3.43/1.76	YES
3.43/1.77	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.43/1.77	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.43/1.77	
3.43/1.77	
3.43/1.77	Left Termination of the query pattern
3.43/1.77	
3.43/1.77	suffix(a,g)
3.43/1.77	
3.43/1.77	w.r.t. the given Prolog program could successfully be proven:
3.43/1.77	
3.43/1.77	(0) Prolog
3.43/1.77	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.43/1.77	(2) PiTRS
3.43/1.77	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.43/1.77	(4) PiDP
3.43/1.77	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.43/1.77	(6) PiDP
3.43/1.77	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.43/1.77	(8) PiDP
3.43/1.77	(9) PiDPToQDPProof [SOUND, 0 ms]
3.43/1.77	(10) QDP
3.43/1.77	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.43/1.77	(12) YES
3.43/1.77	
3.43/1.77	
3.43/1.77	----------------------------------------
3.43/1.77	
3.43/1.77	(0)
3.43/1.77	Obligation:
3.43/1.77	Clauses:
3.43/1.77	
3.43/1.77	suffix(Xs, Ys) :- app(X1, Xs, Ys).
3.43/1.77	app([], X, X).
3.43/1.77	app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
3.43/1.77	
3.43/1.77	
3.43/1.77	Query: suffix(a,g)
3.43/1.77	----------------------------------------
3.43/1.77	
3.43/1.77	(1) PrologToPiTRSProof (SOUND)
3.43/1.77	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.43/1.77	
3.43/1.77	suffix_in_2: (f,b)
3.43/1.77	
3.43/1.77	app_in_3: (f,f,b)
3.43/1.77	
3.43/1.77	Transforming Prolog into the following Term Rewriting System:
3.43/1.78	
3.43/1.78	Pi-finite rewrite system:
3.43/1.78	The TRS R consists of the following rules:
3.43/1.78	
3.43/1.78	   suffix_in_ag(Xs, Ys) -> U1_ag(Xs, Ys, app_in_aag(X1, Xs, Ys))
3.43/1.78	   app_in_aag([], X, X) -> app_out_aag([], X, X)
3.43/1.78	   app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U2_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
3.43/1.78	   U2_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs))
3.43/1.78	   U1_ag(Xs, Ys, app_out_aag(X1, Xs, Ys)) -> suffix_out_ag(Xs, Ys)
3.43/1.78	
3.43/1.78	The argument filtering Pi contains the following mapping:
3.43/1.78	suffix_in_ag(x1, x2)  =  suffix_in_ag(x2)
3.43/1.78	
3.43/1.78	U1_ag(x1, x2, x3)  =  U1_ag(x3)
3.43/1.78	
3.43/1.78	app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
3.43/1.78	
3.43/1.78	app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
3.43/1.78	
3.43/1.78	.(x1, x2)  =  .(x1, x2)
3.43/1.78	
3.43/1.78	U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x5)
3.43/1.78	
3.43/1.78	suffix_out_ag(x1, x2)  =  suffix_out_ag(x1)
3.43/1.78	
3.43/1.78	
3.43/1.78	
3.43/1.78	
3.43/1.78	
3.43/1.78	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.43/1.78	
3.43/1.78	
3.43/1.78	
3.43/1.78	----------------------------------------
3.43/1.78	
3.43/1.78	(2)
3.43/1.78	Obligation:
3.43/1.78	Pi-finite rewrite system:
3.43/1.78	The TRS R consists of the following rules:
3.43/1.78	
3.43/1.78	   suffix_in_ag(Xs, Ys) -> U1_ag(Xs, Ys, app_in_aag(X1, Xs, Ys))
3.43/1.78	   app_in_aag([], X, X) -> app_out_aag([], X, X)
3.43/1.78	   app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U2_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
3.43/1.78	   U2_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs))
3.43/1.78	   U1_ag(Xs, Ys, app_out_aag(X1, Xs, Ys)) -> suffix_out_ag(Xs, Ys)
3.43/1.78	
3.43/1.78	The argument filtering Pi contains the following mapping:
3.43/1.78	suffix_in_ag(x1, x2)  =  suffix_in_ag(x2)
3.43/1.78	
3.43/1.78	U1_ag(x1, x2, x3)  =  U1_ag(x3)
3.43/1.78	
3.43/1.78	app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
3.43/1.78	
3.43/1.78	app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
3.43/1.78	
3.43/1.78	.(x1, x2)  =  .(x1, x2)
3.43/1.78	
3.43/1.78	U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x5)
3.43/1.78	
3.43/1.78	suffix_out_ag(x1, x2)  =  suffix_out_ag(x1)
3.43/1.78	
3.43/1.78	
3.43/1.78	
3.43/1.78	----------------------------------------
3.43/1.78	
