TSTP Solution File: GRP657+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : GRP657+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 : n009.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 01:19:30 EDT 2023
% Result : Theorem 0.19s 0.37s
% Output : Proof 0.19s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12 % Problem : GRP657+1 : TPTP v8.1.2. Released v4.0.0.
% 0.12/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.33 % Computer : n009.cluster.edu
% 0.13/0.33 % Model : x86_64 x86_64
% 0.13/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33 % Memory : 8042.1875MB
% 0.13/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.33 % CPULimit : 300
% 0.13/0.33 % WCLimit : 300
% 0.13/0.33 % DateTime : Mon Aug 28 22:01:05 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.37 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.19/0.37
% 0.19/0.37 % SZS status Theorem
% 0.19/0.37
% 0.19/0.38 % SZS output start Proof
% 0.19/0.38 Take the following subset of the input axioms:
% 0.19/0.38 fof(f01, axiom, ![B, A]: mult(A, ld(A, B))=B).
% 0.19/0.38 fof(f02, axiom, ![B2, A2]: ld(A2, mult(A2, B2))=B2).
% 0.19/0.38 fof(f04, axiom, ![B2, A2]: rd(mult(A2, B2), B2)=A2).
% 0.19/0.38 fof(f05, axiom, ![C, B2, A2]: mult(mult(A2, B2), mult(C, A2))=mult(A2, mult(mult(B2, C), A2))).
% 0.19/0.38 fof(goals, conjecture, ?[X0]: ![X1]: (mult(X1, X0)=X1 & mult(X0, X1)=X1)).
% 0.19/0.38
% 0.19/0.38 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.38 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.38 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.38 fresh(y, y, x1...xn) = u
% 0.19/0.38 C => fresh(s, t, x1...xn) = v
% 0.19/0.38 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.38 variables of u and v.
% 0.19/0.38 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.38 input problem has no model of domain size 1).
% 0.19/0.38
% 0.19/0.38 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.38
% 0.19/0.38 Axiom 1 (f01): mult(X, ld(X, Y)) = Y.
% 0.19/0.38 Axiom 2 (f02): ld(X, mult(X, Y)) = Y.
% 0.19/0.38 Axiom 3 (f04): rd(mult(X, Y), Y) = X.
% 0.19/0.38 Axiom 4 (f05): mult(mult(X, Y), mult(Z, X)) = mult(X, mult(mult(Y, Z), X)).
% 0.19/0.38
% 0.19/0.38 Lemma 5: mult(ld(X, X), Y) = Y.
% 0.19/0.38 Proof:
% 0.19/0.38 mult(ld(X, X), Y)
% 0.19/0.38 = { by axiom 3 (f04) R->L }
% 0.19/0.38 rd(mult(mult(ld(X, X), Y), X), X)
% 0.19/0.38 = { by axiom 2 (f02) R->L }
% 0.19/0.38 rd(ld(X, mult(X, mult(mult(ld(X, X), Y), X))), X)
% 0.19/0.38 = { by axiom 4 (f05) R->L }
% 0.19/0.38 rd(ld(X, mult(mult(X, ld(X, X)), mult(Y, X))), X)
% 0.19/0.38 = { by axiom 1 (f01) }
% 0.19/0.38 rd(ld(X, mult(X, mult(Y, X))), X)
% 0.19/0.38 = { by axiom 2 (f02) }
% 0.19/0.38 rd(mult(Y, X), X)
% 0.19/0.38 = { by axiom 3 (f04) }
% 0.19/0.38 Y
% 0.19/0.38
% 0.19/0.38 Goal 1 (goals): tuple(mult(X, x1(X)), mult(x1_2(X), X)) = tuple(x1(X), x1_2(X)).
% 0.19/0.38 The goal is true when:
% 0.19/0.38 X = ld(X, X)
% 0.19/0.38
% 0.19/0.38 Proof:
% 0.19/0.38 tuple(mult(ld(X, X), x1(ld(X, X))), mult(x1_2(ld(X, X)), ld(X, X)))
% 0.19/0.38 = { by axiom 3 (f04) R->L }
% 0.19/0.38 tuple(mult(ld(X, X), x1(ld(X, X))), mult(x1_2(ld(X, X)), rd(mult(ld(X, X), Y), Y)))
% 0.19/0.38 = { by lemma 5 }
% 0.19/0.38 tuple(mult(ld(X, X), x1(ld(X, X))), mult(x1_2(ld(X, X)), rd(Y, Y)))
% 0.19/0.38 = { by lemma 5 R->L }
% 0.19/0.38 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)))
% 0.19/0.38 = { by axiom 3 (f04) }
% 0.19/0.38 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)))))
% 0.19/0.38 = { by axiom 1 (f01) }
% 0.19/0.38 tuple(mult(ld(X, X), x1(ld(X, X))), x1_2(ld(X, X)))
% 0.19/0.38 = { by lemma 5 }
% 0.19/0.38 tuple(x1(ld(X, X)), x1_2(ld(X, X)))
% 0.19/0.38 % SZS output end Proof
% 0.19/0.38
% 0.19/0.38 RESULT: Theorem (the conjecture is true).
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