TSTP Solution File: BOO038-1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : BOO038-1 : TPTP v8.1.2. Released v2.5.0.
% 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 : n008.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 18:11:31 EDT 2023
% Result : Unsatisfiable 4.98s 1.02s
% Output : Proof 5.82s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : BOO038-1 : TPTP v8.1.2. Released v2.5.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34 % Computer : n008.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 : Sun Aug 27 08:19:32 EDT 2023
% 0.13/0.35 % CPUTime :
% 4.98/1.02 Command-line arguments: --no-flatten-goal
% 4.98/1.02
% 4.98/1.02 % SZS status Unsatisfiable
% 4.98/1.02
% 5.24/1.09 % SZS output start Proof
% 5.24/1.09 Take the following subset of the input axioms:
% 5.24/1.09 fof(dn1, axiom, ![A, B, C, D]: inverse(add(inverse(add(inverse(add(A, B)), C)), inverse(add(A, inverse(add(inverse(C), inverse(add(C, D))))))))=C).
% 5.24/1.09 fof(huntinton, negated_conjecture, add(b, a)!=add(a, b) | (add(add(a, b), c)!=add(a, add(b, c)) | add(inverse(add(inverse(a), b)), inverse(add(inverse(a), inverse(b))))!=a)).
% 5.24/1.09
% 5.24/1.09 Now clausify the problem and encode Horn clauses using encoding 3 of
% 5.24/1.09 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 5.24/1.09 We repeatedly replace C & s=t => u=v by the two clauses:
% 5.24/1.09 fresh(y, y, x1...xn) = u
% 5.24/1.09 C => fresh(s, t, x1...xn) = v
% 5.24/1.09 where fresh is a fresh function symbol and x1..xn are the free
% 5.24/1.09 variables of u and v.
% 5.24/1.09 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 5.24/1.09 input problem has no model of domain size 1).
% 5.24/1.09
% 5.24/1.09 The encoding turns the above axioms into the following unit equations and goals:
% 5.24/1.09
% 5.24/1.09 Axiom 1 (dn1): inverse(add(inverse(add(inverse(add(X, Y)), Z)), inverse(add(X, inverse(add(inverse(Z), inverse(add(Z, W)))))))) = Z.
% 5.24/1.09
% 5.24/1.09 Lemma 2: inverse(add(inverse(X), inverse(add(X, inverse(add(inverse(X), inverse(add(X, Y)))))))) = X.
% 5.24/1.09 Proof:
% 5.24/1.09 inverse(add(inverse(X), inverse(add(X, inverse(add(inverse(X), inverse(add(X, Y))))))))
% 5.24/1.09 = { by axiom 1 (dn1) R->L }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(inverse(add(inverse(add(inverse(add(inverse(inverse(X)), Z)), X)), inverse(add(inverse(inverse(X)), inverse(add(inverse(X), inverse(add(X, W)))))))), inverse(X))), inverse(add(inverse(add(inverse(add(inverse(inverse(X)), Z)), X)), inverse(add(inverse(inverse(X)), inverse(add(inverse(X), inverse(add(X, V)))))))))), inverse(add(X, inverse(add(inverse(X), inverse(add(X, Y))))))))
% 5.24/1.09 = { by axiom 1 (dn1) }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(inverse(add(inverse(add(inverse(add(inverse(inverse(X)), Z)), X)), inverse(add(inverse(inverse(X)), inverse(add(inverse(X), inverse(add(X, W)))))))), inverse(X))), X)), inverse(add(X, inverse(add(inverse(X), inverse(add(X, Y))))))))
% 5.24/1.09 = { by axiom 1 (dn1) }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(X, inverse(X))), X)), inverse(add(X, inverse(add(inverse(X), inverse(add(X, Y))))))))
% 5.24/1.09 = { by axiom 1 (dn1) }
% 5.24/1.09 X
% 5.24/1.09
% 5.24/1.09 Lemma 3: inverse(add(inverse(add(inverse(add(X, Y)), Z)), inverse(add(X, Z)))) = Z.
% 5.24/1.09 Proof:
% 5.24/1.09 inverse(add(inverse(add(inverse(add(X, Y)), Z)), inverse(add(X, Z))))
% 5.24/1.09 = { by lemma 2 R->L }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(X, Y)), Z)), inverse(add(X, inverse(add(inverse(Z), inverse(add(Z, inverse(add(inverse(Z), inverse(add(Z, W))))))))))))
% 5.24/1.09 = { by axiom 1 (dn1) }
% 5.24/1.09 Z
% 5.24/1.09
% 5.24/1.09 Lemma 4: inverse(add(inverse(add(X, Y)), inverse(add(inverse(X), Y)))) = Y.
% 5.24/1.09 Proof:
% 5.24/1.09 inverse(add(inverse(add(X, Y)), inverse(add(inverse(X), Y))))
% 5.24/1.09 = { by lemma 2 R->L }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(inverse(X), inverse(add(X, inverse(add(inverse(X), inverse(add(X, Z)))))))), Y)), inverse(add(inverse(X), Y))))
% 5.24/1.09 = { by lemma 3 }
% 5.24/1.09 Y
% 5.24/1.09
% 5.24/1.09 Lemma 5: inverse(add(X, inverse(add(Y, inverse(add(inverse(Y), X)))))) = inverse(add(inverse(Y), X)).
% 5.24/1.09 Proof:
% 5.24/1.09 inverse(add(X, inverse(add(Y, inverse(add(inverse(Y), X))))))
% 5.24/1.09 = { by lemma 4 R->L }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(Y, X)), inverse(add(inverse(Y), X)))), inverse(add(Y, inverse(add(inverse(Y), X))))))
% 5.24/1.09 = { by lemma 3 }
% 5.24/1.09 inverse(add(inverse(Y), X))
% 5.24/1.09
% 5.24/1.09 Lemma 6: inverse(add(inverse(X), inverse(X))) = X.
