0.00/0.03	% Problem    : theBenchmark.p : TPTP v0.0.0. Released v0.0.0.
0.00/0.04	% Command    : twee %s --tstp --casc --quiet --conditional-encoding if --smaller --drop-non-horn
0.03/0.23	% Computer   : n123.star.cs.uiowa.edu
0.03/0.23	% Model      : x86_64 x86_64
0.03/0.23	% CPU        : Intel(R) Xeon(R) CPU E5-2609 0 @ 2.40GHz
0.03/0.23	% Memory     : 32218.625MB
0.03/0.23	% OS         : Linux 3.10.0-693.2.2.el7.x86_64
0.03/0.23	% CPULimit   : 300
0.03/0.23	% DateTime   : Sat Jul 14 04:24:11 CDT 2018
0.03/0.23	% CPUTime    : 
0.07/0.44	% SZS status Theorem
0.07/0.44	
0.07/0.44	% SZS output start Proof
0.07/0.44	Take the following subset of the input axioms:
0.07/0.44	  fof(aSatz7_7, axiom, ![Xa, Xp]: s(Xa, s(Xa, Xp))=Xp).
0.07/0.44	  fof(aSatz7_8, conjecture,
0.07/0.44	      ![Xa, Xp, Xq, Xr]: (s(Xa, Xp)!=Xr | (Xr!=s(Xa, Xq) | Xp=Xq))).
0.07/0.44	
0.07/0.44	Now clausify the problem and encode Horn clauses using encoding 3 of
0.07/0.44	http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
0.07/0.44	We repeatedly replace C & s=t => u=v by the two clauses:
0.07/0.44	  $$fresh(y, y, x1...xn) = u
0.07/0.44	  C => $$fresh(s, t, x1...xn) = v
0.07/0.44	where $$fresh is a fresh function symbol and x1..xn are the free
0.07/0.44	variables of u and v.
0.07/0.44	A predicate p(X) is encoded as p(X)=$$true (this is sound, because the
0.07/0.44	input problem has no model of domain size 1).
0.07/0.44	
0.07/0.44	The encoding turns the above axioms into the following unit equations and goals:
0.07/0.44	
0.07/0.44	Axiom 212 (aSatz7_7): s(X, s(X, Y)) = Y.
0.07/0.44	Axiom 279 (aSatz7_8): s(sK2_aSatz7_8_Xa, sK4_aSatz7_8_Xp) = sK1_aSatz7_8_Xr.
0.07/0.44	Axiom 280 (aSatz7_8_1): sK1_aSatz7_8_Xr = s(sK2_aSatz7_8_Xa, sK3_aSatz7_8_Xq).
0.07/0.44	
0.07/0.44	Goal 1 (aSatz7_8_2): sK4_aSatz7_8_Xp = sK3_aSatz7_8_Xq.
0.07/0.44	Proof:
0.07/0.44	  sK4_aSatz7_8_Xp
0.07/0.44	= { by axiom 212 (aSatz7_7) }
0.07/0.44	  s(sK2_aSatz7_8_Xa, s(sK2_aSatz7_8_Xa, sK4_aSatz7_8_Xp))
0.07/0.44	= { by axiom 279 (aSatz7_8) }
0.07/0.44	  s(sK2_aSatz7_8_Xa, sK1_aSatz7_8_Xr)
0.07/0.44	= { by axiom 280 (aSatz7_8_1) }
0.07/0.44	  s(sK2_aSatz7_8_Xa, s(sK2_aSatz7_8_Xa, sK3_aSatz7_8_Xq))
0.07/0.44	= { by axiom 212 (aSatz7_7) }
0.07/0.44	  sK3_aSatz7_8_Xq
0.07/0.44	% SZS output end Proof
0.07/0.44	
0.07/0.44	RESULT: Theorem (the conjecture is true).
0.07/0.45	EOF