TSTP Solution File: GEO499+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : GEO499+1 : TPTP v8.1.2. Released v7.0.0.
% 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 : n004.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 : Wed Aug 30 23:29:06 EDT 2023
% Result : Theorem 5.27s 1.05s
% Output : Proof 5.27s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.14 % Problem : GEO499+1 : TPTP v8.1.2. Released v7.0.0.
% 0.07/0.15 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.15/0.36 % Computer : n004.cluster.edu
% 0.15/0.36 % Model : x86_64 x86_64
% 0.15/0.36 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.36 % Memory : 8042.1875MB
% 0.15/0.36 % OS : Linux 3.10.0-693.el7.x86_64
% 0.15/0.36 % CPULimit : 300
% 0.15/0.36 % WCLimit : 300
% 0.15/0.36 % DateTime : Tue Aug 29 22:26:22 EDT 2023
% 0.15/0.36 % CPUTime :
% 5.27/1.05 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 5.27/1.05
% 5.27/1.05 % SZS status Theorem
% 5.27/1.05
% 5.27/1.05 % SZS output start Proof
% 5.27/1.05 Take the following subset of the input axioms:
% 5.27/1.05 fof(aSatz7_15a, axiom, ![Xa, Xp, Xq, Xr]: (~s_t(Xp, Xq, Xr) | s_t(s(Xa, Xp), s(Xa, Xq), s(Xa, Xr)))).
% 5.27/1.05 fof(aSatz7_15b, conjecture, ![Xa2, Xp2, Xq2, Xr2]: (s_t(Xp2, Xq2, Xr2) | ~s_t(s(Xa2, Xp2), s(Xa2, Xq2), s(Xa2, Xr2)))).
% 5.27/1.05 fof(aSatz7_7, axiom, ![Xa2, Xp2]: s(Xa2, s(Xa2, Xp2))=Xp2).
% 5.27/1.05
% 5.27/1.05 Now clausify the problem and encode Horn clauses using encoding 3 of
% 5.27/1.05 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 5.27/1.05 We repeatedly replace C & s=t => u=v by the two clauses:
% 5.27/1.05 fresh(y, y, x1...xn) = u
% 5.27/1.05 C => fresh(s, t, x1...xn) = v
% 5.27/1.05 where fresh is a fresh function symbol and x1..xn are the free
% 5.27/1.05 variables of u and v.
% 5.27/1.05 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 5.27/1.05 input problem has no model of domain size 1).
% 5.27/1.05
% 5.27/1.05 The encoding turns the above axioms into the following unit equations and goals:
% 5.27/1.05
% 5.27/1.05 Axiom 1 (aSatz7_7): s(X, s(X, Y)) = Y.
% 5.27/1.05 Axiom 2 (aSatz7_15a): fresh57(X, X, Y, Z, W, V) = true2.
% 5.27/1.05 Axiom 3 (aSatz7_15b): s_t(s(xa, xp), s(xa, xq), s(xa, xr)) = true2.
% 5.27/1.05 Axiom 4 (aSatz7_15a): fresh57(s_t(X, Y, Z), true2, X, Y, Z, W) = s_t(s(W, X), s(W, Y), s(W, Z)).
% 5.27/1.05
% 5.27/1.05 Goal 1 (aSatz7_15b_1): s_t(xp, xq, xr) = true2.
% 5.27/1.05 Proof:
% 5.27/1.05 s_t(xp, xq, xr)
% 5.27/1.05 = { by axiom 1 (aSatz7_7) R->L }
% 5.27/1.05 s_t(xp, xq, s(xa, s(xa, xr)))
% 5.27/1.05 = { by axiom 1 (aSatz7_7) R->L }
% 5.27/1.05 s_t(xp, s(xa, s(xa, xq)), s(xa, s(xa, xr)))
% 5.27/1.05 = { by axiom 1 (aSatz7_7) R->L }
% 5.27/1.05 s_t(s(xa, s(xa, xp)), s(xa, s(xa, xq)), s(xa, s(xa, xr)))
% 5.27/1.05 = { by axiom 4 (aSatz7_15a) R->L }
% 5.27/1.05 fresh57(s_t(s(xa, xp), s(xa, xq), s(xa, xr)), true2, s(xa, xp), s(xa, xq), s(xa, xr), xa)
% 5.27/1.05 = { by axiom 3 (aSatz7_15b) }
% 5.27/1.05 fresh57(true2, true2, s(xa, xp), s(xa, xq), s(xa, xr), xa)
% 5.27/1.05 = { by axiom 2 (aSatz7_15a) }
% 5.27/1.05 true2
% 5.27/1.05 % SZS output end Proof
% 5.27/1.05
% 5.27/1.05 RESULT: Theorem (the conjecture is true).
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