TSTP Solution File: GRP715+1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : GRP715+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 : n014.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 01:19:52 EDT 2023
% Result : Theorem 0.20s 0.44s
% Output : Proof 0.20s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : GRP715+1 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35 % Computer : n014.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Mon Aug 28 20:58:15 EDT 2023
% 0.13/0.35 % CPUTime :
% 0.20/0.44 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.20/0.44
% 0.20/0.44 % SZS status Theorem
% 0.20/0.44
% 0.20/0.45 % SZS output start Proof
% 0.20/0.45 Take the following subset of the input axioms:
% 0.20/0.45 fof(f06, axiom, ![C, B, A]: mult(mult(mult(A, B), C), B)=mult(A, mult(mult(B, C), B))).
% 0.20/0.45 fof(f07, axiom, ![A2, B2]: mult(A2, mult(B2, B2))=mult(mult(A2, B2), B2)).
% 0.20/0.45 fof(f08, axiom, ![A2]: mult(A2, unit)=A2).
% 0.20/0.45 fof(f09, axiom, ![A2]: mult(unit, A2)=A2).
% 0.20/0.45 fof(f10, axiom, mult(op_a, op_b)=unit).
% 0.20/0.45 fof(f11, axiom, mult(op_b, op_a)=unit).
% 0.20/0.45 fof(goals, conjecture, ![X0]: (mult(mult(X0, op_a), op_b)=X0 & mult(mult(X0, op_b), op_a)=X0)).
% 0.20/0.45
% 0.20/0.45 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.45 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.45 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.45 fresh(y, y, x1...xn) = u
% 0.20/0.45 C => fresh(s, t, x1...xn) = v
% 0.20/0.45 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.45 variables of u and v.
% 0.20/0.45 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.45 input problem has no model of domain size 1).
% 0.20/0.45
% 0.20/0.45 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.45
% 0.20/0.45 Axiom 1 (f08): mult(X, unit) = X.
% 0.20/0.45 Axiom 2 (f10): mult(op_a, op_b) = unit.
% 0.20/0.45 Axiom 3 (f11): mult(op_b, op_a) = unit.
% 0.20/0.45 Axiom 4 (f09): mult(unit, X) = X.
% 0.20/0.45 Axiom 5 (f07): mult(X, mult(Y, Y)) = mult(mult(X, Y), Y).
% 0.20/0.45 Axiom 6 (f06): mult(mult(mult(X, Y), Z), Y) = mult(X, mult(mult(Y, Z), Y)).
% 0.20/0.46
% 0.20/0.46 Lemma 7: mult(op_a, mult(op_b, op_b)) = op_b.
% 0.20/0.46 Proof:
% 0.20/0.46 mult(op_a, mult(op_b, op_b))
% 0.20/0.46 = { by axiom 5 (f07) }
% 0.20/0.46 mult(mult(op_a, op_b), op_b)
% 0.20/0.46 = { by axiom 2 (f10) }
% 0.20/0.46 mult(unit, op_b)
% 0.20/0.46 = { by axiom 4 (f09) }
% 0.20/0.46 op_b
% 0.20/0.46
% 0.20/0.46 Lemma 8: mult(op_b, mult(op_a, op_a)) = op_a.
% 0.20/0.46 Proof:
% 0.20/0.46 mult(op_b, mult(op_a, op_a))
% 0.20/0.46 = { by axiom 5 (f07) }
% 0.20/0.46 mult(mult(op_b, op_a), op_a)
% 0.20/0.46 = { by axiom 3 (f11) }
% 0.20/0.46 mult(unit, op_a)
% 0.20/0.46 = { by axiom 4 (f09) }
% 0.20/0.46 op_a
% 0.20/0.46
% 0.20/0.46 Lemma 9: mult(mult(mult(mult(X, op_a), op_b), op_b), op_a) = X.
% 0.20/0.46 Proof:
% 0.20/0.46 mult(mult(mult(mult(X, op_a), op_b), op_b), op_a)
% 0.20/0.46 = { by axiom 5 (f07) R->L }
% 0.20/0.46 mult(mult(mult(X, op_a), mult(op_b, op_b)), op_a)
% 0.20/0.46 = { by axiom 6 (f06) }
% 0.20/0.46 mult(X, mult(mult(op_a, mult(op_b, op_b)), op_a))
% 0.20/0.46 = { by lemma 7 }
% 0.20/0.46 mult(X, mult(op_b, op_a))
% 0.20/0.46 = { by axiom 3 (f11) }
% 0.20/0.46 mult(X, unit)
% 0.20/0.46 = { by axiom 1 (f08) }
% 0.20/0.46 X
% 0.20/0.46
% 0.20/0.46 Lemma 10: mult(mult(mult(mult(X, op_b), op_a), op_a), op_b) = X.