% 5.24/1.09 Proof:
% 5.24/1.09 inverse(add(inverse(X), inverse(X)))
% 5.24/1.09 = { by lemma 5 R->L }
% 5.24/1.09 inverse(add(inverse(X), inverse(add(X, inverse(add(inverse(X), inverse(X)))))))
% 5.24/1.09 = { by lemma 5 R->L }
% 5.24/1.09 inverse(add(inverse(X), inverse(add(X, inverse(add(inverse(X), inverse(add(X, inverse(add(inverse(X), inverse(X)))))))))))
% 5.24/1.09 = { by lemma 2 }
% 5.24/1.09 X
% 5.24/1.09
% 5.24/1.09 Lemma 7: inverse(add(inverse(X), inverse(add(X, X)))) = X.
% 5.24/1.09 Proof:
% 5.24/1.09 inverse(add(inverse(X), inverse(add(X, X))))
% 5.24/1.09 = { by lemma 2 R->L }
% 5.24/1.09 inverse(add(inverse(X), inverse(add(X, inverse(add(inverse(X), inverse(add(X, inverse(add(inverse(X), inverse(add(X, Y))))))))))))
% 5.24/1.09 = { by lemma 2 }
% 5.24/1.09 X
% 5.24/1.09
% 5.24/1.09 Lemma 8: inverse(add(inverse(add(X, X)), inverse(add(X, X)))) = X.
% 5.24/1.09 Proof:
% 5.24/1.09 inverse(add(inverse(add(X, X)), inverse(add(X, X))))
% 5.24/1.09 = { by lemma 4 R->L }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(X, inverse(add(X, X)))), inverse(add(inverse(X), inverse(add(X, X)))))), inverse(add(X, X))))
% 5.24/1.09 = { by lemma 7 }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(X, inverse(add(X, X)))), X)), inverse(add(X, X))))
% 5.24/1.09 = { by lemma 3 }
% 5.24/1.09 X
% 5.24/1.09
% 5.24/1.09 Lemma 9: add(X, X) = X.
% 5.24/1.09 Proof:
% 5.24/1.09 add(X, X)
% 5.24/1.09 = { by lemma 6 R->L }
% 5.24/1.09 inverse(add(inverse(add(X, X)), inverse(add(X, X))))
% 5.24/1.09 = { by lemma 8 }
% 5.24/1.09 X
% 5.24/1.09
% 5.24/1.09 Lemma 10: add(X, inverse(add(inverse(Y), inverse(add(Y, X))))) = add(Y, X).
% 5.24/1.09 Proof:
% 5.24/1.09 add(X, inverse(add(inverse(Y), inverse(add(Y, X)))))
% 5.24/1.09 = { by lemma 9 R->L }
% 5.24/1.09 add(X, inverse(add(add(inverse(Y), inverse(Y)), inverse(add(Y, X)))))
% 5.24/1.09 = { by lemma 9 R->L }
% 5.24/1.09 add(X, inverse(add(add(inverse(Y), inverse(add(Y, Y))), inverse(add(Y, X)))))
% 5.24/1.09 = { by lemma 7 R->L }
% 5.24/1.09 add(X, inverse(add(add(inverse(Y), inverse(add(Y, Y))), inverse(add(inverse(add(inverse(Y), inverse(add(Y, Y)))), X)))))
% 5.24/1.09 = { by lemma 6 R->L }
% 5.24/1.09 inverse(add(inverse(add(X, inverse(add(add(inverse(Y), inverse(add(Y, Y))), inverse(add(inverse(add(inverse(Y), inverse(add(Y, Y)))), X)))))), inverse(add(X, inverse(add(add(inverse(Y), inverse(add(Y, Y))), inverse(add(inverse(add(inverse(Y), inverse(add(Y, Y)))), X))))))))
% 5.24/1.09 = { by lemma 5 }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(inverse(Y), inverse(add(Y, Y)))), X)), inverse(add(X, inverse(add(add(inverse(Y), inverse(add(Y, Y))), inverse(add(inverse(add(inverse(Y), inverse(add(Y, Y)))), X))))))))
% 5.24/1.09 = { by lemma 5 }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(inverse(Y), inverse(add(Y, Y)))), X)), inverse(add(inverse(add(inverse(Y), inverse(add(Y, Y)))), X))))
% 5.24/1.09 = { by lemma 6 }
% 5.24/1.09 add(inverse(add(inverse(Y), inverse(add(Y, Y)))), X)
% 5.24/1.09 = { by lemma 7 }
% 5.24/1.09 add(Y, X)
% 5.24/1.09
% 5.24/1.09 Lemma 11: inverse(inverse(X)) = X.
% 5.24/1.09 Proof:
% 5.24/1.09 inverse(inverse(X))
% 5.24/1.09 = { by lemma 9 R->L }
% 5.24/1.09 inverse(inverse(add(X, X)))
% 5.24/1.09 = { by lemma 9 R->L }
% 5.24/1.09 inverse(add(inverse(add(X, X)), inverse(add(X, X))))
% 5.24/1.09 = { by lemma 8 }
% 5.24/1.09 X
% 5.24/1.09
% 5.24/1.09 Lemma 12: add(inverse(add(X, Y)), inverse(add(inverse(X), Y))) = inverse(Y).
