1.15 Axiom 12 (clause116): fresh258(ssItem(X), true2, X, Y) = fresh78(ssList(Y), true2, X, Y).
% 5.94/1.15
% 5.94/1.15 Lemma 13: neq(sk4, nil) = true2.
% 5.94/1.15 Proof:
% 5.94/1.15 neq(sk4, nil)
% 5.94/1.15 = { by axiom 1 (co1_5) R->L }
% 5.94/1.15 neq(sk2, nil)
% 5.94/1.15 = { by axiom 4 (co1_7) }
% 5.94/1.15 true2
% 5.94/1.15
% 5.94/1.15 Goal 1 (co1_18): tuple2(singletonP(sk1), neq(sk4, nil)) = tuple2(true2, true2).
% 5.94/1.15 Proof:
% 5.94/1.15 tuple2(singletonP(sk1), neq(sk4, nil))
% 5.94/1.15 = { by lemma 13 }
% 5.94/1.15 tuple2(singletonP(sk1), true2)
% 5.94/1.15 = { by axiom 11 (clause116) R->L }
% 5.94/1.15 tuple2(fresh78(true2, true2, sk5, sk1), true2)
% 5.94/1.15 = { by axiom 3 (co1_1) R->L }
% 5.94/1.15 tuple2(fresh78(ssList(sk1), true2, sk5, sk1), true2)
% 5.94/1.15 = { by axiom 12 (clause116) R->L }
% 5.94/1.15 tuple2(fresh258(ssItem(sk5), true2, sk5, sk1), true2)
% 5.94/1.15 = { by axiom 8 (co1_14) R->L }
% 5.94/1.15 tuple2(fresh258(fresh18(neq(sk4, nil), true2), true2, sk5, sk1), true2)
% 5.94/1.15 = { by lemma 13 }
% 5.94/1.15 tuple2(fresh258(fresh18(true2, true2), true2, sk5, sk1), true2)
% 5.94/1.15 = { by axiom 5 (co1_14) }
% 5.94/1.15 tuple2(fresh258(true2, true2, sk5, sk1), true2)
% 5.94/1.15 = { by axiom 10 (clause116) }
% 5.94/1.15 tuple2(fresh259(cons(sk5, nil), sk1, sk1), true2)
% 5.94/1.15 = { by axiom 9 (co1_16) R->L }
% 5.94/1.15 tuple2(fresh259(fresh16(neq(sk4, nil), true2), sk1, sk1), true2)
% 5.94/1.15 = { by lemma 13 }
% 5.94/1.15 tuple2(fresh259(fresh16(true2, true2), sk1, sk1), true2)
% 5.94/1.15 = { by axiom 6 (co1_16) }
% 5.94/1.15 tuple2(fresh259(sk3, sk1, sk1), true2)
% 5.94/1.15 = { by axiom 2 (co1_6) R->L }
% 5.94/1.15 tuple2(fresh259(sk1, sk1, sk1), true2)
% 5.94/1.15 = { by axiom 7 (clause116) }
% 5.94/1.15 tuple2(true2, true2)
% 5.94/1.15 % SZS output end Proof
% 5.94/1.15
% 5.94/1.15 RESULT: Unsatisfiable (the axioms are contradictory).
%------------------------------------------------------------------------------