0.03/0.12	% Problem    : theBenchmark.p : TPTP v0.0.0. Released v0.0.0.
0.03/0.12	% Command    : twee %s --tstp --casc --quiet --explain-encoding --conditional-encoding if --smaller --drop-non-horn
0.12/0.34	% Computer   : n023.cluster.edu
0.12/0.34	% Model      : x86_64 x86_64
0.12/0.34	% CPU        : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
0.12/0.34	% Memory     : 8042.1875MB
0.12/0.34	% OS         : Linux 3.10.0-693.el7.x86_64
0.12/0.34	% CPULimit   : 180
0.12/0.34	% DateTime   : Thu Aug 29 13:07:07 EDT 2019
0.12/0.34	% CPUTime    : 
0.70/0.88	% SZS status Unsatisfiable
0.70/0.88	
0.70/0.88	% SZS output start Proof
0.70/0.88	Take the following subset of the input axioms:
0.70/0.88	  fof(c01, axiom, ![A]: mult(unit, A)=A).
0.70/0.88	  fof(c02, axiom, ![A]: mult(A, unit)=A).
0.70/0.88	  fof(c03, axiom, ![A]: unit=mult(A, i(A))).
0.70/0.88	  fof(c04, axiom, ![A]: mult(i(A), A)=unit).
0.70/0.88	  fof(c05, axiom, ![A, B]: mult(i(A), i(B))=i(mult(A, B))).
0.70/0.88	  fof(c06, axiom, ![A, B]: mult(i(A), mult(A, B))=B).
0.70/0.88	  fof(c07, axiom, ![A, B]: rd(mult(A, B), B)=A).
0.70/0.88	  fof(c08, axiom, ![A, B]: A=mult(rd(A, B), B)).
0.70/0.88	  fof(c09, axiom, ![A, B, C]: mult(A, mult(B, mult(A, C)))=mult(mult(A, mult(B, A)), C)).
0.70/0.88	  fof(c10, axiom, ![A, B, C]: mult(mult(A, B), C)=mult(mult(A, mult(B, C)), asoc(A, B, C))).
0.70/0.88	  fof(c11, axiom, ![A, B]: mult(A, B)=mult(mult(B, A), op_k(A, B))).
0.70/0.88	  fof(c12, axiom, ![A, B, C]: op_l(A, B, C)=mult(i(mult(C, B)), mult(C, mult(B, A)))).
0.70/0.88	  fof(c22, axiom, ![A, B, C, D, E]: asoc(A, B, asoc(C, D, E))=unit).
0.70/0.88	  fof(goals, negated_conjecture, asoc(a, b, op_k(c, d))!=unit).
0.70/0.88	
0.70/0.88	Now clausify the problem and encode Horn clauses using encoding 3 of
0.70/0.88	http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
0.70/0.88	We repeatedly replace C & s=t => u=v by the two clauses:
0.70/0.88	  fresh(y, y, x1...xn) = u
0.70/0.88	  C => fresh(s, t, x1...xn) = v
0.70/0.88	where fresh is a fresh function symbol and x1..xn are the free
0.70/0.88	variables of u and v.
0.70/0.88	A predicate p(X) is encoded as p(X)=true (this is sound, because the
0.70/0.88	input problem has no model of domain size 1).
0.70/0.88	
0.70/0.88	The encoding turns the above axioms into the following unit equations and goals:
0.70/0.88	
0.70/0.88	Axiom 1 (c06): mult(i(X), mult(X, Y)) = Y.
0.70/0.88	Axiom 2 (c03): unit = mult(X, i(X)).
0.70/0.88	Axiom 3 (c02): mult(X, unit) = X.
0.70/0.88	Axiom 4 (c01): mult(unit, X) = X.
0.70/0.88	Axiom 5 (c11): mult(X, Y) = mult(mult(Y, X), op_k(X, Y)).
0.70/0.88	Axiom 6 (c12): op_l(X, Y, Z) = mult(i(mult(Z, Y)), mult(Z, mult(Y, X))).
0.70/0.88	Axiom 7 (c05): mult(i(X), i(Y)) = i(mult(X, Y)).
0.70/0.88	Axiom 8 (c22): asoc(X, Y, asoc(Z, W, V)) = unit.
0.70/0.88	Axiom 9 (c08): X = mult(rd(X, Y), Y).
0.70/0.88	Axiom 10 (c07): rd(mult(X, Y), Y) = X.