% 5.24/1.09 Proof:
% 5.24/1.09 add(inverse(add(X, Y)), inverse(add(inverse(X), Y)))
% 5.24/1.09 = { by lemma 6 R->L }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(X, Y)), inverse(add(inverse(X), Y)))), inverse(add(inverse(add(X, Y)), inverse(add(inverse(X), Y))))))
% 5.24/1.09 = { by lemma 4 }
% 5.24/1.09 inverse(add(Y, inverse(add(inverse(add(X, Y)), inverse(add(inverse(X), Y))))))
% 5.24/1.09 = { by lemma 4 }
% 5.24/1.09 inverse(add(Y, Y))
% 5.24/1.09 = { by lemma 9 }
% 5.24/1.09 inverse(Y)
% 5.24/1.09
% 5.24/1.09 Lemma 13: add(inverse(add(inverse(X), Y)), inverse(add(X, Y))) = inverse(Y).
% 5.24/1.09 Proof:
% 5.24/1.09 add(inverse(add(inverse(X), Y)), inverse(add(X, Y)))
% 5.24/1.09 = { by lemma 9 R->L }
% 5.24/1.09 add(inverse(add(add(inverse(X), inverse(X)), Y)), inverse(add(X, Y)))
% 5.24/1.09 = { by lemma 9 R->L }
% 5.24/1.09 add(inverse(add(add(inverse(X), inverse(add(X, X))), Y)), inverse(add(X, Y)))
% 5.24/1.09 = { by lemma 7 R->L }
% 5.24/1.09 add(inverse(add(add(inverse(X), inverse(add(X, X))), Y)), inverse(add(inverse(add(inverse(X), inverse(add(X, X)))), Y)))
% 5.24/1.09 = { by lemma 12 }
% 5.24/1.09 inverse(Y)
% 5.24/1.09
% 5.24/1.09 Lemma 14: inverse(add(X, add(Y, X))) = inverse(add(Y, X)).
% 5.24/1.09 Proof:
% 5.24/1.09 inverse(add(X, add(Y, X)))
% 5.24/1.09 = { by lemma 3 R->L }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(inverse(add(Y, X)), X)), inverse(add(Y, X)))), add(Y, X)))
% 5.24/1.09 = { by lemma 6 R->L }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(inverse(add(Y, X)), X)), inverse(add(Y, X)))), inverse(add(inverse(add(Y, X)), inverse(add(Y, X))))))
% 5.24/1.09 = { by lemma 3 }
% 5.24/1.09 inverse(add(Y, X))
% 5.24/1.09
% 5.24/1.09 Lemma 15: inverse(add(X, inverse(add(Y, inverse(X))))) = inverse(X).
% 5.24/1.09 Proof:
% 5.24/1.09 inverse(add(X, inverse(add(Y, inverse(X)))))
% 5.24/1.09 = { by lemma 11 R->L }
% 5.24/1.09 inverse(add(inverse(inverse(X)), inverse(add(Y, inverse(X)))))
% 5.24/1.09 = { by lemma 13 R->L }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(inverse(Y), X)), inverse(add(Y, X)))), inverse(add(Y, inverse(X)))))
% 5.24/1.09 = { by lemma 6 R->L }
% 5.24/1.09 inverse(add(inverse(inverse(add(inverse(add(inverse(add(inverse(Y), X)), inverse(add(Y, X)))), inverse(add(inverse(add(inverse(Y), X)), inverse(add(Y, X))))))), inverse(add(Y, inverse(X)))))
% 5.24/1.09 = { by lemma 14 R->L }
% 5.24/1.09 inverse(add(inverse(inverse(add(inverse(add(inverse(add(inverse(Y), X)), inverse(add(Y, X)))), inverse(add(inverse(add(Y, X)), add(inverse(add(inverse(Y), X)), inverse(add(Y, X)))))))), inverse(add(Y, inverse(X)))))
% 5.24/1.09 = { by lemma 9 R->L }
% 5.24/1.09 inverse(add(inverse(inverse(add(inverse(add(inverse(add(inverse(Y), X)), inverse(add(Y, X)))), inverse(add(add(inverse(add(Y, X)), add(inverse(add(inverse(Y), X)), inverse(add(Y, X)))), add(inverse(add(Y, X)), add(inverse(add(inverse(Y), X)), inverse(add(Y, X))))))))), inverse(add(Y, inverse(X)))))
% 5.24/1.09 = { by lemma 14 R->L }
% 5.24/1.09 inverse(add(inverse(inverse(add(inverse(add(inverse(add(Y, X)), add(inverse(add(inverse(Y), X)), inverse(add(Y, X))))), inverse(add(add(inverse(add(Y, X)), add(inverse(add(inverse(Y), X)), inverse(add(Y, X)))), add(inverse(add(Y, X)), add(inverse(add(inverse(Y), X)), inverse(add(Y, X))))))))), inverse(add(Y, inverse(X)))))
% 5.24/1.09 = { by lemma 7 }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(Y, X)), add(inverse(add(inverse(Y), X)), inverse(add(Y, X))))), inverse(add(Y, inverse(X)))))
% 5.24/1.09 = { by lemma 13 }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(Y, X)), inverse(X))), inverse(add(Y, inverse(X)))))
% 5.24/1.09 = { by lemma 3 }
% 5.24/1.09 inverse(X)
% 5.24/1.09
% 5.24/1.09 Lemma 16: inverse(add(X, inverse(add(inverse(add(Y, Z)), inverse(add(Y, X)))))) = inverse(add(Y, X)).
