TSTP Solution File: SYN231-1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SYN231-1 : 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 : n028.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 03:33:42 EDT 2023
% Result : Unsatisfiable 17.40s 2.58s
% Output : Proof 17.40s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SYN231-1 : TPTP v8.1.2. Released v1.1.0.
% 0.00/0.13 % 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 : n028.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 : Sat Aug 26 17:58:52 EDT 2023
% 0.12/0.34 % CPUTime :
% 17.40/2.58 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 17.40/2.58
% 17.40/2.58 % SZS status Unsatisfiable
% 17.40/2.58
% 17.40/2.58 % SZS output start Proof
% 17.40/2.58 Take the following subset of the input axioms:
% 17.40/2.58 fof(axiom_11, axiom, n0(e, b)).
% 17.40/2.58 fof(axiom_17, axiom, ![X]: q0(X, d)).
% 17.40/2.58 fof(prove_this, negated_conjecture, ![X2]: ~l3(d, X2)).
% 17.40/2.58 fof(rule_001, axiom, ![I, J]: (k1(I) | ~n0(J, I))).
% 17.40/2.58 fof(rule_067, axiom, ![E, F]: (p1(E, E, E) | ~q0(F, E))).
% 17.40/2.58 fof(rule_136, axiom, m2(b) | ~k1(b)).
% 17.40/2.58 fof(rule_137, axiom, ![C, B, A2]: (n2(A2) | ~p1(B, C, A2))).
% 17.40/2.58 fof(rule_217, axiom, ![C2]: (l3(C2, C2) | (~n2(C2) | ~m2(b)))).
% 17.40/2.58
% 17.40/2.58 Now clausify the problem and encode Horn clauses using encoding 3 of
% 17.40/2.58 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 17.40/2.58 We repeatedly replace C & s=t => u=v by the two clauses:
% 17.40/2.58 fresh(y, y, x1...xn) = u
% 17.40/2.58 C => fresh(s, t, x1...xn) = v
% 17.40/2.58 where fresh is a fresh function symbol and x1..xn are the free
% 17.40/2.58 variables of u and v.
% 17.40/2.58 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 17.40/2.58 input problem has no model of domain size 1).
% 17.40/2.58
% 17.40/2.58 The encoding turns the above axioms into the following unit equations and goals:
% 17.40/2.58
% 17.40/2.58 Axiom 1 (axiom_17): q0(X, d) = true2.
% 17.40/2.58 Axiom 2 (axiom_11): n0(e, b) = true2.
% 17.40/2.58 Axiom 3 (rule_136): fresh263(X, X) = true2.
% 17.40/2.58 Axiom 4 (rule_001): fresh440(X, X, Y) = true2.
% 17.40/2.58 Axiom 5 (rule_067): fresh351(X, X, Y) = true2.
% 17.40/2.58 Axiom 6 (rule_136): fresh263(k1(b), true2) = m2(b).
% 17.40/2.58 Axiom 7 (rule_137): fresh262(X, X, Y) = true2.
% 17.40/2.58 Axiom 8 (rule_217): fresh156(X, X, Y) = l3(Y, Y).
% 17.40/2.58 Axiom 9 (rule_217): fresh155(X, X, Y) = true2.
% 17.40/2.58 Axiom 10 (rule_001): fresh440(n0(X, Y), true2, Y) = k1(Y).
% 17.40/2.58 Axiom 11 (rule_067): fresh351(q0(X, Y), true2, Y) = p1(Y, Y, Y).
% 17.40/2.59 Axiom 12 (rule_217): fresh156(n2(X), true2, X) = fresh155(m2(b), true2, X).
% 17.40/2.59 Axiom 13 (rule_137): fresh262(p1(X, Y, Z), true2, Z) = n2(Z).
% 17.40/2.59
% 17.40/2.59 Goal 1 (prove_this): l3(d, X) = true2.
% 17.40/2.59 The goal is true when:
% 17.40/2.59 X = d
% 17.40/2.59
% 17.40/2.59 Proof:
% 17.40/2.59 l3(d, d)
% 17.40/2.59 = { by axiom 8 (rule_217) R->L }
% 17.40/2.59 fresh156(true2, true2, d)
% 17.40/2.59 = { by axiom 7 (rule_137) R->L }
% 17.40/2.59 fresh156(fresh262(true2, true2, d), true2, d)
% 17.40/2.59 = { by axiom 5 (rule_067) R->L }
% 17.40/2.59 fresh156(fresh262(fresh351(true2, true2, d), true2, d), true2, d)
% 17.40/2.59 = { by axiom 1 (axiom_17) R->L }
% 17.40/2.59 fresh156(fresh262(fresh351(q0(X, d), true2, d), true2, d), true2, d)
% 17.40/2.59 = { by axiom 11 (rule_067) }
% 17.40/2.59 fresh156(fresh262(p1(d, d, d), true2, d), true2, d)
% 17.40/2.59 = { by axiom 13 (rule_137) }
% 17.40/2.59 fresh156(n2(d), true2, d)
% 17.40/2.59 = { by axiom 12 (rule_217) }
% 17.40/2.59 fresh155(m2(b), true2, d)
% 17.40/2.59 = { by axiom 6 (rule_136) R->L }
% 17.40/2.59 fresh155(fresh263(k1(b), true2), true2, d)
% 17.40/2.59 = { by axiom 10 (rule_001) R->L }
% 17.40/2.59 fresh155(fresh263(fresh440(n0(e, b), true2, b), true2), true2, d)
% 17.40/2.59 = { by axiom 2 (axiom_11) }
% 17.40/2.59 fresh155(fresh263(fresh440(true2, true2, b), true2), true2, d)
% 17.40/2.59 = { by axiom 4 (rule_001) }
% 17.40/2.59 fresh155(fresh263(true2, true2), true2, d)
% 17.40/2.59 = { by axiom 3 (rule_136) }
% 17.40/2.59 fresh155(true2, true2, d)
% 17.40/2.59 = { by axiom 9 (rule_217) }
% 17.40/2.59 true2
% 17.40/2.59 % SZS output end Proof
% 17.40/2.59
% 17.40/2.59 RESULT: Unsatisfiable (the axioms are contradictory).
%------------------------------------------------------------------------------