% 0.20/0.46 Proof:
% 0.20/0.46 mult(mult(mult(mult(X, op_b), op_a), op_a), op_b)
% 0.20/0.46 = { by axiom 5 (f07) R->L }
% 0.20/0.46 mult(mult(mult(X, op_b), mult(op_a, op_a)), op_b)
% 0.20/0.46 = { by axiom 6 (f06) }
% 0.20/0.46 mult(X, mult(mult(op_b, mult(op_a, op_a)), op_b))
% 0.20/0.46 = { by lemma 8 }
% 0.20/0.46 mult(X, mult(op_a, op_b))
% 0.20/0.46 = { by axiom 2 (f10) }
% 0.20/0.46 mult(X, unit)
% 0.20/0.46 = { by axiom 1 (f08) }
% 0.20/0.46 X
% 0.20/0.46
% 0.20/0.46 Goal 1 (goals): tuple(mult(mult(x0_2, op_a), op_b), mult(mult(x0, op_b), op_a)) = tuple(x0_2, x0).
% 0.20/0.46 Proof:
% 0.20/0.46 tuple(mult(mult(x0_2, op_a), op_b), mult(mult(x0, op_b), op_a))
% 0.20/0.46 = { by lemma 10 R->L }
% 0.20/0.46 tuple(mult(mult(x0_2, op_a), op_b), mult(mult(mult(mult(mult(mult(x0, op_b), op_a), op_a), op_b), op_b), op_a))
% 0.20/0.46 = { by axiom 5 (f07) R->L }
% 0.20/0.46 tuple(mult(mult(x0_2, op_a), op_b), mult(mult(mult(mult(mult(x0, op_b), op_a), op_a), mult(op_b, op_b)), op_a))
% 0.20/0.46 = { by axiom 4 (f09) R->L }
% 0.20/0.46 tuple(mult(mult(x0_2, op_a), op_b), mult(mult(mult(mult(mult(x0, op_b), op_a), mult(unit, op_a)), mult(op_b, op_b)), op_a))
% 0.20/0.46 = { by axiom 2 (f10) R->L }
% 0.20/0.46 tuple(mult(mult(x0_2, op_a), op_b), mult(mult(mult(mult(mult(x0, op_b), op_a), mult(mult(op_a, op_b), op_a)), mult(op_b, op_b)), op_a))
% 0.20/0.46 = { by axiom 6 (f06) R->L }
% 0.20/0.46 tuple(mult(mult(x0_2, op_a), op_b), mult(mult(mult(mult(mult(mult(mult(x0, op_b), op_a), op_a), op_b), op_a), mult(op_b, op_b)), op_a))
% 0.20/0.46 = { by lemma 10 }
% 0.20/0.46 tuple(mult(mult(x0_2, op_a), op_b), mult(mult(mult(x0, op_a), mult(op_b, op_b)), op_a))
% 0.20/0.46 = { by axiom 6 (f06) }
% 0.20/0.46 tuple(mult(mult(x0_2, op_a), op_b), mult(x0, mult(mult(op_a, mult(op_b, op_b)), op_a)))
% 0.20/0.46 = { by lemma 7 }
% 0.20/0.46 tuple(mult(mult(x0_2, op_a), op_b), mult(x0, mult(op_b, op_a)))
% 0.20/0.46 = { by axiom 3 (f11) }
% 0.20/0.46 tuple(mult(mult(x0_2, op_a), op_b), mult(x0, unit))
% 0.20/0.46 = { by axiom 1 (f08) }
% 0.20/0.46 tuple(mult(mult(x0_2, op_a), op_b), x0)
% 0.20/0.46 = { by lemma 9 R->L }
% 0.20/0.46 tuple(mult(mult(mult(mult(mult(mult(x0_2, op_a), op_b), op_b), op_a), op_a), op_b), x0)
% 0.20/0.46 = { by axiom 5 (f07) R->L }
% 0.20/0.46 tuple(mult(mult(mult(mult(mult(x0_2, op_a), op_b), op_b), mult(op_a, op_a)), op_b), x0)
% 0.20/0.46 = { by axiom 4 (f09) R->L }
% 0.20/0.46 tuple(mult(mult(mult(mult(mult(x0_2, op_a), op_b), mult(unit, op_b)), mult(op_a, op_a)), op_b), x0)
% 0.20/0.46 = { by axiom 3 (f11) R->L }
% 0.20/0.46 tuple(mult(mult(mult(mult(mult(x0_2, op_a), op_b), mult(mult(op_b, op_a), op_b)), mult(op_a, op_a)), op_b), x0)
% 0.20/0.46 = { by axiom 6 (f06) R->L }
% 0.20/0.46 tuple(mult(mult(mult(mult(mult(mult(mult(x0_2, op_a), op_b), op_b), op_a), op_b), mult(op_a, op_a)), op_b), x0)
% 0.20/0.46 = { by lemma 9 }
% 0.20/0.46 tuple(mult(mult(mult(x0_2, op_b), mult(op_a, op_a)), op_b), x0)
% 0.20/0.46 = { by axiom 6 (f06) }
% 0.20/0.46 tuple(mult(x0_2, mult(mult(op_b, mult(op_a, op_a)), op_b)), x0)
% 0.20/0.46 = { by lemma 8 }
% 0.20/0.46 tuple(mult(x0_2, mult(op_a, op_b)), x0)
% 0.20/0.46 = { by axiom 2 (f10) }
% 0.20/0.46 tuple(mult(x0_2, unit), x0)
% 0.20/0.46 = { by axiom 1 (f08) }
% 0.20/0.46 tuple(x0_2, x0)
% 0.20/0.46 % SZS output end Proof
% 0.20/0.46
% 0.20/0.46 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------