0.07/0.12	% Problem    : theBenchmark.p : TPTP v0.0.0. Released v0.0.0.
0.07/0.13	% Command    : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
0.12/0.34	% Computer   : n005.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   : 1200
0.12/0.34	% WCLimit    : 120
0.12/0.34	% DateTime   : Tue Jul 13 13:57:33 EDT 2021
0.12/0.34	% CPUTime    : 
5.13/1.00	% SZS status Theorem
5.13/1.00	
5.13/1.01	% SZS output start Proof
5.13/1.01	Take the following subset of the input axioms:
5.13/1.01	  fof(aSatz7_15a, axiom, ![Xa, Xp, Xq, Xr]: (s_t(s(Xa, Xp), s(Xa, Xq), s(Xa, Xr)) | ~s_t(Xp, Xq, Xr))).
5.13/1.01	  fof(aSatz7_15b, conjecture, ![Xa, Xp, Xq, Xr]: (~s_t(s(Xa, Xp), s(Xa, Xq), s(Xa, Xr)) | s_t(Xp, Xq, Xr))).
5.13/1.01	  fof(aSatz7_7, axiom, ![Xa, Xp]: s(Xa, s(Xa, Xp))=Xp).
5.13/1.01	
5.13/1.01	Now clausify the problem and encode Horn clauses using encoding 3 of
5.13/1.01	http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
5.13/1.01	We repeatedly replace C & s=t => u=v by the two clauses:
5.13/1.01	  fresh(y, y, x1...xn) = u
5.13/1.01	  C => fresh(s, t, x1...xn) = v
5.13/1.01	where fresh is a fresh function symbol and x1..xn are the free
5.13/1.01	variables of u and v.
5.13/1.01	A predicate p(X) is encoded as p(X)=true (this is sound, because the
5.13/1.01	input problem has no model of domain size 1).
5.13/1.01	
5.13/1.01	The encoding turns the above axioms into the following unit equations and goals:
5.13/1.01	
5.13/1.01	Axiom 1 (aSatz7_7): s(X, s(X, Y)) = Y.
5.13/1.01	Axiom 2 (aSatz7_15a): fresh58(X, X, Y, Z, W, V) = true2.
5.13/1.01	Axiom 3 (aSatz7_15b): s_t(s(xa, xp), s(xa, xq), s(xa, xr)) = true2.
5.13/1.01	Axiom 4 (aSatz7_15a): fresh58(s_t(X, Y, Z), true2, X, Y, Z, W) = s_t(s(W, X), s(W, Y), s(W, Z)).
5.13/1.01	
5.13/1.01	Goal 1 (aSatz7_15b_1): s_t(xp, xq, xr) = true2.
5.13/1.01	Proof:
5.13/1.01	  s_t(xp, xq, xr)
5.13/1.01	= { by axiom 1 (aSatz7_7) R->L }
5.13/1.01	  s_t(xp, xq, s(xa, s(xa, xr)))
5.13/1.01	= { by axiom 1 (aSatz7_7) R->L }
5.13/1.01	  s_t(xp, s(xa, s(xa, xq)), s(xa, s(xa, xr)))
5.13/1.01	= { by axiom 1 (aSatz7_7) R->L }
5.13/1.01	  s_t(s(xa, s(xa, xp)), s(xa, s(xa, xq)), s(xa, s(xa, xr)))
5.13/1.01	= { by axiom 4 (aSatz7_15a) R->L }
5.13/1.01	  fresh58(s_t(s(xa, xp), s(xa, xq), s(xa, xr)), true2, s(xa, xp), s(xa, xq), s(xa, xr), xa)
5.13/1.01	= { by axiom 3 (aSatz7_15b) }
5.13/1.01	  fresh58(true2, true2, s(xa, xp), s(xa, xq), s(xa, xr), xa)
5.13/1.01	= { by axiom 2 (aSatz7_15a) }
5.13/1.01	  true2
5.13/1.01	% SZS output end Proof
5.13/1.01	
5.13/1.01	RESULT: Theorem (the conjecture is true).
5.13/1.01	EOF