0.70/0.88	Axiom 11 (c09): mult(X, mult(Y, mult(X, Z))) = mult(mult(X, mult(Y, X)), Z).
0.70/0.88	Axiom 12 (c04): mult(i(X), X) = unit.
0.70/0.88	Axiom 13 (c10): mult(mult(X, Y), Z) = mult(mult(X, mult(Y, Z)), asoc(X, Y, Z)).
0.70/0.88	
0.70/0.88	Lemma 14: mult(i(mult(Y, X)), Y) = op_l(i(X), X, Y).
0.70/0.88	Proof:
0.70/0.88	  mult(i(mult(Y, X)), Y)
0.70/0.88	= { by axiom 3 (c02) }
0.70/0.88	  mult(i(mult(Y, X)), mult(Y, unit))
0.70/0.88	= { by axiom 2 (c03) }
0.70/0.88	  mult(i(mult(Y, X)), mult(Y, mult(X, i(X))))
0.70/0.88	= { by axiom 6 (c12) }
0.70/0.88	  op_l(i(X), X, Y)
0.70/0.88	
0.70/0.88	Lemma 15: i(i(X)) = X.
0.70/0.88	Proof:
0.70/0.88	  i(i(X))
0.70/0.88	= { by axiom 3 (c02) }
0.70/0.88	  mult(i(i(X)), unit)
0.70/0.88	= { by axiom 12 (c04) }
0.70/0.88	  mult(i(i(X)), mult(i(X), X))
0.70/0.88	= { by axiom 1 (c06) }
0.70/0.88	  X
0.70/0.88	
0.70/0.88	Lemma 16: i(mult(X, i(Y))) = mult(i(X), Y).
0.70/0.88	Proof:
0.70/0.88	  i(mult(X, i(Y)))
0.70/0.88	= { by axiom 7 (c05) }
0.70/0.88	  mult(i(X), i(i(Y)))
0.70/0.88	= { by lemma 15 }
0.70/0.88	  mult(i(X), Y)
0.70/0.88	
0.70/0.88	Lemma 17: mult(i(mult(Y, X)), mult(X, Y)) = op_k(X, Y).
0.70/0.88	Proof:
0.70/0.88	  mult(i(mult(Y, X)), mult(X, Y))
0.70/0.88	= { by axiom 5 (c11) }
0.70/0.89	  mult(i(mult(Y, X)), mult(mult(Y, X), op_k(X, Y)))
0.70/0.89	= { by axiom 1 (c06) }
0.70/0.89	  op_k(X, Y)
0.70/0.89	
0.70/0.89	Lemma 18: mult(mult(Y, X), i(mult(X, Y))) = op_k(i(X), i(Y)).
0.70/0.89	Proof:
0.70/0.89	  mult(mult(Y, X), i(mult(X, Y)))
0.70/0.89	= { by lemma 15 }
0.70/0.89	  mult(i(i(mult(Y, X))), i(mult(X, Y)))
0.70/0.89	= { by axiom 7 (c05) }
0.70/0.89	  mult(i(mult(i(Y), i(X))), i(mult(X, Y)))
0.70/0.89	= { by axiom 7 (c05) }
0.70/0.89	  mult(i(mult(i(Y), i(X))), mult(i(X), i(Y)))
0.70/0.89	= { by lemma 17 }
0.70/0.89	  op_k(i(X), i(Y))
0.70/0.89	
0.70/0.89	Lemma 19: i(op_k(i(X), i(Y))) = op_k(X, Y).
0.70/0.89	Proof:
0.70/0.89	  i(op_k(i(X), i(Y)))
0.70/0.89	= { by lemma 18 }
0.70/0.89	  i(mult(mult(Y, X), i(mult(X, Y))))
0.70/0.89	= { by lemma 16 }
0.70/0.89	  mult(i(mult(Y, X)), mult(X, Y))
0.70/0.89	= { by lemma 17 }
0.70/0.89	  op_k(X, Y)
0.70/0.89	
0.70/0.89	Lemma 20: i(mult(X, rd(Y, Z))) = mult(i(X), rd(Z, Y)).
0.70/0.89	Proof:
0.70/0.89	  i(mult(X, rd(Y, Z)))
0.70/0.89	= { by axiom 7 (c05) }
0.70/0.89	  mult(i(X), i(rd(Y, Z)))
0.70/0.89	= { by axiom 10 (c07) }
0.70/0.89	  mult(i(X), rd(mult(i(rd(Y, Z)), mult(rd(Y, Z), Z)), mult(rd(Y, Z), Z)))
0.70/0.89	= { by axiom 1 (c06) }
0.70/0.89	  mult(i(X), rd(Z, mult(rd(Y, Z), Z)))
0.70/0.89	= { by axiom 9 (c08) }
0.70/0.89	  mult(i(X), rd(Z, Y))
0.70/0.89	
0.70/0.89	Goal 1 (goals): asoc(a, b, op_k(c, d)) = unit.