% 5.24/1.09 Proof:
% 5.24/1.09 inverse(add(X, inverse(add(inverse(add(Y, Z)), inverse(add(Y, X))))))
% 5.24/1.09 = { by lemma 3 R->L }
% 5.24/1.09 inverse(add(inverse(add(inverse(add(inverse(add(Y, Z)), X)), inverse(add(Y, X)))), inverse(add(inverse(add(Y, Z)), inverse(add(Y, X))))))
% 5.24/1.09 = { by lemma 3 }
% 5.24/1.09 inverse(add(Y, X))
% 5.24/1.09
% 5.24/1.09 Lemma 17: add(X, inverse(add(inverse(add(Y, Z)), inverse(add(Y, X))))) = add(Y, X).
% 5.24/1.09 Proof:
% 5.24/1.09 add(X, inverse(add(inverse(add(Y, Z)), inverse(add(Y, X)))))
% 5.24/1.09 = { by lemma 7 R->L }
% 5.24/1.09 inverse(add(inverse(add(X, inverse(add(inverse(add(Y, Z)), inverse(add(Y, X)))))), inverse(add(add(X, inverse(add(inverse(add(Y, Z)), inverse(add(Y, X))))), add(X, inverse(add(inverse(add(Y, Z)), inverse(add(Y, X)))))))))
% 5.24/1.10 = { by lemma 16 }
% 5.24/1.10 inverse(add(inverse(add(Y, X)), inverse(add(add(X, inverse(add(inverse(add(Y, Z)), inverse(add(Y, X))))), add(X, inverse(add(inverse(add(Y, Z)), inverse(add(Y, X)))))))))
% 5.24/1.10 = { by lemma 9 }
% 5.24/1.10 inverse(add(inverse(add(Y, X)), inverse(add(X, inverse(add(inverse(add(Y, Z)), inverse(add(Y, X))))))))
% 5.24/1.10 = { by lemma 16 }
% 5.24/1.10 inverse(add(inverse(add(Y, X)), inverse(add(Y, X))))
% 5.24/1.10 = { by lemma 6 }
% 5.24/1.10 add(Y, X)
% 5.24/1.10
% 5.24/1.10 Lemma 18: add(X, inverse(add(inverse(X), inverse(add(X, Y))))) = inverse(add(inverse(X), inverse(add(X, Y)))).
% 5.24/1.10 Proof:
% 5.24/1.10 add(X, inverse(add(inverse(X), inverse(add(X, Y)))))
% 5.24/1.10 = { by lemma 17 R->L }
% 5.24/1.10 add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(inverse(add(X, Y)), inverse(add(X, inverse(add(inverse(X), inverse(add(X, Y)))))))))
% 5.24/1.10 = { by lemma 5 }
% 5.24/1.10 add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(inverse(X), inverse(add(X, Y)))))
% 5.24/1.10 = { by lemma 9 }
% 5.24/1.10 inverse(add(inverse(X), inverse(add(X, Y))))
% 5.24/1.10
% 5.24/1.10 Lemma 19: inverse(add(inverse(X), inverse(add(X, Y)))) = X.
% 5.24/1.10 Proof:
% 5.24/1.10 inverse(add(inverse(X), inverse(add(X, Y))))
% 5.24/1.10 = { by lemma 11 R->L }
% 5.24/1.10 inverse(inverse(inverse(add(inverse(X), inverse(add(X, Y))))))
% 5.24/1.10 = { by lemma 9 R->L }
% 5.24/1.10 inverse(inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(inverse(X), inverse(add(X, Y)))))))
% 5.24/1.10 = { by lemma 9 R->L }
% 5.24/1.10 inverse(inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y))))))))
% 5.24/1.10 = { by lemma 6 R->L }
% 5.24/1.10 inverse(inverse(inverse(add(inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y))))))), inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))))))))
% 5.24/1.10 = { by lemma 15 R->L }
% 5.24/1.10 inverse(inverse(inverse(add(inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y))))))), inverse(add(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))), inverse(add(inverse(X), inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))))))))))))
% 5.24/1.10 = { by lemma 9 R->L }
% 5.24/1.10 inverse(inverse(inverse(add(inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y))))))), inverse(add(add(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))), inverse(add(inverse(X), inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))))))), add(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))), inverse(add(inverse(X), inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y))))))))))))))))
% 5.24/1.10 = { by lemma 15 R->L }
% 5.24/1.10 inverse(inverse(inverse(add(inverse(add(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))), inverse(add(inverse(X), inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y))))))))))), inverse(add(add(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))), inverse(add(inverse(X), inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))))))), add(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))), inverse(add(inverse(X), inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y))))))))))))))))
% 5.24/1.10 = { by lemma 7 }
% 5.24/1.10 inverse(inverse(add(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))), inverse(add(inverse(X), inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y))))))))))))
% 5.24/1.10 = { by lemma 7 }
% 5.24/1.10 inverse(inverse(add(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(add(inverse(X), inverse(add(X, Y))), add(inverse(X), inverse(add(X, Y)))))), inverse(add(inverse(X), add(inverse(X), inverse(add(X, Y))))))))
% 5.24/1.10 = { by lemma 9 }
% 5.24/1.10 inverse(inverse(add(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(inverse(X), inverse(add(X, Y))))), inverse(add(inverse(X), add(inverse(X), inverse(add(X, Y))))))))
% 5.24/1.10 = { by lemma 9 }
% 5.24/1.10 inverse(inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(inverse(X), add(inverse(X), inverse(add(X, Y))))))))
% 5.24/1.10 = { by lemma 5 R->L }
% 5.24/1.10 inverse(inverse(add(inverse(add(inverse(X), add(inverse(X), inverse(add(X, Y))))), inverse(add(add(inverse(X), inverse(add(X, Y))), inverse(add(inverse(add(inverse(X), inverse(add(X, Y)))), inverse(add(inverse(X), add(inverse(X), inverse(add(X, Y))))))))))))
% 5.24/1.10 = { by lemma 16 }
% 5.24/1.10 inverse(inverse(add(inverse(add(inverse(X), add(inverse(X), inverse(add(X, Y))))), inverse(add(inverse(X), add(inverse(X), inverse(add(X, Y))))))))
% 5.24/1.10 = { by lemma 6 }
% 5.24/1.10 inverse(add(inverse(X), add(inverse(X), inverse(add(X, Y)))))
% 5.24/1.10 = { by lemma 11 R->L }
% 5.24/1.10 inverse(add(inverse(X), inverse(inverse(add(inverse(X), inverse(add(X, Y)))))))
% 5.24/1.10 = { by lemma 18 R->L }
% 5.24/1.10 inverse(add(inverse(X), inverse(add(X, inverse(add(inverse(X), inverse(add(X, Y))))))))
% 5.24/1.10 = { by lemma 2 }
% 5.24/1.10 X
% 5.24/1.10
% 5.24/1.10 Lemma 20: add(X, Y) = add(Y, X).