3.43/1.78	(3) DependencyPairsProof (EQUIVALENT)
3.43/1.78	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.43/1.78	Pi DP problem:
3.43/1.78	The TRS P consists of the following rules:
3.43/1.78	
3.43/1.78	   SUFFIX_IN_AG(Xs, Ys) -> U1_AG(Xs, Ys, app_in_aag(X1, Xs, Ys))
3.43/1.78	   SUFFIX_IN_AG(Xs, Ys) -> APP_IN_AAG(X1, Xs, Ys)
3.43/1.78	   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U2_AAG(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
3.43/1.78	   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs)
3.43/1.78	
3.43/1.78	The TRS R consists of the following rules:
3.43/1.78	
3.43/1.78	   suffix_in_ag(Xs, Ys) -> U1_ag(Xs, Ys, app_in_aag(X1, Xs, Ys))
3.43/1.78	   app_in_aag([], X, X) -> app_out_aag([], X, X)
3.43/1.78	   app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U2_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
3.43/1.78	   U2_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs))
3.43/1.78	   U1_ag(Xs, Ys, app_out_aag(X1, Xs, Ys)) -> suffix_out_ag(Xs, Ys)
3.43/1.78	
3.43/1.78	The argument filtering Pi contains the following mapping:
3.43/1.78	suffix_in_ag(x1, x2)  =  suffix_in_ag(x2)
3.43/1.78	
3.43/1.78	U1_ag(x1, x2, x3)  =  U1_ag(x3)
3.43/1.78	
3.43/1.78	app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
3.43/1.78	
3.43/1.78	app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
3.43/1.78	
3.43/1.78	.(x1, x2)  =  .(x1, x2)
3.43/1.78	
3.43/1.78	U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x5)
3.43/1.78	
3.43/1.78	suffix_out_ag(x1, x2)  =  suffix_out_ag(x1)
3.43/1.78	
3.43/1.78	SUFFIX_IN_AG(x1, x2)  =  SUFFIX_IN_AG(x2)
3.43/1.78	
3.43/1.78	U1_AG(x1, x2, x3)  =  U1_AG(x3)
3.43/1.78	
3.43/1.78	APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)
3.43/1.78	
3.43/1.78	U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x1, x5)
3.43/1.78	
3.43/1.78	
3.43/1.78	We have to consider all (P,R,Pi)-chains
3.43/1.78	----------------------------------------
3.43/1.78	
3.43/1.78	(4)
3.43/1.78	Obligation:
3.43/1.78	Pi DP problem:
3.43/1.78	The TRS P consists of the following rules:
3.43/1.78	
3.43/1.78	   SUFFIX_IN_AG(Xs, Ys) -> U1_AG(Xs, Ys, app_in_aag(X1, Xs, Ys))
3.43/1.78	   SUFFIX_IN_AG(Xs, Ys) -> APP_IN_AAG(X1, Xs, Ys)
3.43/1.78	   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U2_AAG(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
3.43/1.78	   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs)
3.43/1.78	
3.43/1.78	The TRS R consists of the following rules:
3.43/1.78	
3.43/1.78	   suffix_in_ag(Xs, Ys) -> U1_ag(Xs, Ys, app_in_aag(X1, Xs, Ys))
3.43/1.78	   app_in_aag([], X, X) -> app_out_aag([], X, X)
3.43/1.78	   app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U2_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
3.43/1.78	   U2_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs))
3.43/1.78	   U1_ag(Xs, Ys, app_out_aag(X1, Xs, Ys)) -> suffix_out_ag(Xs, Ys)
3.43/1.78	
3.43/1.78	The argument filtering Pi contains the following mapping:
3.43/1.78	suffix_in_ag(x1, x2)  =  suffix_in_ag(x2)
3.43/1.78	
3.43/1.78	U1_ag(x1, x2, x3)  =  U1_ag(x3)
3.43/1.78	
3.43/1.78	app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
3.43/1.78	
3.43/1.78	app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
3.43/1.78	
3.43/1.78	.(x1, x2)  =  .(x1, x2)
3.43/1.78	
3.43/1.78	U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x5)
3.43/1.78	
3.43/1.78	suffix_out_ag(x1, x2)  =  suffix_out_ag(x1)
3.43/1.78	
3.43/1.78	SUFFIX_IN_AG(x1, x2)  =  SUFFIX_IN_AG(x2)
3.43/1.78	
3.43/1.78	U1_AG(x1, x2, x3)  =  U1_AG(x3)
3.43/1.78	
3.43/1.78	APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)
3.43/1.78	
3.43/1.78	U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x1, x5)
3.43/1.78	
3.43/1.78	
3.43/1.78	We have to consider all (P,R,Pi)-chains
3.43/1.78	----------------------------------------
3.43/1.78	
3.43/1.78	(5) DependencyGraphProof (EQUIVALENT)
3.43/1.78	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.
