TSTP Solution File: GEO628+1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : GEO628+1 : TPTP v8.1.2. Released v7.5.0.
% 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 : n021.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 23:29:36 EDT 2023
% Result : Theorem 11.34s 1.86s
% Output : Proof 11.83s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.08/0.12 % Problem : GEO628+1 : TPTP v8.1.2. Released v7.5.0.
% 0.08/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.35 % Computer : n021.cluster.edu
% 0.12/0.35 % Model : x86_64 x86_64
% 0.12/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.35 % Memory : 8042.1875MB
% 0.12/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.35 % CPULimit : 300
% 0.12/0.35 % WCLimit : 300
% 0.12/0.35 % DateTime : Tue Aug 29 20:26:14 EDT 2023
% 0.12/0.35 % CPUTime :
% 11.34/1.86 Command-line arguments: --no-flatten-goal
% 11.34/1.86
% 11.34/1.86 % SZS status Theorem
% 11.34/1.86
% 11.83/1.87 % SZS output start Proof
% 11.83/1.87 Take the following subset of the input axioms:
% 11.83/1.87 fof(exemplo6GDDFULL8110991, conjecture, ![A, B, C, F, O, P, G, P1, G1, F1, K, NWPNT1, NWPNT2]: ((circle(O, A, B, C) & (circle(O, A, P1, NWPNT1) & (perp(F, P1, A, C) & (coll(F, A, C) & (perp(G, P1, A, B) & (coll(G, A, B) & (circle(O, A, P, NWPNT2) & (perp(G1, P, A, B) & (coll(G1, A, B) & (perp(F1, P, A, C) & (coll(F1, A, C) & (perp(F, G, P, K) & perp(G1, F1, P1, K))))))))))))) => cyclic(A, P1, P, K))).
% 11.83/1.87 fof(ruleD1, axiom, ![A2, B2, C2]: (coll(A2, B2, C2) => coll(A2, C2, B2))).
% 11.83/1.87 fof(ruleD17, axiom, ![D, E, B2, C2, A2_2]: ((cyclic(A2_2, B2, C2, D) & cyclic(A2_2, B2, C2, E)) => cyclic(B2, C2, D, E))).
% 11.83/1.87 fof(ruleD19, axiom, ![Q, U, V, B2, C2, D2, A2_2, P2]: (eqangle(A2_2, B2, C2, D2, P2, Q, U, V) => eqangle(C2, D2, A2_2, B2, U, V, P2, Q))).
% 11.83/1.87 fof(ruleD2, axiom, ![B2, C2, A2_2]: (coll(A2_2, B2, C2) => coll(B2, A2_2, C2))).
% 11.83/1.87 fof(ruleD21, axiom, ![B2, C2, D2, A2_2, P2, Q2, U2, V2]: (eqangle(A2_2, B2, C2, D2, P2, Q2, U2, V2) => eqangle(A2_2, B2, P2, Q2, C2, D2, U2, V2))).
% 11.83/1.87 fof(ruleD3, axiom, ![B2, C2, D2, A2_2]: ((coll(A2_2, B2, C2) & coll(A2_2, B2, D2)) => coll(C2, D2, A2_2))).
% 11.83/1.87 fof(ruleD39, axiom, ![B2, C2, D2, A2_2, P2, Q2]: (eqangle(A2_2, B2, P2, Q2, C2, D2, P2, Q2) => para(A2_2, B2, C2, D2))).
% 11.83/1.87 fof(ruleD40, axiom, ![B2, C2, D2, A2_2, P2, Q2]: (para(A2_2, B2, C2, D2) => eqangle(A2_2, B2, P2, Q2, C2, D2, P2, Q2))).
% 11.83/1.87 fof(ruleD42b, axiom, ![B2, A2_2, P2, Q2]: ((eqangle(P2, A2_2, P2, B2, Q2, A2_2, Q2, B2) & coll(P2, Q2, B2)) => cyclic(A2_2, B2, P2, Q2))).
