TSTP Solution File: GRP659+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : GRP659+1 : TPTP v8.1.2. Released v4.0.0.
% 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 : n018.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 : Thu Aug 31 01:19:31 EDT 2023
% Result : Theorem 0.19s 0.45s
% Output : Proof 0.19s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : GRP659+1 : TPTP v8.1.2. Released v4.0.0.
% 0.07/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34 % Computer : n018.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 : Mon Aug 28 21:02:02 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.45 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.19/0.45
% 0.19/0.45 % SZS status Theorem
% 0.19/0.45
% 0.19/0.47 % SZS output start Proof
% 0.19/0.47 Take the following subset of the input axioms:
% 0.19/0.48 fof(f01, axiom, ![B, A]: mult(A, ld(A, B))=B).
% 0.19/0.48 fof(f02, axiom, ![B2, A2]: ld(A2, mult(A2, B2))=B2).
% 0.19/0.48 fof(f03, axiom, ![B2, A2]: mult(rd(A2, B2), B2)=A2).
% 0.19/0.48 fof(f04, axiom, ![B2, A2]: rd(mult(A2, B2), B2)=A2).
% 0.19/0.48 fof(f05, axiom, ![C, B2, A2]: mult(mult(A2, mult(B2, C)), B2)=mult(mult(A2, B2), mult(C, B2))).
% 0.19/0.48 fof(goals, conjecture, ?[X0]: ![X1]: (mult(X1, X0)=X1 & mult(X0, X1)=X1)).
% 0.19/0.48
% 0.19/0.48 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.48 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.48 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.48 fresh(y, y, x1...xn) = u
% 0.19/0.48 C => fresh(s, t, x1...xn) = v
% 0.19/0.48 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.48 variables of u and v.
% 0.19/0.48 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.48 input problem has no model of domain size 1).
% 0.19/0.48
% 0.19/0.48 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.48
% 0.19/0.48 Axiom 1 (f01): mult(X, ld(X, Y)) = Y.
% 0.19/0.48 Axiom 2 (f03): mult(rd(X, Y), Y) = X.
% 0.19/0.48 Axiom 3 (f02): ld(X, mult(X, Y)) = Y.
% 0.19/0.48 Axiom 4 (f04): rd(mult(X, Y), Y) = X.
% 0.19/0.48 Axiom 5 (f05): mult(mult(X, mult(Y, Z)), Y) = mult(mult(X, Y), mult(Z, Y)).
% 0.19/0.48
% 0.19/0.48 Lemma 6: mult(mult(X, Y), mult(ld(Y, Z), Y)) = mult(mult(X, Z), Y).
% 0.19/0.48 Proof:
% 0.19/0.48 mult(mult(X, Y), mult(ld(Y, Z), Y))
% 0.19/0.48 = { by axiom 5 (f05) R->L }
% 0.19/0.48 mult(mult(X, mult(Y, ld(Y, Z))), Y)
% 0.19/0.48 = { by axiom 1 (f01) }
% 0.19/0.48 mult(mult(X, Z), Y)
% 0.19/0.48
% 0.19/0.48 Lemma 7: mult(X, mult(ld(Y, Y), Y)) = mult(X, Y).
% 0.19/0.48 Proof:
% 0.19/0.48 mult(X, mult(ld(Y, Y), Y))
% 0.19/0.48 = { by axiom 2 (f03) R->L }
% 0.19/0.48 mult(mult(rd(X, Y), Y), mult(ld(Y, Y), Y))
% 0.19/0.48 = { by lemma 6 }
% 0.19/0.48 mult(mult(rd(X, Y), Y), Y)
% 0.19/0.48 = { by axiom 2 (f03) }
% 0.19/0.48 mult(X, Y)
% 0.19/0.48
% 0.19/0.48 Lemma 8: mult(ld(X, X), X) = X.
% 0.19/0.48 Proof:
% 0.19/0.48 mult(ld(X, X), X)
% 0.19/0.48 = { by axiom 3 (f02) R->L }
% 0.19/0.48 ld(Y, mult(Y, mult(ld(X, X), X)))
% 0.19/0.48 = { by lemma 7 }
% 0.19/0.48 ld(Y, mult(Y, X))
% 0.19/0.48 = { by axiom 3 (f02) }
% 0.19/0.48 X
% 0.19/0.48
% 0.19/0.48 Lemma 9: rd(X, X) = ld(X, X).