% 5.24/1.10 Proof:
% 5.24/1.10 add(X, Y)
% 5.24/1.10 = { by lemma 10 R->L }
% 5.24/1.10 add(Y, inverse(add(inverse(X), inverse(add(X, Y)))))
% 5.24/1.10 = { by lemma 19 }
% 5.24/1.10 add(Y, X)
% 5.24/1.10
% 5.24/1.10 Lemma 21: add(X, add(Y, inverse(add(Z, inverse(add(Y, X)))))) = add(Y, X).
% 5.24/1.10 Proof:
% 5.24/1.10 add(X, add(Y, inverse(add(Z, inverse(add(Y, X))))))
% 5.24/1.10 = { by lemma 11 R->L }
% 5.24/1.10 add(X, inverse(inverse(add(Y, inverse(add(Z, inverse(add(Y, X))))))))
% 5.24/1.10 = { by lemma 20 R->L }
% 5.24/1.10 add(X, inverse(inverse(add(Y, inverse(add(inverse(add(Y, X)), Z))))))
% 5.24/1.10 = { by lemma 9 R->L }
% 5.24/1.10 add(X, inverse(add(inverse(add(Y, inverse(add(inverse(add(Y, X)), Z)))), inverse(add(Y, inverse(add(inverse(add(Y, X)), Z)))))))
% 5.24/1.10 = { by lemma 16 R->L }
% 5.24/1.10 add(X, inverse(add(inverse(add(Y, inverse(add(inverse(add(Y, X)), Z)))), inverse(add(inverse(add(inverse(add(Y, X)), Z)), inverse(add(inverse(add(Y, X)), inverse(add(Y, inverse(add(inverse(add(Y, X)), Z)))))))))))
% 5.24/1.10 = { by lemma 17 }
% 5.24/1.10 add(X, inverse(add(inverse(add(Y, X)), inverse(add(Y, inverse(add(inverse(add(Y, X)), Z)))))))
% 5.24/1.10 = { by lemma 20 }
% 5.24/1.10 add(X, inverse(add(inverse(add(Y, X)), inverse(add(Y, inverse(add(Z, inverse(add(Y, X)))))))))
% 5.24/1.10 = { by lemma 20 R->L }
% 5.24/1.10 add(X, inverse(add(inverse(add(Y, X)), inverse(add(inverse(add(Z, inverse(add(Y, X)))), Y)))))
% 5.24/1.10 = { by lemma 20 R->L }
% 5.24/1.10 add(X, inverse(add(inverse(add(inverse(add(Z, inverse(add(Y, X)))), Y)), inverse(add(Y, X)))))
% 5.24/1.10 = { by lemma 16 R->L }
% 5.24/1.10 add(X, inverse(add(inverse(add(Y, inverse(add(inverse(add(inverse(add(Z, inverse(add(Y, X)))), W)), inverse(add(inverse(add(Z, inverse(add(Y, X)))), Y)))))), inverse(add(Y, X)))))
% 5.24/1.10 = { by lemma 17 }
% 5.24/1.10 add(Y, X)
% 5.24/1.10
% 5.24/1.10 Lemma 22: add(X, inverse(add(Y, inverse(add(X, inverse(add(Y, Z))))))) = add(X, inverse(add(Y, Z))).
% 5.24/1.10 Proof:
% 5.24/1.10 add(X, inverse(add(Y, inverse(add(X, inverse(add(Y, Z)))))))
% 5.24/1.10 = { by lemma 20 R->L }
% 5.24/1.10 add(X, inverse(add(Y, inverse(add(X, inverse(add(Z, Y)))))))
% 5.24/1.10 = { by lemma 20 R->L }
% 5.24/1.10 add(X, inverse(add(Y, inverse(add(inverse(add(Z, Y)), X)))))
% 5.24/1.10 = { by lemma 4 R->L }
% 5.24/1.10 add(X, inverse(add(inverse(add(inverse(add(Z, Y)), inverse(add(inverse(Z), Y)))), inverse(add(inverse(add(Z, Y)), X)))))
% 5.24/1.10 = { by lemma 17 }
% 5.24/1.10 add(inverse(add(Z, Y)), X)
% 5.24/1.10 = { by lemma 20 }
% 5.24/1.10 add(X, inverse(add(Z, Y)))
% 5.24/1.10 = { by lemma 20 }
% 5.24/1.10 add(X, inverse(add(Y, Z)))
% 5.24/1.10
% 5.24/1.10 Lemma 23: add(inverse(add(X, Y)), inverse(add(X, inverse(add(Z, Y))))) = inverse(X).