% 11.83/1.87 fof(ruleD66, axiom, ![B2, C2, A2_2]: (para(A2_2, B2, A2_2, C2) => coll(A2_2, B2, C2))).
% 11.83/1.87 fof(ruleD8, axiom, ![B2, C2, D2, A2_2]: (perp(A2_2, B2, C2, D2) => perp(C2, D2, A2_2, B2))).
% 11.83/1.87 fof(ruleD9, axiom, ![B2, C2, D2, E2, F2, A2_2]: ((perp(A2_2, B2, C2, D2) & perp(C2, D2, E2, F2)) => para(A2_2, B2, E2, F2))).
% 11.83/1.87
% 11.83/1.87 Now clausify the problem and encode Horn clauses using encoding 3 of
% 11.83/1.87 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 11.83/1.87 We repeatedly replace C & s=t => u=v by the two clauses:
% 11.83/1.87 fresh(y, y, x1...xn) = u
% 11.83/1.87 C => fresh(s, t, x1...xn) = v
% 11.83/1.87 where fresh is a fresh function symbol and x1..xn are the free
% 11.83/1.87 variables of u and v.
% 11.83/1.87 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 11.83/1.87 input problem has no model of domain size 1).
% 11.83/1.87
% 11.83/1.87 The encoding turns the above axioms into the following unit equations and goals:
% 11.83/1.87
% 11.83/1.87 Axiom 1 (exemplo6GDDFULL8110991_5): perp(f, g, p, k) = true.
% 11.83/1.87 Axiom 2 (ruleD1): fresh146(X, X, Y, Z, W) = true.
% 11.83/1.87 Axiom 3 (ruleD2): fresh133(X, X, Y, Z, W) = true.
% 11.83/1.87 Axiom 4 (ruleD3): fresh119(X, X, Y, Z, W) = true.
% 11.83/1.87 Axiom 5 (ruleD66): fresh66(X, X, Y, Z, W) = true.
% 11.83/1.87 Axiom 6 (ruleD17): fresh136(X, X, Y, Z, W, V) = true.
% 11.83/1.87 Axiom 7 (ruleD3): fresh120(X, X, Y, Z, W, V) = coll(W, V, Y).
% 11.83/1.87 Axiom 8 (ruleD39): fresh106(X, X, Y, Z, W, V) = true.
% 11.83/1.87 Axiom 9 (ruleD42b): fresh102(X, X, Y, Z, W, V) = cyclic(Y, Z, W, V).
% 11.83/1.87 Axiom 10 (ruleD42b): fresh101(X, X, Y, Z, W, V) = true.
% 11.83/1.87 Axiom 11 (ruleD8): fresh52(X, X, Y, Z, W, V) = true.
% 11.83/1.87 Axiom 12 (ruleD9): fresh50(X, X, Y, Z, W, V) = true.
% 11.83/1.88 Axiom 13 (ruleD17): fresh137(X, X, Y, Z, W, V, U) = cyclic(Z, W, V, U).
% 11.83/1.88 Axiom 14 (ruleD1): fresh146(coll(X, Y, Z), true, X, Y, Z) = coll(X, Z, Y).
% 11.83/1.88 Axiom 15 (ruleD2): fresh133(coll(X, Y, Z), true, X, Y, Z) = coll(Y, X, Z).
% 11.83/1.88 Axiom 16 (ruleD40): fresh104(X, X, Y, Z, W, V, U, T) = true.
% 11.83/1.88 Axiom 17 (ruleD9): fresh51(X, X, Y, Z, W, V, U, T) = para(Y, Z, U, T).
% 11.83/1.88 Axiom 18 (ruleD3): fresh120(coll(X, Y, Z), true, X, Y, W, Z) = fresh119(coll(X, Y, W), true, X, W, Z).
% 11.83/1.88 Axiom 19 (ruleD66): fresh66(para(X, Y, X, Z), true, X, Y, Z) = coll(X, Y, Z).
