3.71/1.70	YES
3.71/1.71	proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl
3.71/1.71	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.71/1.71	
3.71/1.71	
3.71/1.71	Left Termination of the query pattern
3.71/1.71	
3.71/1.71	q(g)
3.71/1.71	
3.71/1.71	w.r.t. the given Prolog program could successfully be proven:
3.71/1.71	
3.71/1.71	(0) Prolog
3.71/1.71	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.71/1.71	(2) PiTRS
3.71/1.71	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.71/1.71	(4) PiDP
3.71/1.71	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.71/1.71	(6) PiDP
3.71/1.71	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.71/1.71	(8) PiDP
3.71/1.71	(9) PiDPToQDPProof [EQUIVALENT, 0 ms]
3.71/1.71	(10) QDP
3.71/1.71	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.71/1.71	(12) YES
3.71/1.71	
3.71/1.71	
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(0)
3.71/1.71	Obligation:
3.71/1.71	Clauses:
3.71/1.71	
3.71/1.71	q(X) :- p(X, 0).
3.71/1.71	p(0, X1).
3.71/1.71	p(s(X), Y) :- p(X, s(Y)).
3.71/1.71	
3.71/1.71	
3.71/1.71	Query: q(g)
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(1) PrologToPiTRSProof (SOUND)
3.71/1.71	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.71/1.71	
3.71/1.71	q_in_1: (b)
3.71/1.71	
3.71/1.71	p_in_2: (b,b)
3.71/1.71	
3.71/1.71	Transforming Prolog into the following Term Rewriting System:
3.71/1.71	
3.71/1.71	Pi-finite rewrite system:
3.71/1.71	The TRS R consists of the following rules:
3.71/1.71	
3.71/1.71	   q_in_g(X) -> U1_g(X, p_in_gg(X, 0))
3.71/1.71	   p_in_gg(0, X1) -> p_out_gg(0, X1)
3.71/1.71	   p_in_gg(s(X), Y) -> U2_gg(X, Y, p_in_gg(X, s(Y)))
3.71/1.71	   U2_gg(X, Y, p_out_gg(X, s(Y))) -> p_out_gg(s(X), Y)
3.71/1.71	   U1_g(X, p_out_gg(X, 0)) -> q_out_g(X)
3.71/1.71	
3.71/1.71	The argument filtering Pi contains the following mapping:
3.71/1.71	q_in_g(x1)  =  q_in_g(x1)
3.71/1.71	
3.71/1.71	U1_g(x1, x2)  =  U1_g(x2)
3.71/1.71	
3.71/1.71	p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
3.71/1.71	
3.71/1.71	0  =  0
3.71/1.71	
3.71/1.71	p_out_gg(x1, x2)  =  p_out_gg
3.71/1.71	
3.71/1.71	s(x1)  =  s(x1)
3.71/1.71	
3.71/1.71	U2_gg(x1, x2, x3)  =  U2_gg(x3)
3.71/1.71	
3.71/1.71	q_out_g(x1)  =  q_out_g
3.71/1.71	
3.71/1.71	
3.71/1.71	
3.71/1.71	
3.71/1.71	
3.71/1.71	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.71/1.71	
3.71/1.71	
3.71/1.71	
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(2)
3.71/1.71	Obligation:
3.71/1.71	Pi-finite rewrite system:
3.71/1.71	The TRS R consists of the following rules:
3.71/1.71	
3.71/1.71	   q_in_g(X) -> U1_g(X, p_in_gg(X, 0))
3.71/1.71	   p_in_gg(0, X1) -> p_out_gg(0, X1)
3.71/1.71	   p_in_gg(s(X), Y) -> U2_gg(X, Y, p_in_gg(X, s(Y)))
3.71/1.71	   U2_gg(X, Y, p_out_gg(X, s(Y))) -> p_out_gg(s(X), Y)
3.71/1.71	   U1_g(X, p_out_gg(X, 0)) -> q_out_g(X)
3.71/1.71	
3.71/1.71	The argument filtering Pi contains the following mapping:
3.71/1.71	q_in_g(x1)  =  q_in_g(x1)
