TSTP Solution File: SYN248-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN248-1 : 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 : n006.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 03:33:46 EDT 2023
% Result : Unsatisfiable 21.30s 3.15s
% Output : Proof 21.30s
% 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 : SYN248-1 : 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
% 0.13/0.34 % Computer : n006.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 : Sat Aug 26 20:13:07 EDT 2023
% 0.13/0.34 % CPUTime :
% 21.30/3.15 Command-line arguments: --no-flatten-goal
% 21.30/3.15
% 21.30/3.15 % SZS status Unsatisfiable
% 21.30/3.15
% 21.30/3.15 % SZS output start Proof
% 21.30/3.15 Take the following subset of the input axioms:
% 21.30/3.15 fof(axiom_17, axiom, ![X]: q0(X, d)).
% 21.30/3.15 fof(axiom_26, axiom, n0(d, c)).
% 21.30/3.15 fof(axiom_37, axiom, n0(b, a)).
% 21.30/3.15 fof(prove_this, negated_conjecture, ~m1(d, c, c)).
% 21.30/3.15 fof(rule_001, axiom, ![I, J]: (k1(I) | ~n0(J, I))).
% 21.30/3.15 fof(rule_034, axiom, ![B, A2]: (m1(A2, B, B) | (~k1(a) | (~k1(B) | ~q0(A2, A2))))).
% 21.30/3.15
% 21.30/3.15 Now clausify the problem and encode Horn clauses using encoding 3 of
% 21.30/3.15 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 21.30/3.15 We repeatedly replace C & s=t => u=v by the two clauses:
% 21.30/3.15 fresh(y, y, x1...xn) = u
% 21.30/3.15 C => fresh(s, t, x1...xn) = v
% 21.30/3.15 where fresh is a fresh function symbol and x1..xn are the free
% 21.30/3.15 variables of u and v.
% 21.30/3.15 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 21.30/3.15 input problem has no model of domain size 1).
% 21.30/3.15
% 21.30/3.15 The encoding turns the above axioms into the following unit equations and goals:
% 21.30/3.15
% 21.30/3.15 Axiom 1 (axiom_26): n0(d, c) = true.
% 21.30/3.15 Axiom 2 (axiom_37): n0(b, a) = true.
% 21.30/3.15 Axiom 3 (axiom_17): q0(X, d) = true.
% 21.30/3.15 Axiom 4 (rule_001): fresh440(X, X, Y) = true.
% 21.30/3.15 Axiom 5 (rule_034): fresh653(X, X, Y, Z) = m1(Y, Z, Z).
% 21.30/3.15 Axiom 6 (rule_034): fresh397(X, X, Y, Z) = true.
% 21.30/3.15 Axiom 7 (rule_034): fresh652(X, X, Y, Z) = fresh653(k1(Z), true, Y, Z).
% 21.30/3.15 Axiom 8 (rule_001): fresh440(n0(X, Y), true, Y) = k1(Y).
% 21.30/3.15 Axiom 9 (rule_034): fresh652(k1(a), true, X, Y) = fresh397(q0(X, X), true, X, Y).
% 21.30/3.15
% 21.30/3.15 Goal 1 (prove_this): m1(d, c, c) = true.
% 21.30/3.15 Proof:
% 21.30/3.15 m1(d, c, c)
% 21.30/3.15 = { by axiom 5 (rule_034) R->L }
% 21.30/3.15 fresh653(true, true, d, c)
% 21.30/3.15 = { by axiom 4 (rule_001) R->L }
% 21.30/3.15 fresh653(fresh440(true, true, c), true, d, c)
% 21.30/3.15 = { by axiom 1 (axiom_26) R->L }
% 21.30/3.15 fresh653(fresh440(n0(d, c), true, c), true, d, c)
% 21.30/3.15 = { by axiom 8 (rule_001) }
% 21.30/3.15 fresh653(k1(c), true, d, c)
% 21.30/3.15 = { by axiom 7 (rule_034) R->L }
% 21.30/3.15 fresh652(true, true, d, c)
% 21.30/3.15 = { by axiom 4 (rule_001) R->L }
% 21.30/3.15 fresh652(fresh440(true, true, a), true, d, c)
% 21.30/3.15 = { by axiom 2 (axiom_37) R->L }
% 21.30/3.15 fresh652(fresh440(n0(b, a), true, a), true, d, c)
% 21.30/3.15 = { by axiom 8 (rule_001) }
% 21.30/3.15 fresh652(k1(a), true, d, c)
% 21.30/3.15 = { by axiom 9 (rule_034) }
% 21.30/3.15 fresh397(q0(d, d), true, d, c)
% 21.30/3.15 = { by axiom 3 (axiom_17) }
% 21.30/3.15 fresh397(true, true, d, c)
% 21.30/3.15 = { by axiom 6 (rule_034) }
% 21.30/3.15 true
% 21.30/3.15 % SZS output end Proof
% 21.30/3.15
% 21.30/3.15 RESULT: Unsatisfiable (the axioms are contradictory).
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