% 11.83/1.88 Axiom 20 (ruleD19): fresh134(X, X, Y, Z, W, V, U, T, S, X2) = true.
% 11.83/1.88 Axiom 21 (ruleD21): fresh131(X, X, Y, Z, W, V, U, T, S, X2) = true.
% 11.83/1.88 Axiom 22 (ruleD8): fresh52(perp(X, Y, Z, W), true, X, Y, Z, W) = perp(Z, W, X, Y).
% 11.83/1.88 Axiom 23 (ruleD17): fresh137(cyclic(X, Y, Z, W), true, X, Y, Z, V, W) = fresh136(cyclic(X, Y, Z, V), true, Y, Z, V, W).
% 11.83/1.88 Axiom 24 (ruleD40): fresh104(para(X, Y, Z, W), true, X, Y, Z, W, V, U) = eqangle(X, Y, V, U, Z, W, V, U).
% 11.83/1.88 Axiom 25 (ruleD9): fresh51(perp(X, Y, Z, W), true, V, U, X, Y, Z, W) = fresh50(perp(V, U, X, Y), true, V, U, Z, W).
% 11.83/1.88 Axiom 26 (ruleD39): fresh106(eqangle(X, Y, Z, W, V, U, Z, W), true, X, Y, V, U) = para(X, Y, V, U).
% 11.83/1.88 Axiom 27 (ruleD42b): fresh102(eqangle(X, Y, X, Z, W, Y, W, Z), true, Y, Z, X, W) = fresh101(coll(X, W, Z), true, Y, Z, X, W).
% 11.83/1.88 Axiom 28 (ruleD19): fresh134(eqangle(X, Y, Z, W, V, U, T, S), true, X, Y, Z, W, V, U, T, S) = eqangle(Z, W, X, Y, T, S, V, U).
% 11.83/1.88 Axiom 29 (ruleD21): fresh131(eqangle(X, Y, Z, W, V, U, T, S), true, X, Y, Z, W, V, U, T, S) = eqangle(X, Y, V, U, Z, W, T, S).
% 11.83/1.88
% 11.83/1.88 Lemma 30: eqangle(p, k, X, Y, p, k, X, Y) = true.
% 11.83/1.88 Proof:
% 11.83/1.88 eqangle(p, k, X, Y, p, k, X, Y)
% 11.83/1.88 = { by axiom 24 (ruleD40) R->L }
% 11.83/1.88 fresh104(para(p, k, p, k), true, p, k, p, k, X, Y)
% 11.83/1.88 = { by axiom 17 (ruleD9) R->L }
% 11.83/1.88 fresh104(fresh51(true, true, p, k, f, g, p, k), true, p, k, p, k, X, Y)
% 11.83/1.88 = { by axiom 1 (exemplo6GDDFULL8110991_5) R->L }
% 11.83/1.88 fresh104(fresh51(perp(f, g, p, k), true, p, k, f, g, p, k), true, p, k, p, k, X, Y)
% 11.83/1.88 = { by axiom 25 (ruleD9) }
% 11.83/1.88 fresh104(fresh50(perp(p, k, f, g), true, p, k, p, k), true, p, k, p, k, X, Y)
% 11.83/1.88 = { by axiom 22 (ruleD8) R->L }
% 11.83/1.88 fresh104(fresh50(fresh52(perp(f, g, p, k), true, f, g, p, k), true, p, k, p, k), true, p, k, p, k, X, Y)
% 11.83/1.88 = { by axiom 1 (exemplo6GDDFULL8110991_5) }
% 11.83/1.88 fresh104(fresh50(fresh52(true, true, f, g, p, k), true, p, k, p, k), true, p, k, p, k, X, Y)
% 11.83/1.88 = { by axiom 11 (ruleD8) }
% 11.83/1.88 fresh104(fresh50(true, true, p, k, p, k), true, p, k, p, k, X, Y)
% 11.83/1.88 = { by axiom 12 (ruleD9) }
% 11.83/1.88 fresh104(true, true, p, k, p, k, X, Y)
% 11.83/1.88 = { by axiom 16 (ruleD40) }
% 11.83/1.88 true
% 11.83/1.88
% 11.83/1.88 Lemma 31: coll(X, X, Y) = true.