3.71/1.71	
3.71/1.71	U1_g(x1, x2)  =  U1_g(x2)
3.71/1.71	
3.71/1.71	p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
3.71/1.71	
3.71/1.71	0  =  0
3.71/1.71	
3.71/1.71	p_out_gg(x1, x2)  =  p_out_gg
3.71/1.71	
3.71/1.71	s(x1)  =  s(x1)
3.71/1.71	
3.71/1.71	U2_gg(x1, x2, x3)  =  U2_gg(x3)
3.71/1.71	
3.71/1.71	q_out_g(x1)  =  q_out_g
3.71/1.71	
3.71/1.71	
3.71/1.71	
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(3) DependencyPairsProof (EQUIVALENT)
3.71/1.71	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.71/1.71	Pi DP problem:
3.71/1.71	The TRS P consists of the following rules:
3.71/1.71	
3.71/1.71	   Q_IN_G(X) -> U1_G(X, p_in_gg(X, 0))
3.71/1.71	   Q_IN_G(X) -> P_IN_GG(X, 0)
3.71/1.71	   P_IN_GG(s(X), Y) -> U2_GG(X, Y, p_in_gg(X, s(Y)))
3.71/1.71	   P_IN_GG(s(X), Y) -> P_IN_GG(X, s(Y))
3.71/1.71	
3.71/1.71	The TRS R consists of the following rules:
3.71/1.71	
3.71/1.71	   q_in_g(X) -> U1_g(X, p_in_gg(X, 0))
3.71/1.71	   p_in_gg(0, X1) -> p_out_gg(0, X1)
3.71/1.71	   p_in_gg(s(X), Y) -> U2_gg(X, Y, p_in_gg(X, s(Y)))
3.71/1.71	   U2_gg(X, Y, p_out_gg(X, s(Y))) -> p_out_gg(s(X), Y)
3.71/1.71	   U1_g(X, p_out_gg(X, 0)) -> q_out_g(X)
3.71/1.71	
3.71/1.71	The argument filtering Pi contains the following mapping:
3.71/1.71	q_in_g(x1)  =  q_in_g(x1)
3.71/1.71	
3.71/1.71	U1_g(x1, x2)  =  U1_g(x2)
3.71/1.71	
3.71/1.71	p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
3.71/1.71	
3.71/1.71	0  =  0
3.71/1.71	
3.71/1.71	p_out_gg(x1, x2)  =  p_out_gg
3.71/1.71	
3.71/1.71	s(x1)  =  s(x1)
3.71/1.71	
3.71/1.71	U2_gg(x1, x2, x3)  =  U2_gg(x3)
3.71/1.71	
3.71/1.71	q_out_g(x1)  =  q_out_g
3.71/1.71	
3.71/1.71	Q_IN_G(x1)  =  Q_IN_G(x1)
3.71/1.71	
3.71/1.71	U1_G(x1, x2)  =  U1_G(x2)
3.71/1.71	
3.71/1.71	P_IN_GG(x1, x2)  =  P_IN_GG(x1, x2)
3.71/1.71	
3.71/1.71	U2_GG(x1, x2, x3)  =  U2_GG(x3)
3.71/1.71	
3.71/1.71	
3.71/1.71	We have to consider all (P,R,Pi)-chains
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(4)
3.71/1.71	Obligation:
3.71/1.71	Pi DP problem:
3.71/1.71	The TRS P consists of the following rules:
3.71/1.71	
3.71/1.71	   Q_IN_G(X) -> U1_G(X, p_in_gg(X, 0))
3.71/1.71	   Q_IN_G(X) -> P_IN_GG(X, 0)
3.71/1.71	   P_IN_GG(s(X), Y) -> U2_GG(X, Y, p_in_gg(X, s(Y)))
3.71/1.71	   P_IN_GG(s(X), Y) -> P_IN_GG(X, s(Y))
3.71/1.71	
3.71/1.71	The TRS R consists of the following rules:
3.71/1.71	
3.71/1.71	   q_in_g(X) -> U1_g(X, p_in_gg(X, 0))
3.71/1.71	   p_in_gg(0, X1) -> p_out_gg(0, X1)
3.71/1.71	   p_in_gg(s(X), Y) -> U2_gg(X, Y, p_in_gg(X, s(Y)))
3.71/1.71	   U2_gg(X, Y, p_out_gg(X, s(Y))) -> p_out_gg(s(X), Y)
3.71/1.71	   U1_g(X, p_out_gg(X, 0)) -> q_out_g(X)
3.71/1.71	
3.71/1.71	The argument filtering Pi contains the following mapping:
3.71/1.71	q_in_g(x1)  =  q_in_g(x1)
3.71/1.71	
3.71/1.71	U1_g(x1, x2)  =  U1_g(x2)
3.71/1.71	
3.71/1.71	p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
3.71/1.71	
3.71/1.71	0  =  0
3.71/1.71	
3.71/1.71	p_out_gg(x1, x2)  =  p_out_gg
3.71/1.71	