3.43/1.78	----------------------------------------
3.43/1.78	
3.43/1.78	(6)
3.43/1.78	Obligation:
3.43/1.78	Pi DP problem:
3.43/1.78	The TRS P consists of the following rules:
3.43/1.78	
3.43/1.78	   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs)
3.43/1.78	
3.43/1.78	The TRS R consists of the following rules:
3.43/1.78	
3.43/1.78	   suffix_in_ag(Xs, Ys) -> U1_ag(Xs, Ys, app_in_aag(X1, Xs, Ys))
3.43/1.78	   app_in_aag([], X, X) -> app_out_aag([], X, X)
3.43/1.78	   app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U2_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
3.43/1.78	   U2_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs))
3.43/1.78	   U1_ag(Xs, Ys, app_out_aag(X1, Xs, Ys)) -> suffix_out_ag(Xs, Ys)
3.43/1.78	
3.43/1.78	The argument filtering Pi contains the following mapping:
3.43/1.78	suffix_in_ag(x1, x2)  =  suffix_in_ag(x2)
3.43/1.78	
3.43/1.78	U1_ag(x1, x2, x3)  =  U1_ag(x3)
3.43/1.78	
3.43/1.78	app_in_aag(x1, x2, x3)  =  app_in_aag(x3)
3.43/1.78	
3.43/1.78	app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2)
3.43/1.78	
3.43/1.78	.(x1, x2)  =  .(x1, x2)
3.43/1.78	
3.43/1.78	U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x5)
3.43/1.78	
3.43/1.78	suffix_out_ag(x1, x2)  =  suffix_out_ag(x1)
3.43/1.78	
3.43/1.78	APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)
3.43/1.78	
3.43/1.78	
3.43/1.78	We have to consider all (P,R,Pi)-chains
3.43/1.78	----------------------------------------
3.43/1.78	
3.43/1.78	(7) UsableRulesProof (EQUIVALENT)
3.43/1.78	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.43/1.78	----------------------------------------
3.43/1.78	
3.43/1.78	(8)
3.43/1.78	Obligation:
3.43/1.78	Pi DP problem:
3.43/1.78	The TRS P consists of the following rules:
3.43/1.78	
3.43/1.78	   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs)
3.43/1.78	
3.43/1.78	R is empty.
3.43/1.78	The argument filtering Pi contains the following mapping:
3.43/1.78	.(x1, x2)  =  .(x1, x2)
3.43/1.78	
3.43/1.78	APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)
3.43/1.78	
3.43/1.78	
3.43/1.78	We have to consider all (P,R,Pi)-chains
3.43/1.78	----------------------------------------
3.43/1.78	
3.43/1.78	(9) PiDPToQDPProof (SOUND)
3.43/1.78	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.43/1.78	----------------------------------------
3.43/1.78	
3.43/1.78	(10)
3.43/1.78	Obligation:
3.43/1.78	Q DP problem:
3.43/1.78	The TRS P consists of the following rules:
3.43/1.78	
3.43/1.78	   APP_IN_AAG(.(X, Zs)) -> APP_IN_AAG(Zs)
3.43/1.78	
3.43/1.78	R is empty.
3.43/1.78	Q is empty.
3.43/1.78	We have to consider all (P,Q,R)-chains.
3.43/1.78	----------------------------------------
3.43/1.78	
3.43/1.78	(11) QDPSizeChangeProof (EQUIVALENT)
3.43/1.78	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. 
3.43/1.78	
3.43/1.78	From the DPs we obtained the following set of size-change graphs:
3.43/1.78	*APP_IN_AAG(.(X, Zs)) -> APP_IN_AAG(Zs)
3.43/1.78	The graph contains the following edges 1 > 1
3.43/1.78	
3.43/1.78	
3.43/1.78	----------------------------------------
3.43/1.78	
3.43/1.78	(12)
3.43/1.78	YES
3.78/1.81	EOF