% 11.83/1.88 Proof:
% 11.83/1.88 coll(X, X, Y)
% 11.83/1.88 = { by axiom 14 (ruleD1) R->L }
% 11.83/1.88 fresh146(coll(X, Y, X), true, X, Y, X)
% 11.83/1.88 = { by axiom 15 (ruleD2) R->L }
% 11.83/1.88 fresh146(fresh133(coll(Y, X, X), true, Y, X, X), true, X, Y, X)
% 11.83/1.88 = { by axiom 19 (ruleD66) R->L }
% 11.83/1.88 fresh146(fresh133(fresh66(para(Y, X, Y, X), true, Y, X, X), true, Y, X, X), true, X, Y, X)
% 11.83/1.88 = { by axiom 26 (ruleD39) R->L }
% 11.83/1.88 fresh146(fresh133(fresh66(fresh106(eqangle(Y, X, p, k, Y, X, p, k), true, Y, X, Y, X), true, Y, X, X), true, Y, X, X), true, X, Y, X)
% 11.83/1.88 = { by axiom 28 (ruleD19) R->L }
% 11.83/1.88 fresh146(fresh133(fresh66(fresh106(fresh134(eqangle(p, k, Y, X, p, k, Y, X), true, p, k, Y, X, p, k, Y, X), true, Y, X, Y, X), true, Y, X, X), true, Y, X, X), true, X, Y, X)
% 11.83/1.88 = { by lemma 30 }
% 11.83/1.88 fresh146(fresh133(fresh66(fresh106(fresh134(true, true, p, k, Y, X, p, k, Y, X), true, Y, X, Y, X), true, Y, X, X), true, Y, X, X), true, X, Y, X)
% 11.83/1.88 = { by axiom 20 (ruleD19) }
% 11.83/1.88 fresh146(fresh133(fresh66(fresh106(true, true, Y, X, Y, X), true, Y, X, X), true, Y, X, X), true, X, Y, X)
% 11.83/1.88 = { by axiom 8 (ruleD39) }
% 11.83/1.88 fresh146(fresh133(fresh66(true, true, Y, X, X), true, Y, X, X), true, X, Y, X)
% 11.83/1.88 = { by axiom 5 (ruleD66) }
% 11.83/1.88 fresh146(fresh133(true, true, Y, X, X), true, X, Y, X)
% 11.83/1.88 = { by axiom 3 (ruleD2) }
% 11.83/1.88 fresh146(true, true, X, Y, X)
% 11.83/1.88 = { by axiom 2 (ruleD1) }
% 11.83/1.88 true
% 11.83/1.88
% 11.83/1.88 Lemma 32: cyclic(k, k, p, X) = true.
