TSTP Solution File: SYN106-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN106-1 : 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 : n027.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 03:33:11 EDT 2023
% Result : Unsatisfiable 12.32s 2.03s
% Output : Proof 12.32s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.15 % Problem : SYN106-1 : TPTP v8.1.2. Released v1.1.0.
% 0.00/0.16 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.15/0.37 % Computer : n027.cluster.edu
% 0.15/0.37 % Model : x86_64 x86_64
% 0.15/0.37 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.37 % Memory : 8042.1875MB
% 0.15/0.37 % OS : Linux 3.10.0-693.el7.x86_64
% 0.15/0.37 % CPULimit : 300
% 0.15/0.37 % WCLimit : 300
% 0.15/0.37 % DateTime : Sat Aug 26 20:28:20 EDT 2023
% 0.15/0.37 % CPUTime :
% 12.32/2.03 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 12.32/2.03
% 12.32/2.03 % SZS status Unsatisfiable
% 12.32/2.03
% 12.32/2.03 % SZS output start Proof
% 12.32/2.03 Take the following subset of the input axioms:
% 12.32/2.03 fof(axiom_30, axiom, n0(e, e)).
% 12.32/2.03 fof(prove_this, negated_conjecture, ~k1(e)).
% 12.32/2.03 fof(rule_001, axiom, ![I, J]: (k1(I) | ~n0(J, I))).
% 12.32/2.03
% 12.32/2.03 Now clausify the problem and encode Horn clauses using encoding 3 of
% 12.32/2.03 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 12.32/2.03 We repeatedly replace C & s=t => u=v by the two clauses:
% 12.32/2.03 fresh(y, y, x1...xn) = u
% 12.32/2.03 C => fresh(s, t, x1...xn) = v
% 12.32/2.03 where fresh is a fresh function symbol and x1..xn are the free
% 12.32/2.03 variables of u and v.
% 12.32/2.03 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 12.32/2.03 input problem has no model of domain size 1).
% 12.32/2.03
% 12.32/2.03 The encoding turns the above axioms into the following unit equations and goals:
% 12.32/2.03
% 12.32/2.03 Axiom 1 (axiom_30): n0(e, e) = true.
% 12.32/2.03 Axiom 2 (rule_001): fresh440(X, X, Y) = true.
% 12.32/2.04 Axiom 3 (rule_001): fresh440(n0(X, Y), true, Y) = k1(Y).
% 12.32/2.04
% 12.32/2.04 Goal 1 (prove_this): k1(e) = true.
% 12.32/2.04 Proof:
% 12.32/2.04 k1(e)
% 12.32/2.04 = { by axiom 3 (rule_001) R->L }
% 12.32/2.04 fresh440(n0(e, e), true, e)
% 12.32/2.04 = { by axiom 1 (axiom_30) }
% 12.32/2.04 fresh440(true, true, e)
% 12.32/2.04 = { by axiom 2 (rule_001) }
% 12.32/2.04 true
% 12.32/2.04 % SZS output end Proof
% 12.32/2.04
% 12.32/2.04 RESULT: Unsatisfiable (the axioms are contradictory).
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