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
0.12/0.33	% Computer   : n019.cluster.edu
0.12/0.33	% Model      : x86_64 x86_64
0.12/0.33	% CPU        : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
0.12/0.33	% Memory     : 8042.1875MB
0.12/0.33	% OS         : Linux 3.10.0-693.el7.x86_64
0.12/0.33	% CPULimit   : 180
0.12/0.33	% DateTime   : Thu Aug 29 09:47:22 EDT 2019
0.12/0.33	% CPUTime    : 
0.19/0.38	% SZS status Unsatisfiable
0.19/0.38	
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(c02, axiom, ![B, A]: B=ld(A, mult(A, B))).
0.19/0.38	  fof(c06, axiom, ![A]: A=mult(unit, A)).
0.19/0.38	  fof(c07, axiom, ![B, A, C]: mult(A, mult(B, mult(A, C)))=mult(mult(A, mult(B, A)), C)).
0.19/0.38	  fof(c08, axiom, ![A]: mult(op_c, A)=mult(A, op_c)).
0.19/0.38	  fof(c09, axiom, ![B, A]: mult(mult(mult(op_c, op_c), A), B)=mult(mult(op_c, op_c), mult(A, B))).
0.19/0.38	  fof(goals, negated_conjecture, mult(mult(a, b), op_c)!=mult(a, mult(b, op_c))).
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 (c09): mult(mult(mult(op_c, op_c), X), Y) = mult(mult(op_c, op_c), mult(X, Y)).
0.19/0.38	Axiom 2 (c02): X = ld(Y, mult(Y, X)).
0.19/0.38	Axiom 3 (c07): mult(X, mult(Y, mult(X, Z))) = mult(mult(X, mult(Y, X)), Z).
0.19/0.38	Axiom 4 (c08): mult(op_c, X) = mult(X, op_c).
0.19/0.38	Axiom 5 (c06): X = mult(unit, X).
0.19/0.38	
0.19/0.38	Lemma 6: mult(mult(X, X), Y) = mult(X, mult(X, Y)).
0.19/0.38	Proof:
0.19/0.38	  mult(mult(X, X), Y)
0.19/0.38	= { by axiom 5 (c06) }
0.19/0.38	  mult(mult(X, mult(unit, X)), Y)
0.19/0.38	= { by axiom 3 (c07) }
0.19/0.38	  mult(X, mult(unit, mult(X, Y)))
0.19/0.38	= { by axiom 5 (c06) }
0.19/0.38	  mult(X, mult(X, Y))
0.19/0.38	
0.19/0.38	Goal 1 (goals): mult(mult(a, b), op_c) = mult(a, mult(b, op_c)).
0.19/0.38	Proof:
0.19/0.38	  mult(mult(a, b), op_c)
0.19/0.38	= { by axiom 4 (c08) }
0.19/0.38	  mult(op_c, mult(a, b))
0.19/0.38	= { by axiom 2 (c02) }
0.19/0.38	  ld(op_c, mult(op_c, mult(op_c, mult(a, b))))
0.19/0.38	= { by lemma 6 }
0.19/0.38	  ld(op_c, mult(mult(op_c, op_c), mult(a, b)))
0.19/0.38	= { by axiom 1 (c09) }
0.19/0.38	  ld(op_c, mult(mult(mult(op_c, op_c), a), b))
0.19/0.38	= { by lemma 6 }
0.19/0.38	  ld(op_c, mult(mult(op_c, mult(op_c, a)), b))
0.19/0.38	= { by axiom 4 (c08) }
0.19/0.38	  ld(op_c, mult(mult(op_c, mult(a, op_c)), b))
0.19/0.38	= { by axiom 3 (c07) }
0.19/0.38	  ld(op_c, mult(op_c, mult(a, mult(op_c, b))))
0.19/0.38	= { by axiom 4 (c08) }
0.19/0.38	  ld(op_c, mult(op_c, mult(a, mult(b, op_c))))
0.19/0.38	= { by axiom 2 (c02) }
0.19/0.38	  mult(a, mult(b, op_c))
0.19/0.38	% SZS output end Proof
0.19/0.38	
0.19/0.38	RESULT: Unsatisfiable (the axioms are contradictory).
0.19/0.38	EOF