TSTP Solution File: SYN559-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN559-1 : TPTP v8.1.2. Released v2.5.0.
% 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 : n015.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 03:35:15 EDT 2023
% Result : Unsatisfiable 0.20s 0.47s
% Output : Proof 0.20s
% 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 : SYN559-1 : TPTP v8.1.2. Released v2.5.0.
% 0.07/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.34 % Computer : n015.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Sat Aug 26 20:51:23 EDT 2023
% 0.14/0.34 % CPUTime :
% 0.20/0.47 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.20/0.47
% 0.20/0.47 % SZS status Unsatisfiable
% 0.20/0.47
% 0.20/0.47 % SZS output start Proof
% 0.20/0.47 Take the following subset of the input axioms:
% 0.20/0.48 fof(not_p8_14, negated_conjecture, ![X26, X25]: (~p8(X26, c9) | (~p7(c10, X26) | (~p2(f6(c9, X26), X25) | ~p2(f5(X25, c10), f3(f4(c11), c12)))))).
% 0.20/0.48 fof(p2_1, negated_conjecture, ![X0]: p2(X0, X0)).
% 0.20/0.48 fof(p2_4, negated_conjecture, p2(f6(c9, c13), c14)).
% 0.20/0.48 fof(p2_5, negated_conjecture, p2(f5(c14, c10), c15)).
% 0.20/0.48 fof(p2_6, negated_conjecture, p2(f3(f4(c11), c12), c15)).
% 0.20/0.48 fof(p2_8, negated_conjecture, ![X1, X2, X0_2]: (p2(X1, X2) | (~p2(X0_2, X1) | ~p2(X0_2, X2)))).
% 0.20/0.48 fof(p7_3, negated_conjecture, p7(c10, c13)).
% 0.20/0.48 fof(p8_2, negated_conjecture, p8(c13, c9)).
% 0.20/0.48
% 0.20/0.48 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.48 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.48 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.48 fresh(y, y, x1...xn) = u
% 0.20/0.48 C => fresh(s, t, x1...xn) = v
% 0.20/0.48 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.48 variables of u and v.
% 0.20/0.48 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.48 input problem has no model of domain size 1).
% 0.20/0.48
% 0.20/0.48 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.48
% 0.20/0.48 Axiom 1 (p2_1): p2(X, X) = true2.
% 0.20/0.48 Axiom 2 (p8_2): p8(c13, c9) = true2.
% 0.20/0.48 Axiom 3 (p7_3): p7(c10, c13) = true2.
% 0.20/0.48 Axiom 4 (p2_4): p2(f6(c9, c13), c14) = true2.
% 0.20/0.48 Axiom 5 (p2_5): p2(f5(c14, c10), c15) = true2.
% 0.20/0.48 Axiom 6 (p2_8): fresh3(X, X, Y, Z) = true2.
% 0.20/0.48 Axiom 7 (p2_6): p2(f3(f4(c11), c12), c15) = true2.
% 0.20/0.48 Axiom 8 (p2_8): fresh4(X, X, Y, Z, W) = p2(Y, Z).
% 0.20/0.48 Axiom 9 (p2_8): fresh4(p2(X, Y), true2, Z, Y, X) = fresh3(p2(X, Z), true2, Z, Y).
% 0.20/0.48
% 0.20/0.48 Lemma 10: fresh3(p2(X, Y), true2, Y, X) = p2(Y, X).
% 0.20/0.48 Proof:
% 0.20/0.48 fresh3(p2(X, Y), true2, Y, X)
% 0.20/0.48 = { by axiom 9 (p2_8) R->L }
% 0.20/0.48 fresh4(p2(X, X), true2, Y, X, X)
% 0.20/0.48 = { by axiom 1 (p2_1) }
% 0.20/0.48 fresh4(true2, true2, Y, X, X)
% 0.20/0.48 = { by axiom 8 (p2_8) }
% 0.20/0.48 p2(Y, X)
% 0.20/0.48
% 0.20/0.48 Goal 1 (not_p8_14): tuple(p2(f6(c9, X), Y), p2(f5(Y, c10), f3(f4(c11), c12)), p8(X, c9), p7(c10, X)) = tuple(true2, true2, true2, true2).
% 0.20/0.48 The goal is true when:
% 0.20/0.48 X = c13
% 0.20/0.48 Y = c14
% 0.20/0.48
% 0.20/0.48 Proof:
% 0.20/0.48 tuple(p2(f6(c9, c13), c14), p2(f5(c14, c10), f3(f4(c11), c12)), p8(c13, c9), p7(c10, c13))
% 0.20/0.48 = { by axiom 4 (p2_4) }
% 0.20/0.48 tuple(true2, p2(f5(c14, c10), f3(f4(c11), c12)), p8(c13, c9), p7(c10, c13))
% 0.20/0.48 = { by axiom 2 (p8_2) }
% 0.20/0.48 tuple(true2, p2(f5(c14, c10), f3(f4(c11), c12)), true2, p7(c10, c13))
% 0.20/0.48 = { by axiom 3 (p7_3) }
% 0.20/0.48 tuple(true2, p2(f5(c14, c10), f3(f4(c11), c12)), true2, true2)
% 0.20/0.48 = { by axiom 8 (p2_8) R->L }
% 0.20/0.48 tuple(true2, fresh4(true2, true2, f5(c14, c10), f3(f4(c11), c12), c15), true2, true2)
% 0.20/0.48 = { by axiom 6 (p2_8) R->L }
% 0.20/0.48 tuple(true2, fresh4(fresh3(true2, true2, c15, f3(f4(c11), c12)), true2, f5(c14, c10), f3(f4(c11), c12), c15), true2, true2)
% 0.20/0.48 = { by axiom 7 (p2_6) R->L }
% 0.20/0.48 tuple(true2, fresh4(fresh3(p2(f3(f4(c11), c12), c15), true2, c15, f3(f4(c11), c12)), true2, f5(c14, c10), f3(f4(c11), c12), c15), true2, true2)
% 0.20/0.48 = { by lemma 10 }
% 0.20/0.48 tuple(true2, fresh4(p2(c15, f3(f4(c11), c12)), true2, f5(c14, c10), f3(f4(c11), c12), c15), true2, true2)
% 0.20/0.48 = { by axiom 9 (p2_8) }
% 0.20/0.48 tuple(true2, fresh3(p2(c15, f5(c14, c10)), true2, f5(c14, c10), f3(f4(c11), c12)), true2, true2)
% 0.20/0.48 = { by lemma 10 R->L }
% 0.20/0.48 tuple(true2, fresh3(fresh3(p2(f5(c14, c10), c15), true2, c15, f5(c14, c10)), true2, f5(c14, c10), f3(f4(c11), c12)), true2, true2)
% 0.20/0.48 = { by axiom 5 (p2_5) }
% 0.20/0.48 tuple(true2, fresh3(fresh3(true2, true2, c15, f5(c14, c10)), true2, f5(c14, c10), f3(f4(c11), c12)), true2, true2)
% 0.20/0.48 = { by axiom 6 (p2_8) }
% 0.20/0.48 tuple(true2, fresh3(true2, true2, f5(c14, c10), f3(f4(c11), c12)), true2, true2)
% 0.20/0.48 = { by axiom 6 (p2_8) }
% 0.20/0.48 tuple(true2, true2, true2, true2)
% 0.20/0.48 % SZS output end Proof
% 0.20/0.48
% 0.20/0.48 RESULT: Unsatisfiable (the axioms are contradictory).
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