0.70/0.89	Proof:
0.70/0.89	  asoc(a, b, op_k(c, d))
0.70/0.89	= { by axiom 10 (c07) }
0.70/0.89	  asoc(a, b, op_k(rd(mult(c, d), d), d))
0.70/0.89	= { by lemma 17 }
0.70/0.89	  asoc(a, b, mult(i(mult(d, rd(mult(c, d), d))), mult(rd(mult(c, d), d), d)))
0.70/0.89	= { by axiom 3 (c02) }
0.70/0.89	  asoc(a, b, mult(i(mult(d, mult(rd(mult(c, d), d), unit))), mult(rd(mult(c, d), d), d)))
0.70/0.89	= { by axiom 2 (c03) }
0.70/0.89	  asoc(a, b, mult(i(mult(d, mult(rd(mult(c, d), d), mult(d, i(d))))), mult(rd(mult(c, d), d), d)))
0.70/0.89	= { by axiom 11 (c09) }
0.70/0.89	  asoc(a, b, mult(i(mult(mult(d, mult(rd(mult(c, d), d), d)), i(d))), mult(rd(mult(c, d), d), d)))
0.70/0.89	= { by axiom 9 (c08) }
0.70/0.89	  asoc(a, b, mult(i(mult(mult(d, mult(c, d)), i(d))), mult(rd(mult(c, d), d), d)))
0.70/0.89	= { by axiom 9 (c08) }
0.70/0.89	  asoc(a, b, mult(i(mult(mult(d, mult(c, d)), i(d))), mult(c, d)))
0.70/0.89	= { by axiom 1 (c06) }
0.70/0.89	  asoc(a, b, mult(i(mult(mult(d, mult(c, d)), i(d))), mult(i(d), mult(d, mult(c, d)))))
0.70/0.89	= { by lemma 17 }
0.70/0.89	  asoc(a, b, op_k(i(d), mult(d, mult(c, d))))
0.70/0.89	= { by lemma 19 }
0.70/0.89	  asoc(a, b, i(op_k(i(i(d)), i(mult(d, mult(c, d))))))
0.70/0.89	= { by lemma 15 }
0.70/0.89	  asoc(a, b, i(op_k(d, i(mult(d, mult(c, d))))))
0.70/0.89	= { by lemma 19 }
0.70/0.89	  asoc(a, b, i(i(op_k(i(d), i(i(mult(d, mult(c, d))))))))
0.70/0.89	= { by lemma 15 }
0.70/0.89	  asoc(a, b, op_k(i(d), i(i(mult(d, mult(c, d))))))
0.70/0.89	= { by lemma 18 }
0.70/0.89	  asoc(a, b, mult(mult(i(mult(d, mult(c, d))), d), i(mult(d, i(mult(d, mult(c, d)))))))
0.70/0.89	= { by lemma 16 }
0.70/0.89	  asoc(a, b, mult(mult(i(mult(d, mult(c, d))), d), mult(i(d), mult(d, mult(c, d)))))
0.70/0.89	= { by axiom 1 (c06) }
0.70/0.89	  asoc(a, b, mult(mult(i(mult(d, mult(c, d))), d), mult(c, d)))
0.70/0.89	= { by axiom 13 (c10) }
0.70/0.89	  asoc(a, b, mult(mult(i(mult(d, mult(c, d))), mult(d, mult(c, d))), asoc(i(mult(d, mult(c, d))), d, mult(c, d))))
0.70/0.89	= { by axiom 12 (c04) }
0.70/0.89	  asoc(a, b, mult(unit, asoc(i(mult(d, mult(c, d))), d, mult(c, d))))
0.70/0.89	= { by axiom 4 (c01) }
0.70/0.89	  asoc(a, b, asoc(i(mult(d, mult(c, d))), d, mult(c, d)))
0.70/0.89	= { by axiom 8 (c22) }
0.70/0.89	  unit
0.70/0.89	% SZS output end Proof
0.70/0.89	
0.70/0.89	RESULT: Unsatisfiable (the axioms are contradictory).
0.70/0.89	EOF