% 0.19/0.48 Proof:
% 0.19/0.48 rd(X, X)
% 0.19/0.48 = { by lemma 8 R->L }
% 0.19/0.48 rd(mult(ld(X, X), X), X)
% 0.19/0.48 = { by axiom 4 (f04) }
% 0.19/0.48 ld(X, X)
% 0.19/0.48
% 0.19/0.48 Lemma 10: ld(mult(X, Y), mult(mult(X, Z), Y)) = mult(ld(Y, Z), Y).
% 0.19/0.48 Proof:
% 0.19/0.48 ld(mult(X, Y), mult(mult(X, Z), Y))
% 0.19/0.48 = { by lemma 6 R->L }
% 0.19/0.48 ld(mult(X, Y), mult(mult(X, Y), mult(ld(Y, Z), Y)))
% 0.19/0.48 = { by axiom 3 (f02) }
% 0.19/0.48 mult(ld(Y, Z), Y)
% 0.19/0.48
% 0.19/0.48 Lemma 11: ld(mult(X, Y), mult(Z, Y)) = mult(ld(Y, ld(X, Z)), Y).
% 0.19/0.48 Proof:
% 0.19/0.48 ld(mult(X, Y), mult(Z, Y))
% 0.19/0.48 = { by axiom 1 (f01) R->L }
% 0.19/0.48 ld(mult(X, Y), mult(mult(X, ld(X, Z)), Y))
% 0.19/0.48 = { by lemma 10 }
% 0.19/0.48 mult(ld(Y, ld(X, Z)), Y)
% 0.19/0.48
% 0.19/0.48 Lemma 12: ld(ld(X, ld(Y, Y)), Y) = mult(X, Y).
% 0.19/0.48 Proof:
% 0.19/0.48 ld(ld(X, ld(Y, Y)), Y)
% 0.19/0.48 = { by axiom 1 (f01) R->L }
% 0.19/0.48 mult(X, ld(X, ld(ld(X, ld(Y, Y)), Y)))
% 0.19/0.48 = { by axiom 4 (f04) R->L }
% 0.19/0.48 mult(X, ld(X, ld(ld(X, ld(Y, rd(mult(Y, X), X))), Y)))
% 0.19/0.48 = { by axiom 4 (f04) R->L }
% 0.19/0.48 mult(X, ld(X, ld(rd(mult(ld(X, ld(Y, rd(mult(Y, X), X))), X), X), Y)))
% 0.19/0.48 = { by lemma 11 R->L }
% 0.19/0.48 mult(X, ld(X, ld(rd(ld(mult(Y, X), mult(rd(mult(Y, X), X), X)), X), Y)))
% 0.19/0.48 = { by axiom 2 (f03) }
% 0.19/0.48 mult(X, ld(X, ld(rd(ld(mult(Y, X), mult(Y, X)), X), Y)))
% 0.19/0.48 = { by axiom 4 (f04) R->L }
% 0.19/0.48 mult(X, rd(mult(ld(X, ld(rd(ld(mult(Y, X), mult(Y, X)), X), Y)), X), X))
% 0.19/0.48 = { by lemma 11 R->L }
% 0.19/0.48 mult(X, rd(ld(mult(rd(ld(mult(Y, X), mult(Y, X)), X), X), mult(Y, X)), X))
% 0.19/0.48 = { by axiom 2 (f03) }
% 0.19/0.48 mult(X, rd(ld(ld(mult(Y, X), mult(Y, X)), mult(Y, X)), X))
% 0.19/0.48 = { by lemma 8 R->L }
% 0.19/0.48 mult(X, rd(ld(ld(mult(Y, X), mult(Y, X)), mult(ld(mult(Y, X), mult(Y, X)), mult(Y, X))), X))
% 0.19/0.48 = { by axiom 3 (f02) }
% 0.19/0.48 mult(X, rd(mult(Y, X), X))
% 0.19/0.48 = { by axiom 4 (f04) }
% 0.19/0.48 mult(X, Y)
% 0.19/0.48
% 0.19/0.48 Lemma 13: rd(mult(X, mult(Y, Z)), Z) = mult(rd(X, Z), mult(Z, Y)).