% 5.24/1.10 Proof:
% 5.24/1.10 add(inverse(add(X, Y)), inverse(add(X, inverse(add(Z, Y)))))
% 5.82/1.10 = { by lemma 20 R->L }
% 5.82/1.10 add(inverse(add(X, Y)), inverse(add(X, inverse(add(Y, Z)))))
% 5.82/1.10 = { by lemma 20 R->L }
% 5.82/1.10 add(inverse(add(X, Y)), inverse(add(inverse(add(Y, Z)), X)))
% 5.82/1.10 = { by lemma 20 R->L }
% 5.82/1.10 add(inverse(add(Y, X)), inverse(add(inverse(add(Y, Z)), X)))
% 5.82/1.10 = { by lemma 20 R->L }
% 5.82/1.10 add(inverse(add(inverse(add(Y, Z)), X)), inverse(add(Y, X)))
% 5.82/1.10 = { by lemma 6 R->L }
% 5.82/1.10 inverse(add(inverse(add(inverse(add(inverse(add(Y, Z)), X)), inverse(add(Y, X)))), inverse(add(inverse(add(inverse(add(Y, Z)), X)), inverse(add(Y, X))))))
% 5.82/1.10 = { by lemma 3 }
% 5.82/1.10 inverse(add(X, inverse(add(inverse(add(inverse(add(Y, Z)), X)), inverse(add(Y, X))))))
% 5.82/1.10 = { by lemma 3 }
% 5.82/1.10 inverse(add(X, X))
% 5.82/1.10 = { by lemma 9 }
% 5.82/1.10 inverse(X)
% 5.82/1.10
% 5.82/1.10 Lemma 24: add(X, inverse(add(Y, inverse(add(X, Y))))) = X.
% 5.82/1.10 Proof:
% 5.82/1.10 add(X, inverse(add(Y, inverse(add(X, Y)))))
% 5.82/1.10 = { by lemma 20 R->L }
% 5.82/1.10 add(X, inverse(add(inverse(add(X, Y)), Y)))
% 5.82/1.10 = { by lemma 22 R->L }
% 5.82/1.10 add(X, inverse(add(inverse(add(X, Y)), inverse(add(X, inverse(add(inverse(add(X, Y)), Y)))))))
% 5.82/1.10 = { by lemma 23 }
% 5.82/1.10 add(X, inverse(inverse(X)))
% 5.82/1.10 = { by lemma 11 }
% 5.82/1.10 add(X, X)
% 5.82/1.10 = { by lemma 9 }
% 5.82/1.10 X
% 5.82/1.10
% 5.82/1.10 Lemma 25: add(X, inverse(add(X, Y))) = add(X, inverse(Y)).
% 5.82/1.10 Proof:
% 5.82/1.10 add(X, inverse(add(X, Y)))
% 5.82/1.10 = { by lemma 20 R->L }
% 5.82/1.10 add(X, inverse(add(Y, X)))
% 5.82/1.10 = { by lemma 22 R->L }
% 5.82/1.10 add(X, inverse(add(Y, inverse(add(X, inverse(add(Y, X)))))))
% 5.82/1.10 = { by lemma 24 }
% 5.82/1.10 add(X, inverse(Y))
% 5.82/1.10
% 5.82/1.10 Lemma 26: add(inverse(X), add(X, Y)) = add(X, inverse(X)).
% 5.82/1.10 Proof:
% 5.82/1.10 add(inverse(X), add(X, Y))
% 5.82/1.10 = { by lemma 11 R->L }
% 5.82/1.10 add(inverse(X), inverse(inverse(add(X, Y))))
% 5.82/1.10 = { by lemma 25 R->L }
% 5.82/1.10 add(inverse(X), inverse(add(inverse(X), inverse(add(X, Y)))))
% 5.82/1.10 = { by lemma 18 R->L }
% 5.82/1.10 add(inverse(X), add(X, inverse(add(inverse(X), inverse(add(X, Y))))))
% 5.82/1.10 = { by lemma 11 R->L }
% 5.82/1.10 add(inverse(X), inverse(inverse(add(X, inverse(add(inverse(X), inverse(add(X, Y))))))))
% 5.82/1.11 = { by lemma 25 R->L }
% 5.82/1.11 add(inverse(X), inverse(add(inverse(X), inverse(add(X, inverse(add(inverse(X), inverse(add(X, Y)))))))))
% 5.82/1.11 = { by lemma 2 }
% 5.82/1.11 add(inverse(X), X)
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 add(X, inverse(X))
% 5.82/1.11
% 5.82/1.11 Lemma 27: add(inverse(add(X, inverse(X))), Y) = Y.
% 5.82/1.11 Proof:
% 5.82/1.11 add(inverse(add(X, inverse(X))), Y)
% 5.82/1.11 = { by lemma 20 R->L }
% 5.82/1.11 add(Y, inverse(add(X, inverse(X))))
% 5.82/1.11 = { by lemma 26 R->L }
% 5.82/1.11 add(Y, inverse(add(inverse(X), add(X, Y))))
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 add(Y, inverse(add(inverse(X), add(Y, X))))
% 5.82/1.11 = { by lemma 20 R->L }
% 5.82/1.11 add(Y, inverse(add(add(Y, X), inverse(X))))
% 5.82/1.11 = { by lemma 11 R->L }
% 5.82/1.11 add(Y, inverse(add(inverse(inverse(add(Y, X))), inverse(X))))
% 5.82/1.11 = { by lemma 12 R->L }
% 5.82/1.11 add(Y, inverse(add(inverse(inverse(add(Y, X))), add(inverse(add(Y, X)), inverse(add(inverse(Y), X))))))
% 5.82/1.11 = { by lemma 26 }
% 5.82/1.11 add(Y, inverse(add(inverse(add(Y, X)), inverse(inverse(add(Y, X))))))
% 5.82/1.11 = { by lemma 11 }
% 5.82/1.11 add(Y, inverse(add(inverse(add(Y, X)), add(Y, X))))
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 add(Y, inverse(add(add(Y, X), inverse(add(Y, X)))))
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 add(Y, inverse(add(add(Y, X), inverse(add(X, Y)))))
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 add(Y, inverse(add(add(X, Y), inverse(add(X, Y)))))
% 5.82/1.11 = { by lemma 14 R->L }
% 5.82/1.11 add(Y, inverse(add(add(X, Y), inverse(add(Y, add(X, Y))))))
% 5.82/1.11 = { by lemma 24 }
% 5.82/1.11 Y
% 5.82/1.11
% 5.82/1.11 Lemma 28: add(inverse(X), inverse(add(Y, add(X, Z)))) = inverse(X).
