0.07/0.12	% Problem    : theBenchmark.p : TPTP v0.0.0. Released v0.0.0.
0.07/0.13	% Command    : twee %s --tstp --casc --quiet --explain-encoding --conditional-encoding if --smaller --drop-non-horn
0.12/0.34	% Computer   : n016.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 11:47:53 EDT 2019
0.12/0.34	% CPUTime    : 
0.20/0.44	% SZS status Unsatisfiable
0.20/0.44	
0.20/0.44	% SZS output start Proof
0.20/0.44	Take the following subset of the input axioms:
0.20/0.44	  fof(c01, axiom, ![A, B]: B=mult(A, ld(A, B))).
0.20/0.44	  fof(c02, axiom, ![A, B]: B=ld(A, mult(A, B))).
0.20/0.44	  fof(c03, axiom, ![A, B]: A=mult(rd(A, B), B)).
0.20/0.44	  fof(c05, axiom, ![A]: mult(A, unit)=A).
0.20/0.44	  fof(c07, axiom, ![A, B, C]: mult(mult(mult(A, B), A), mult(A, C))=mult(A, mult(mult(mult(B, A), A), C))).
0.20/0.45	  fof(goals, negated_conjecture, mult(a, mult(b, c))!=mult(rd(mult(a, b), a), mult(a, c))).
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 (c05): mult(X, unit) = X.
0.20/0.45	Axiom 2 (c03): X = mult(rd(X, Y), Y).
0.20/0.45	Axiom 3 (c01): X = mult(Y, ld(Y, X)).
0.20/0.45	Axiom 4 (c02): X = ld(Y, mult(Y, X)).
0.20/0.45	Axiom 5 (c07): mult(mult(mult(X, Y), X), mult(X, Z)) = mult(X, mult(mult(mult(Y, X), X), Z)).
0.20/0.45	
0.20/0.45	Goal 1 (goals): mult(a, mult(b, c)) = mult(rd(mult(a, b), a), mult(a, c)).
0.20/0.45	Proof:
0.20/0.45	  mult(a, mult(b, c))
0.20/0.45	= { by axiom 4 (c02) }
0.20/0.45	  mult(a, mult(ld(a, mult(a, b)), c))
0.20/0.45	= { by axiom 2 (c03) }
0.20/0.45	  mult(a, mult(ld(a, mult(rd(mult(a, b), a), a)), c))
0.20/0.45	= { by axiom 2 (c03) }
0.20/0.45	  mult(a, mult(ld(a, mult(mult(rd(rd(mult(a, b), a), a), a), a)), c))
0.20/0.45	= { by axiom 3 (c01) }
0.20/0.45	  mult(a, mult(ld(a, mult(mult(mult(a, ld(a, rd(rd(mult(a, b), a), a))), a), a)), c))
0.20/0.45	= { by axiom 1 (c05) }
0.20/0.45	  mult(a, mult(ld(a, mult(mult(mult(a, ld(a, rd(rd(mult(a, b), a), a))), a), mult(a, unit))), c))
0.20/0.45	= { by axiom 5 (c07) }
0.20/0.45	  mult(a, mult(ld(a, mult(a, mult(mult(mult(ld(a, rd(rd(mult(a, b), a), a)), a), a), unit))), c))
0.20/0.45	= { by axiom 1 (c05) }
0.20/0.45	  mult(a, mult(ld(a, mult(a, mult(mult(ld(a, rd(rd(mult(a, b), a), a)), a), a))), c))
0.20/0.45	= { by axiom 4 (c02) }
0.20/0.45	  mult(a, mult(mult(mult(ld(a, rd(rd(mult(a, b), a), a)), a), a), c))
0.20/0.45	= { by axiom 4 (c02) }
0.20/0.45	  mult(a, ld(a, mult(a, mult(mult(mult(ld(a, rd(rd(mult(a, b), a), a)), a), a), c))))
0.20/0.45	= { by axiom 5 (c07) }
0.20/0.45	  mult(a, ld(a, mult(mult(mult(a, ld(a, rd(rd(mult(a, b), a), a))), a), mult(a, c))))
0.20/0.45	= { by axiom 3 (c01) }
0.20/0.45	  mult(a, ld(a, mult(mult(rd(rd(mult(a, b), a), a), a), mult(a, c))))
0.20/0.45	= { by axiom 2 (c03) }
0.20/0.45	  mult(a, ld(a, mult(rd(mult(a, b), a), mult(a, c))))
0.20/0.45	= { by axiom 3 (c01) }
0.20/0.45	  mult(rd(mult(a, b), a), mult(a, c))
0.20/0.45	% SZS output end Proof
0.20/0.45	
0.20/0.45	RESULT: Unsatisfiable (the axioms are contradictory).
0.20/0.45	EOF