0.11/0.12	% Problem    : theBenchmark.p : TPTP v0.0.0. Released v0.0.0.
0.11/0.12	% Command    : twee %s --tstp --casc --quiet --explain-encoding --conditional-encoding if --smaller --drop-non-horn
0.12/0.33	% Computer   : n017.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 11:40:19 EDT 2019
0.12/0.33	% CPUTime    : 
0.56/0.73	% SZS status Unsatisfiable
0.56/0.73	
0.56/0.73	% SZS output start Proof
0.56/0.73	Take the following subset of the input axioms:
0.56/0.73	  fof(c01, axiom, ![A, B]: B=mult(A, ld(A, B))).
0.56/0.73	  fof(c02, axiom, ![A, B]: B=ld(A, mult(A, B))).
0.56/0.73	  fof(c06, axiom, ![A]: mult(unit, A)=A).
0.56/0.73	  fof(c08, axiom, ![A, B, C]: mult(mult(A, mult(B, mult(B, C))), B)=mult(mult(A, B), mult(B, mult(C, B)))).
0.56/0.73	  fof(goals, negated_conjecture, mult(mult(a, c), ld(c, mult(b, c)))!=mult(mult(a, b), c)).
0.56/0.73	
0.56/0.73	Now clausify the problem and encode Horn clauses using encoding 3 of
0.56/0.73	http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
0.56/0.73	We repeatedly replace C & s=t => u=v by the two clauses:
0.56/0.73	  fresh(y, y, x1...xn) = u
0.56/0.73	  C => fresh(s, t, x1...xn) = v
0.56/0.73	where fresh is a fresh function symbol and x1..xn are the free
0.56/0.73	variables of u and v.
0.56/0.73	A predicate p(X) is encoded as p(X)=true (this is sound, because the
0.56/0.73	input problem has no model of domain size 1).
0.56/0.73	
0.56/0.73	The encoding turns the above axioms into the following unit equations and goals:
0.56/0.73	
0.56/0.73	Axiom 1 (c01): X = mult(Y, ld(Y, X)).
0.56/0.73	Axiom 2 (c02): X = ld(Y, mult(Y, X)).
0.56/0.73	Axiom 3 (c06): mult(unit, X) = X.
0.56/0.73	Axiom 4 (c08): mult(mult(X, mult(Y, mult(Y, Z))), Y) = mult(mult(X, Y), mult(Y, mult(Z, Y))).
0.56/0.73	
0.56/0.73	Goal 1 (goals): mult(mult(a, c), ld(c, mult(b, c))) = mult(mult(a, b), c).
0.56/0.73	Proof:
0.56/0.73	  mult(mult(a, c), ld(c, mult(b, c)))
0.56/0.73	= { by axiom 1 (c01) }
0.56/0.73	  mult(mult(a, c), ld(c, mult(mult(c, ld(c, b)), c)))
0.56/0.73	= { by axiom 1 (c01) }
0.56/0.73	  mult(mult(a, c), ld(c, mult(mult(c, mult(c, ld(c, ld(c, b)))), c)))
0.56/0.73	= { by axiom 3 (c06) }
0.56/0.73	  mult(mult(a, c), ld(c, mult(mult(unit, mult(c, mult(c, ld(c, ld(c, b))))), c)))
0.56/0.73	= { by axiom 4 (c08) }
0.56/0.73	  mult(mult(a, c), ld(c, mult(mult(unit, c), mult(c, mult(ld(c, ld(c, b)), c)))))
0.56/0.73	= { by axiom 3 (c06) }
0.56/0.73	  mult(mult(a, c), ld(c, mult(c, mult(c, mult(ld(c, ld(c, b)), c)))))
0.56/0.73	= { by axiom 2 (c02) }
0.56/0.73	  mult(mult(a, c), mult(c, mult(ld(c, ld(c, b)), c)))
0.56/0.73	= { by axiom 4 (c08) }
0.56/0.73	  mult(mult(a, mult(c, mult(c, ld(c, ld(c, b))))), c)
0.56/0.73	= { by axiom 1 (c01) }
0.56/0.73	  mult(mult(a, mult(c, ld(c, b))), c)
0.56/0.73	= { by axiom 1 (c01) }
0.56/0.73	  mult(mult(a, b), c)
0.56/0.73	% SZS output end Proof
0.56/0.73	
0.56/0.73	RESULT: Unsatisfiable (the axioms are contradictory).
0.56/0.73	EOF