% 5.82/1.11 Proof:
% 5.82/1.11 add(inverse(X), inverse(add(Y, add(X, Z))))
% 5.82/1.11 = { by lemma 25 R->L }
% 5.82/1.11 add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z)))))
% 5.82/1.11 = { by lemma 11 R->L }
% 5.82/1.11 inverse(inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z)))))))
% 5.82/1.11 = { by lemma 27 R->L }
% 5.82/1.11 inverse(add(inverse(add(X, inverse(X))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 26 R->L }
% 5.82/1.11 inverse(add(inverse(add(inverse(X), add(X, Z))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 11 R->L }
% 5.82/1.11 inverse(add(inverse(inverse(inverse(add(inverse(X), add(X, Z))))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 23 R->L }
% 5.82/1.11 inverse(add(inverse(add(inverse(add(inverse(add(inverse(X), add(X, Z))), inverse(add(inverse(X), inverse(add(Y, add(X, Z))))))), inverse(add(inverse(add(inverse(X), add(X, Z))), inverse(add(inverse(add(inverse(X), add(X, Z))), inverse(add(inverse(X), inverse(add(Y, add(X, Z))))))))))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 23 }
% 5.82/1.11 inverse(add(inverse(add(inverse(inverse(inverse(X))), inverse(add(inverse(add(inverse(X), add(X, Z))), inverse(add(inverse(add(inverse(X), add(X, Z))), inverse(add(inverse(X), inverse(add(Y, add(X, Z))))))))))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 11 }
% 5.82/1.11 inverse(add(inverse(add(inverse(X), inverse(add(inverse(add(inverse(X), add(X, Z))), inverse(add(inverse(add(inverse(X), add(X, Z))), inverse(add(inverse(X), inverse(add(Y, add(X, Z))))))))))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 25 }
% 5.82/1.11 inverse(add(inverse(add(inverse(X), inverse(add(inverse(add(inverse(X), add(X, Z))), inverse(inverse(add(inverse(X), inverse(add(Y, add(X, Z)))))))))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 11 }
% 5.82/1.11 inverse(add(inverse(add(inverse(X), inverse(add(inverse(add(inverse(X), add(X, Z))), add(inverse(X), inverse(add(Y, add(X, Z)))))))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 inverse(add(inverse(add(inverse(X), inverse(add(inverse(add(inverse(X), add(X, Z))), add(inverse(X), inverse(add(add(X, Z), Y))))))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 26 }
% 5.82/1.11 inverse(add(inverse(add(inverse(X), inverse(add(inverse(add(X, inverse(X))), add(inverse(X), inverse(add(add(X, Z), Y))))))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 27 }
% 5.82/1.11 inverse(add(inverse(add(inverse(X), inverse(add(inverse(X), inverse(add(add(X, Z), Y)))))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 25 }
% 5.82/1.11 inverse(add(inverse(add(inverse(X), inverse(inverse(add(add(X, Z), Y))))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 11 }
% 5.82/1.11 inverse(add(inverse(add(inverse(X), add(add(X, Z), Y))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 inverse(add(inverse(add(inverse(X), add(Y, add(X, Z)))), inverse(add(inverse(X), inverse(add(inverse(X), add(Y, add(X, Z))))))))
% 5.82/1.11 = { by lemma 20 R->L }
% 5.82/1.11 inverse(add(inverse(add(inverse(X), add(Y, add(X, Z)))), inverse(add(inverse(add(inverse(X), add(Y, add(X, Z)))), inverse(X)))))
% 5.82/1.11 = { by lemma 20 R->L }
% 5.82/1.11 inverse(add(inverse(add(inverse(add(inverse(X), add(Y, add(X, Z)))), inverse(X))), inverse(add(inverse(X), add(Y, add(X, Z))))))
% 5.82/1.11 = { by lemma 19 R->L }
% 5.82/1.11 inverse(add(inverse(add(inverse(add(inverse(X), add(Y, add(X, Z)))), inverse(add(inverse(inverse(X)), inverse(add(inverse(X), add(Y, add(X, Z)))))))), inverse(add(inverse(X), add(Y, add(X, Z))))))
% 5.82/1.11 = { by lemma 3 R->L }
% 5.82/1.11 inverse(add(inverse(add(inverse(add(inverse(X), add(Y, add(X, Z)))), inverse(add(inverse(inverse(X)), inverse(add(inverse(X), add(Y, add(X, Z)))))))), inverse(add(inverse(add(inverse(add(inverse(inverse(X)), W)), inverse(add(inverse(X), add(Y, add(X, Z)))))), inverse(add(inverse(inverse(X)), inverse(add(inverse(X), add(Y, add(X, Z))))))))))
% 5.82/1.11 = { by lemma 4 R->L }
% 5.82/1.11 inverse(add(inverse(add(inverse(add(inverse(add(inverse(add(inverse(inverse(X)), W)), inverse(add(inverse(X), add(Y, add(X, Z)))))), inverse(add(inverse(inverse(add(inverse(inverse(X)), W))), inverse(add(inverse(X), add(Y, add(X, Z)))))))), inverse(add(inverse(inverse(X)), inverse(add(inverse(X), add(Y, add(X, Z)))))))), inverse(add(inverse(add(inverse(add(inverse(inverse(X)), W)), inverse(add(inverse(X), add(Y, add(X, Z)))))), inverse(add(inverse(inverse(X)), inverse(add(inverse(X), add(Y, add(X, Z))))))))))
% 5.82/1.11 = { by lemma 3 }
% 5.82/1.11 inverse(add(inverse(inverse(X)), inverse(add(inverse(X), add(Y, add(X, Z))))))
% 5.82/1.11 = { by lemma 19 }
% 5.82/1.11 inverse(X)
% 5.82/1.11
% 5.82/1.11 Lemma 29: add(Y, add(Z, X)) = add(X, add(Y, Z)).
