TSTP Solution File: SYN175-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN175-1 : 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 : n005.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 03:33:28 EDT 2023
% Result : Unsatisfiable 15.96s 2.55s
% Output : Proof 16.74s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SYN175-1 : TPTP v8.1.2. Released v1.1.0.
% 0.07/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.33 % Computer : n005.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Sat Aug 26 17:06:53 EDT 2023
% 0.12/0.34 % CPUTime :
% 15.96/2.55 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 15.96/2.55
% 15.96/2.55 % SZS status Unsatisfiable
% 15.96/2.55
% 15.96/2.55 % SZS output start Proof
% 15.96/2.55 Take the following subset of the input axioms:
% 16.74/2.55 fof(axiom_34, axiom, n0(c, d)).
% 16.74/2.55 fof(prove_this, negated_conjecture, ![X, Y]: ~q2(X, d, Y)).
% 16.74/2.55 fof(rule_002, axiom, ![G, H]: (l1(G, G) | ~n0(H, G))).
% 16.74/2.55 fof(rule_186, axiom, ![G2, H2]: (q2(G2, G2, H2) | ~l1(H2, G2))).
% 16.74/2.55
% 16.74/2.55 Now clausify the problem and encode Horn clauses using encoding 3 of
% 16.74/2.55 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 16.74/2.55 We repeatedly replace C & s=t => u=v by the two clauses:
% 16.74/2.55 fresh(y, y, x1...xn) = u
% 16.74/2.55 C => fresh(s, t, x1...xn) = v
% 16.74/2.55 where fresh is a fresh function symbol and x1..xn are the free
% 16.74/2.55 variables of u and v.
% 16.74/2.55 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 16.74/2.55 input problem has no model of domain size 1).
% 16.74/2.55
% 16.74/2.55 The encoding turns the above axioms into the following unit equations and goals:
% 16.74/2.55
% 16.74/2.55 Axiom 1 (axiom_34): n0(c, d) = true2.
% 16.74/2.55 Axiom 2 (rule_002): fresh441(X, X, Y) = true2.
% 16.74/2.55 Axiom 3 (rule_002): fresh441(n0(X, Y), true2, Y) = l1(Y, Y).
% 16.74/2.55 Axiom 4 (rule_186): fresh195(X, X, Y, Z) = true2.
% 16.74/2.55 Axiom 5 (rule_186): fresh195(l1(X, Y), true2, Y, X) = q2(Y, Y, X).
% 16.74/2.55
% 16.74/2.55 Goal 1 (prove_this): q2(X, d, Y) = true2.
% 16.74/2.56 The goal is true when:
% 16.74/2.56 X = d
% 16.74/2.56 Y = d
% 16.74/2.56
% 16.74/2.56 Proof:
% 16.74/2.56 q2(d, d, d)
% 16.74/2.56 = { by axiom 5 (rule_186) R->L }
% 16.74/2.56 fresh195(l1(d, d), true2, d, d)
% 16.74/2.56 = { by axiom 3 (rule_002) R->L }
% 16.74/2.56 fresh195(fresh441(n0(c, d), true2, d), true2, d, d)
% 16.74/2.56 = { by axiom 1 (axiom_34) }
% 16.74/2.56 fresh195(fresh441(true2, true2, d), true2, d, d)
% 16.74/2.56 = { by axiom 2 (rule_002) }
% 16.74/2.56 fresh195(true2, true2, d, d)
% 16.74/2.56 = { by axiom 4 (rule_186) }
% 16.74/2.56 true2
% 16.74/2.56 % SZS output end Proof
% 16.74/2.56
% 16.74/2.56 RESULT: Unsatisfiable (the axioms are contradictory).
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