3.71/1.71	s(x1)  =  s(x1)
3.71/1.71	
3.71/1.71	U2_gg(x1, x2, x3)  =  U2_gg(x3)
3.71/1.71	
3.71/1.71	q_out_g(x1)  =  q_out_g
3.71/1.71	
3.71/1.71	Q_IN_G(x1)  =  Q_IN_G(x1)
3.71/1.71	
3.71/1.71	U1_G(x1, x2)  =  U1_G(x2)
3.71/1.71	
3.71/1.71	P_IN_GG(x1, x2)  =  P_IN_GG(x1, x2)
3.71/1.71	
3.71/1.71	U2_GG(x1, x2, x3)  =  U2_GG(x3)
3.71/1.71	
3.71/1.71	
3.71/1.71	We have to consider all (P,R,Pi)-chains
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(5) DependencyGraphProof (EQUIVALENT)
3.71/1.71	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(6)
3.71/1.71	Obligation:
3.71/1.71	Pi DP problem:
3.71/1.71	The TRS P consists of the following rules:
3.71/1.71	
3.71/1.71	   P_IN_GG(s(X), Y) -> P_IN_GG(X, s(Y))
3.71/1.71	
3.71/1.71	The TRS R consists of the following rules:
3.71/1.71	
3.71/1.71	   q_in_g(X) -> U1_g(X, p_in_gg(X, 0))
3.71/1.71	   p_in_gg(0, X1) -> p_out_gg(0, X1)
3.71/1.71	   p_in_gg(s(X), Y) -> U2_gg(X, Y, p_in_gg(X, s(Y)))
3.71/1.71	   U2_gg(X, Y, p_out_gg(X, s(Y))) -> p_out_gg(s(X), Y)
3.71/1.71	   U1_g(X, p_out_gg(X, 0)) -> q_out_g(X)
3.71/1.71	
3.71/1.71	The argument filtering Pi contains the following mapping:
3.71/1.71	q_in_g(x1)  =  q_in_g(x1)
3.71/1.71	
3.71/1.71	U1_g(x1, x2)  =  U1_g(x2)
3.71/1.71	
3.71/1.71	p_in_gg(x1, x2)  =  p_in_gg(x1, x2)
3.71/1.71	
3.71/1.71	0  =  0
3.71/1.71	
3.71/1.71	p_out_gg(x1, x2)  =  p_out_gg
3.71/1.71	
3.71/1.71	s(x1)  =  s(x1)
3.71/1.71	
3.71/1.71	U2_gg(x1, x2, x3)  =  U2_gg(x3)
3.71/1.71	
3.71/1.71	q_out_g(x1)  =  q_out_g
3.71/1.71	
3.71/1.71	P_IN_GG(x1, x2)  =  P_IN_GG(x1, x2)
3.71/1.71	
3.71/1.71	
3.71/1.71	We have to consider all (P,R,Pi)-chains
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(7) UsableRulesProof (EQUIVALENT)
3.71/1.71	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(8)
3.71/1.71	Obligation:
3.71/1.71	Pi DP problem:
3.71/1.71	The TRS P consists of the following rules:
3.71/1.71	
3.71/1.71	   P_IN_GG(s(X), Y) -> P_IN_GG(X, s(Y))
3.71/1.71	
3.71/1.71	R is empty.
3.71/1.71	Pi is empty.
3.71/1.71	We have to consider all (P,R,Pi)-chains
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(9) PiDPToQDPProof (EQUIVALENT)
3.71/1.71	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(10)
3.71/1.71	Obligation:
3.71/1.71	Q DP problem:
3.71/1.71	The TRS P consists of the following rules:
3.71/1.71	
3.71/1.71	   P_IN_GG(s(X), Y) -> P_IN_GG(X, s(Y))
3.71/1.71	
3.71/1.71	R is empty.
3.71/1.71	Q is empty.
3.71/1.71	We have to consider all (P,Q,R)-chains.
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(11) QDPSizeChangeProof (EQUIVALENT)
3.71/1.71	By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
3.71/1.71	
3.71/1.71	From the DPs we obtained the following set of size-change graphs:
3.71/1.71	*P_IN_GG(s(X), Y) -> P_IN_GG(X, s(Y))
3.71/1.71	The graph contains the following edges 1 > 1
3.71/1.71	
3.71/1.71	
3.71/1.71	----------------------------------------
3.71/1.71	
3.71/1.71	(12)
3.71/1.71	YES
3.71/1.74	EOF