% 0.19/0.48 Proof:
% 0.19/0.48 rd(mult(X, mult(Y, Z)), Z)
% 0.19/0.48 = { by axiom 2 (f03) R->L }
% 0.19/0.48 rd(mult(mult(rd(X, Z), Z), mult(Y, Z)), Z)
% 0.19/0.48 = { by axiom 5 (f05) R->L }
% 0.19/0.48 rd(mult(mult(rd(X, Z), mult(Z, Y)), Z), Z)
% 0.19/0.48 = { by axiom 4 (f04) }
% 0.19/0.48 mult(rd(X, Z), mult(Z, Y))
% 0.19/0.48
% 0.19/0.48 Lemma 14: rd(X, ld(Y, X)) = Y.
% 0.19/0.48 Proof:
% 0.19/0.48 rd(X, ld(Y, X))
% 0.19/0.48 = { by axiom 1 (f01) R->L }
% 0.19/0.48 rd(mult(Y, ld(Y, X)), ld(Y, X))
% 0.19/0.48 = { by axiom 4 (f04) }
% 0.19/0.48 Y
% 0.19/0.48
% 0.19/0.48 Lemma 15: mult(X, rd(ld(X, X), Y)) = rd(X, Y).
% 0.19/0.48 Proof:
% 0.19/0.48 mult(X, rd(ld(X, X), Y))
% 0.19/0.48 = { by axiom 4 (f04) R->L }
% 0.19/0.48 mult(X, rd(mult(rd(ld(X, X), Y), X), X))
% 0.19/0.48 = { by lemma 12 R->L }
% 0.19/0.48 mult(X, rd(ld(ld(rd(ld(X, X), Y), ld(X, X)), X), X))
% 0.19/0.48 = { by axiom 2 (f03) R->L }
% 0.19/0.48 mult(X, rd(ld(ld(rd(ld(X, X), Y), mult(rd(ld(X, X), Y), Y)), X), X))
% 0.19/0.48 = { by axiom 3 (f02) }
% 0.19/0.48 mult(X, rd(ld(Y, X), X))
% 0.19/0.48 = { by axiom 3 (f02) R->L }
% 0.19/0.48 ld(rd(Y, X), mult(rd(Y, X), mult(X, rd(ld(Y, X), X))))
% 0.19/0.48 = { by lemma 13 R->L }
% 0.19/0.48 ld(rd(Y, X), rd(mult(Y, mult(rd(ld(Y, X), X), X)), X))
% 0.19/0.48 = { by axiom 2 (f03) }
% 0.19/0.48 ld(rd(Y, X), rd(mult(Y, ld(Y, X)), X))
% 0.19/0.48 = { by axiom 1 (f01) }
% 0.19/0.48 ld(rd(Y, X), rd(X, X))
% 0.19/0.48 = { by lemma 9 }
% 0.19/0.48 ld(rd(Y, X), ld(X, X))
% 0.19/0.48 = { by lemma 14 R->L }
% 0.19/0.48 rd(X, ld(ld(rd(Y, X), ld(X, X)), X))
% 0.19/0.48 = { by lemma 12 }
% 0.19/0.48 rd(X, mult(rd(Y, X), X))
% 0.19/0.48 = { by axiom 2 (f03) }
% 0.19/0.48 rd(X, Y)
% 0.19/0.48
% 0.19/0.48 Lemma 16: rd(X, ld(Y, ld(X, X))) = mult(X, Y).
% 0.19/0.48 Proof:
% 0.19/0.48 rd(X, ld(Y, ld(X, X)))
% 0.19/0.48 = { by lemma 15 R->L }
% 0.19/0.48 mult(X, rd(ld(X, X), ld(Y, ld(X, X))))
% 0.19/0.48 = { by lemma 14 }
% 0.19/0.48 mult(X, Y)
% 0.19/0.48
% 0.19/0.48 Lemma 17: mult(ld(X, X), Y) = Y.
