TSTP Solution File: LDA007-2 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : LDA007-2 : TPTP v8.1.2. Bugfixed v2.6.0.
% 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 : n025.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 09:04:47 EDT 2023
% Result : Unsatisfiable 0.19s 0.39s
% Output : Proof 0.19s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : LDA007-2 : TPTP v8.1.2. Bugfixed v2.6.0.
% 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 : n025.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 : Sun Aug 27 01:37:53 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.39 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
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% 0.19/0.39 % SZS status Unsatisfiable
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% 0.19/0.40 % SZS output start Proof
% 0.19/0.40 Take the following subset of the input axioms:
% 0.19/0.40 fof(a1, axiom, ![X, Y, Z]: f(X, f(Y, Z))=f(f(X, Y), f(X, Z))).
% 0.19/0.40 fof(clause_1, axiom, tt=f(t, t)).
% 0.19/0.40 fof(clause_3, axiom, ts=f(t, s)).
% 0.19/0.40 fof(clause_4, axiom, tt_ts=f(tt, ts)).
% 0.19/0.40 fof(clause_7, axiom, tk=f(t, k)).
% 0.19/0.40 fof(clause_9, axiom, tsk=f(ts, k)).
% 0.19/0.40 fof(prove_equation, negated_conjecture, f(t, tsk)!=f(tt_ts, tk)).
% 0.19/0.40
% 0.19/0.40 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.40 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.40 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.40 fresh(y, y, x1...xn) = u
% 0.19/0.40 C => fresh(s, t, x1...xn) = v
% 0.19/0.40 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.40 variables of u and v.
% 0.19/0.40 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.40 input problem has no model of domain size 1).
% 0.19/0.40
% 0.19/0.40 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.40
% 0.19/0.40 Axiom 1 (clause_1): tt = f(t, t).
% 0.19/0.40 Axiom 2 (clause_7): tk = f(t, k).
% 0.19/0.40 Axiom 3 (clause_3): ts = f(t, s).
% 0.19/0.40 Axiom 4 (clause_4): tt_ts = f(tt, ts).
% 0.19/0.40 Axiom 5 (clause_9): tsk = f(ts, k).
% 0.19/0.40 Axiom 6 (a1): f(X, f(Y, Z)) = f(f(X, Y), f(X, Z)).
% 0.19/0.40
% 0.19/0.40 Goal 1 (prove_equation): f(t, tsk) = f(tt_ts, tk).
% 0.19/0.40 Proof:
% 0.19/0.40 f(t, tsk)
% 0.19/0.40 = { by axiom 5 (clause_9) }
% 0.19/0.40 f(t, f(ts, k))
% 0.19/0.40 = { by axiom 6 (a1) }
% 0.19/0.40 f(f(t, ts), f(t, k))
% 0.19/0.40 = { by axiom 2 (clause_7) R->L }
% 0.19/0.40 f(f(t, ts), tk)
% 0.19/0.40 = { by axiom 3 (clause_3) }
% 0.19/0.40 f(f(t, f(t, s)), tk)
% 0.19/0.40 = { by axiom 6 (a1) }
% 0.19/0.40 f(f(f(t, t), f(t, s)), tk)
% 0.19/0.40 = { by axiom 1 (clause_1) R->L }
% 0.19/0.40 f(f(tt, f(t, s)), tk)
% 0.19/0.40 = { by axiom 3 (clause_3) R->L }
% 0.19/0.40 f(f(tt, ts), tk)
% 0.19/0.40 = { by axiom 4 (clause_4) R->L }
% 0.19/0.40 f(tt_ts, tk)
% 0.19/0.40 % SZS output end Proof
% 0.19/0.40
% 0.19/0.40 RESULT: Unsatisfiable (the axioms are contradictory).
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