% 11.83/1.88 Proof:
% 11.83/1.88 cyclic(k, k, p, X)
% 11.83/1.88 = { by axiom 9 (ruleD42b) R->L }
% 11.83/1.88 fresh102(true, true, k, k, p, X)
% 11.83/1.88 = { by axiom 21 (ruleD21) R->L }
% 11.83/1.88 fresh102(fresh131(true, true, p, k, X, k, p, k, X, k), true, k, k, p, X)
% 11.83/1.88 = { by lemma 30 R->L }
% 11.83/1.88 fresh102(fresh131(eqangle(p, k, X, k, p, k, X, k), true, p, k, X, k, p, k, X, k), true, k, k, p, X)
% 11.83/1.88 = { by axiom 29 (ruleD21) }
% 11.83/1.88 fresh102(eqangle(p, k, p, k, X, k, X, k), true, k, k, p, X)
% 11.83/1.88 = { by axiom 27 (ruleD42b) }
% 11.83/1.88 fresh101(coll(p, X, k), true, k, k, p, X)
% 11.83/1.88 = { by axiom 7 (ruleD3) R->L }
% 11.83/1.88 fresh101(fresh120(true, true, k, k, p, X), true, k, k, p, X)
% 11.83/1.88 = { by lemma 31 R->L }
% 11.83/1.88 fresh101(fresh120(coll(k, k, X), true, k, k, p, X), true, k, k, p, X)
% 11.83/1.88 = { by axiom 18 (ruleD3) }
% 11.83/1.88 fresh101(fresh119(coll(k, k, p), true, k, p, X), true, k, k, p, X)
% 11.83/1.88 = { by lemma 31 }
% 11.83/1.88 fresh101(fresh119(true, true, k, p, X), true, k, k, p, X)
% 11.83/1.88 = { by axiom 4 (ruleD3) }
% 11.83/1.88 fresh101(true, true, k, k, p, X)
% 11.83/1.88 = { by axiom 10 (ruleD42b) }
% 11.83/1.88 true
% 11.83/1.88
% 11.83/1.88 Lemma 33: cyclic(k, p, X, Y) = true.
% 11.83/1.88 Proof:
% 11.83/1.88 cyclic(k, p, X, Y)
% 11.83/1.88 = { by axiom 13 (ruleD17) R->L }
% 11.83/1.88 fresh137(true, true, k, k, p, X, Y)
% 11.83/1.88 = { by lemma 32 R->L }
% 11.83/1.88 fresh137(cyclic(k, k, p, Y), true, k, k, p, X, Y)
% 11.83/1.88 = { by axiom 23 (ruleD17) }
% 11.83/1.88 fresh136(cyclic(k, k, p, X), true, k, p, X, Y)
% 11.83/1.88 = { by lemma 32 }
% 11.83/1.88 fresh136(true, true, k, p, X, Y)
% 11.83/1.88 = { by axiom 6 (ruleD17) }
% 11.83/1.88 true
% 11.83/1.88
% 11.83/1.88 Lemma 34: cyclic(p, X, Y, Z) = true.
% 11.83/1.88 Proof:
% 11.83/1.88 cyclic(p, X, Y, Z)
% 11.83/1.88 = { by axiom 13 (ruleD17) R->L }
% 11.83/1.88 fresh137(true, true, k, p, X, Y, Z)
% 11.83/1.88 = { by lemma 33 R->L }
% 11.83/1.88 fresh137(cyclic(k, p, X, Z), true, k, p, X, Y, Z)
% 11.83/1.88 = { by axiom 23 (ruleD17) }
% 11.83/1.88 fresh136(cyclic(k, p, X, Y), true, p, X, Y, Z)
% 11.83/1.88 = { by lemma 33 }
% 11.83/1.88 fresh136(true, true, p, X, Y, Z)
% 11.83/1.88 = { by axiom 6 (ruleD17) }
% 11.83/1.88 true
% 11.83/1.88
% 11.83/1.88 Goal 1 (exemplo6GDDFULL8110991_13): cyclic(a, p1, p, k) = true.
% 11.83/1.88 Proof:
% 11.83/1.88 cyclic(a, p1, p, k)
% 11.83/1.88 = { by axiom 13 (ruleD17) R->L }
% 11.83/1.88 fresh137(true, true, p, a, p1, p, k)
% 11.83/1.88 = { by lemma 34 R->L }
% 11.83/1.88 fresh137(cyclic(p, a, p1, k), true, p, a, p1, p, k)
% 11.83/1.88 = { by axiom 23 (ruleD17) }
% 11.83/1.88 fresh136(cyclic(p, a, p1, p), true, a, p1, p, k)
% 11.83/1.88 = { by lemma 34 }
% 11.83/1.88 fresh136(true, true, a, p1, p, k)
% 11.83/1.88 = { by axiom 6 (ruleD17) }
% 11.83/1.88 true
% 11.83/1.88 % SZS output end Proof
% 11.83/1.88
% 11.83/1.88 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------