% 0.19/0.48 Proof:
% 0.19/0.48 mult(ld(X, X), Y)
% 0.19/0.48 = { by lemma 16 R->L }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(ld(X, X), ld(X, X))))
% 0.19/0.48 = { by lemma 9 R->L }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(ld(X, X), rd(X, X))))
% 0.19/0.48 = { by axiom 1 (f01) R->L }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(ld(X, X), rd(mult(X, ld(X, X)), X))))
% 0.19/0.48 = { by axiom 3 (f02) R->L }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(ld(X, X), rd(mult(X, ld(mult(Z, X), mult(mult(Z, X), ld(X, X)))), X))))
% 0.19/0.48 = { by lemma 7 R->L }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(ld(X, X), rd(mult(X, ld(mult(Z, X), mult(mult(Z, mult(ld(X, X), X)), ld(X, X)))), X))))
% 0.19/0.48 = { by axiom 5 (f05) }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(ld(X, X), rd(mult(X, ld(mult(Z, X), mult(mult(Z, ld(X, X)), mult(X, ld(X, X))))), X))))
% 0.19/0.48 = { by axiom 1 (f01) }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(ld(X, X), rd(mult(X, ld(mult(Z, X), mult(mult(Z, ld(X, X)), X))), X))))
% 0.19/0.48 = { by lemma 10 }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(ld(X, X), rd(mult(X, mult(ld(X, ld(X, X)), X)), X))))
% 0.19/0.48 = { by lemma 13 }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(ld(X, X), mult(rd(X, X), mult(X, ld(X, ld(X, X)))))))
% 0.19/0.48 = { by axiom 1 (f01) }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(ld(X, X), mult(rd(X, X), ld(X, X)))))
% 0.19/0.48 = { by lemma 9 }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(ld(X, X), mult(ld(X, X), ld(X, X)))))
% 0.19/0.48 = { by axiom 3 (f02) }
% 0.19/0.48 rd(ld(X, X), ld(Y, ld(X, X)))
% 0.19/0.48 = { by axiom 3 (f02) R->L }
% 0.19/0.48 ld(X, mult(X, rd(ld(X, X), ld(Y, ld(X, X)))))
% 0.19/0.48 = { by lemma 15 }
% 0.19/0.48 ld(X, rd(X, ld(Y, ld(X, X))))
% 0.19/0.48 = { by lemma 16 }
% 0.19/0.48 ld(X, mult(X, Y))
% 0.19/0.48 = { by axiom 3 (f02) }
% 0.19/0.48 Y
% 0.19/0.48
% 0.19/0.48 Goal 1 (goals): tuple(mult(X, x1(X)), mult(x1_2(X), X)) = tuple(x1(X), x1_2(X)).
% 0.19/0.48 The goal is true when:
% 0.19/0.48 X = ld(X, X)
% 0.19/0.48
% 0.19/0.48 Proof:
% 0.19/0.48 tuple(mult(ld(X, X), x1(ld(X, X))), mult(x1_2(ld(X, X)), ld(X, X)))
% 0.19/0.48 = { by axiom 4 (f04) R->L }
% 0.19/0.48 tuple(mult(ld(X, X), x1(ld(X, X))), mult(x1_2(ld(X, X)), rd(mult(ld(X, X), Y), Y)))
% 0.19/0.48 = { by lemma 17 }
% 0.19/0.48 tuple(mult(ld(X, X), x1(ld(X, X))), mult(x1_2(ld(X, X)), rd(Y, Y)))
% 0.19/0.48 = { by lemma 17 R->L }
% 0.19/0.48 tuple(mult(ld(X, X), x1(ld(X, X))), mult(x1_2(ld(X, X)), rd(mult(ld(x1_2(ld(X, X)), x1_2(ld(X, X))), Y), Y)))
% 0.19/0.48 = { by axiom 4 (f04) }
% 0.19/0.48 tuple(mult(ld(X, X), x1(ld(X, X))), mult(x1_2(ld(X, X)), ld(x1_2(ld(X, X)), x1_2(ld(X, X)))))
% 0.19/0.48 = { by axiom 1 (f01) }
% 0.19/0.48 tuple(mult(ld(X, X), x1(ld(X, X))), x1_2(ld(X, X)))
% 0.19/0.48 = { by lemma 17 }
% 0.19/0.48 tuple(x1(ld(X, X)), x1_2(ld(X, X)))
% 0.19/0.48 % SZS output end Proof
% 0.19/0.48
% 0.19/0.48 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------