% 5.82/1.11 Proof:
% 5.82/1.11 add(Y, add(Z, X))
% 5.82/1.11 = { by lemma 21 R->L }
% 5.82/1.11 add(add(Z, X), add(Y, inverse(add(inverse(Z), inverse(add(Y, add(Z, X)))))))
% 5.82/1.11 = { by lemma 28 }
% 5.82/1.11 add(add(Z, X), add(Y, inverse(inverse(Z))))
% 5.82/1.11 = { by lemma 11 }
% 5.82/1.11 add(add(Z, X), add(Y, Z))
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 add(add(Y, Z), add(Z, X))
% 5.82/1.11 = { by lemma 10 R->L }
% 5.82/1.11 add(add(Z, inverse(add(inverse(Y), inverse(add(Y, Z))))), add(Z, X))
% 5.82/1.11 = { by lemma 20 R->L }
% 5.82/1.11 add(add(Z, inverse(add(inverse(Y), inverse(add(Y, Z))))), add(X, Z))
% 5.82/1.11 = { by lemma 20 R->L }
% 5.82/1.11 add(add(inverse(add(inverse(Y), inverse(add(Y, Z)))), Z), add(X, Z))
% 5.82/1.11 = { by lemma 11 R->L }
% 5.82/1.11 add(add(inverse(add(inverse(Y), inverse(add(Y, Z)))), Z), add(X, inverse(inverse(Z))))
% 5.82/1.11 = { by lemma 28 R->L }
% 5.82/1.11 add(add(inverse(add(inverse(Y), inverse(add(Y, Z)))), Z), add(X, inverse(add(inverse(Z), inverse(add(X, add(Z, inverse(add(inverse(Y), inverse(add(Y, Z)))))))))))
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 add(add(inverse(add(inverse(Y), inverse(add(Y, Z)))), Z), add(X, inverse(add(inverse(Z), inverse(add(X, add(inverse(add(inverse(Y), inverse(add(Y, Z)))), Z)))))))
% 5.82/1.11 = { by lemma 21 }
% 5.82/1.11 add(X, add(inverse(add(inverse(Y), inverse(add(Y, Z)))), Z))
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 add(X, add(Z, inverse(add(inverse(Y), inverse(add(Y, Z))))))
% 5.82/1.11 = { by lemma 10 }
% 5.82/1.11 add(X, add(Y, Z))
% 5.82/1.11
% 5.82/1.11 Goal 1 (huntinton): tuple(add(inverse(add(inverse(a), b)), inverse(add(inverse(a), inverse(b)))), add(add(a, b), c), add(b, a)) = tuple(a, add(a, add(b, c)), add(a, b)).
% 5.82/1.11 Proof:
% 5.82/1.11 tuple(add(inverse(add(inverse(a), b)), inverse(add(inverse(a), inverse(b)))), add(add(a, b), c), add(b, a))
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 tuple(add(inverse(add(b, inverse(a))), inverse(add(inverse(a), inverse(b)))), add(add(a, b), c), add(b, a))
% 5.82/1.11 = { by lemma 20 }
% 5.82/1.11 tuple(add(inverse(add(b, inverse(a))), inverse(add(inverse(a), inverse(b)))), add(c, add(a, b)), add(b, a))
% 5.82/1.12 = { by lemma 20 }
% 5.82/1.12 tuple(add(inverse(add(b, inverse(a))), inverse(add(inverse(b), inverse(a)))), add(c, add(a, b)), add(b, a))
% 5.82/1.12 = { by lemma 12 }
% 5.82/1.12 tuple(inverse(inverse(a)), add(c, add(a, b)), add(b, a))
% 5.82/1.12 = { by lemma 11 }
% 5.82/1.12 tuple(a, add(c, add(a, b)), add(b, a))
% 5.82/1.12 = { by lemma 29 }
% 5.82/1.12 tuple(a, add(b, add(c, a)), add(b, a))
% 5.82/1.12 = { by lemma 20 R->L }
% 5.82/1.12 tuple(a, add(b, add(a, c)), add(b, a))
% 5.82/1.12 = { by lemma 29 R->L }
% 5.82/1.12 tuple(a, add(a, add(c, b)), add(b, a))
% 5.82/1.12 = { by lemma 20 R->L }
% 5.82/1.12 tuple(a, add(a, add(c, b)), add(a, b))
% 5.82/1.12 = { by lemma 20 R->L }
% 5.82/1.12 tuple(a, add(a, add(b, c)), add(a, b))
% 5.82/1.12 % SZS output end Proof
% 5.82/1.12
% 5.82/1.12 RESULT: Unsatisfiable (the axioms are contradictory).
%------------------------------------------------------------------------------