3.72/1.92	YES
3.72/1.93	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.72/1.93	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.72/1.93	
3.72/1.93	
3.72/1.93	Left Termination of the query pattern
3.72/1.93	
3.72/1.93	app(g,a,a)
3.72/1.93	
3.72/1.93	w.r.t. the given Prolog program could successfully be proven:
3.72/1.93	
3.72/1.93	(0) Prolog
3.72/1.93	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.72/1.93	(2) PiTRS
3.72/1.93	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.72/1.93	(4) PiDP
3.72/1.93	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.72/1.93	(6) PiDP
3.72/1.93	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.72/1.93	(8) PiDP
3.72/1.93	(9) PiDPToQDPProof [SOUND, 0 ms]
3.72/1.93	(10) QDP
3.72/1.93	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.72/1.93	(12) YES
3.72/1.93	
3.72/1.93	
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(0)
3.72/1.93	Obligation:
3.72/1.93	Clauses:
3.72/1.93	
3.72/1.93	app([], X, X).
3.72/1.93	app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
3.72/1.93	
3.72/1.93	
3.72/1.93	Query: app(g,a,a)
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(1) PrologToPiTRSProof (SOUND)
3.72/1.93	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.72/1.93	
3.72/1.93	app_in_3: (b,f,f)
3.72/1.93	
3.72/1.93	Transforming Prolog into the following Term Rewriting System:
3.72/1.93	
3.72/1.93	Pi-finite rewrite system:
3.72/1.93	The TRS R consists of the following rules:
3.72/1.93	
3.72/1.93	   app_in_gaa([], X, X) -> app_out_gaa([], X, X)
3.72/1.93	   app_in_gaa(.(X, Xs), Ys, .(X, Zs)) -> U1_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
3.72/1.93	   U1_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) -> app_out_gaa(.(X, Xs), Ys, .(X, Zs))
3.72/1.93	
3.72/1.93	The argument filtering Pi contains the following mapping:
3.72/1.93	app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
3.72/1.93	
3.72/1.93	[]  =  []
3.72/1.93	
3.72/1.93	app_out_gaa(x1, x2, x3)  =  app_out_gaa
3.72/1.93	
3.72/1.93	.(x1, x2)  =  .(x1, x2)
3.72/1.93	
3.72/1.93	U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
3.72/1.93	
3.72/1.93	
3.72/1.93	
3.72/1.93	
3.72/1.93	
3.72/1.93	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.72/1.93	
3.72/1.93	
3.72/1.93	
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(2)
3.72/1.93	Obligation:
3.72/1.93	Pi-finite rewrite system:
3.72/1.93	The TRS R consists of the following rules:
3.72/1.93	
3.72/1.93	   app_in_gaa([], X, X) -> app_out_gaa([], X, X)
3.72/1.93	   app_in_gaa(.(X, Xs), Ys, .(X, Zs)) -> U1_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
3.72/1.93	   U1_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) -> app_out_gaa(.(X, Xs), Ys, .(X, Zs))
3.72/1.93	
3.72/1.93	The argument filtering Pi contains the following mapping:
3.72/1.93	app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
3.72/1.93	
3.72/1.93	[]  =  []
3.72/1.93	
3.72/1.93	app_out_gaa(x1, x2, x3)  =  app_out_gaa
3.72/1.93	
3.72/1.93	.(x1, x2)  =  .(x1, x2)
3.72/1.93	
3.72/1.93	U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
3.72/1.93	
3.72/1.93	
3.72/1.93	
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(3) DependencyPairsProof (EQUIVALENT)
3.72/1.93	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.72/1.93	Pi DP problem:
3.72/1.93	The TRS P consists of the following rules:
3.72/1.93	
3.72/1.93	   APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) -> U1_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
3.72/1.93	   APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GAA(Xs, Ys, Zs)
3.72/1.93	
3.72/1.93	The TRS R consists of the following rules:
3.72/1.93	
3.72/1.93	   app_in_gaa([], X, X) -> app_out_gaa([], X, X)
3.72/1.93	   app_in_gaa(.(X, Xs), Ys, .(X, Zs)) -> U1_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
3.72/1.93	   U1_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) -> app_out_gaa(.(X, Xs), Ys, .(X, Zs))
3.72/1.93	
3.72/1.93	The argument filtering Pi contains the following mapping:
3.72/1.93	app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
3.72/1.93	
3.72/1.93	[]  =  []
3.72/1.93	
3.72/1.93	app_out_gaa(x1, x2, x3)  =  app_out_gaa
3.72/1.93	
3.72/1.93	.(x1, x2)  =  .(x1, x2)
3.72/1.93	
3.72/1.93	U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
3.72/1.93	
3.72/1.93	APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
3.72/1.93	
3.72/1.93	U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x5)
3.72/1.93	
3.72/1.93	
3.72/1.93	We have to consider all (P,R,Pi)-chains
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(4)
3.72/1.93	Obligation:
3.72/1.93	Pi DP problem:
3.72/1.93	The TRS P consists of the following rules:
3.72/1.93	
3.72/1.93	   APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) -> U1_GAA(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
3.72/1.93	   APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GAA(Xs, Ys, Zs)
3.72/1.93	
3.72/1.93	The TRS R consists of the following rules:
3.72/1.93	
3.72/1.93	   app_in_gaa([], X, X) -> app_out_gaa([], X, X)
3.72/1.93	   app_in_gaa(.(X, Xs), Ys, .(X, Zs)) -> U1_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
3.72/1.93	   U1_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) -> app_out_gaa(.(X, Xs), Ys, .(X, Zs))
3.72/1.93	
3.72/1.93	The argument filtering Pi contains the following mapping:
3.72/1.93	app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
3.72/1.93	
3.72/1.93	[]  =  []
3.72/1.93	
3.72/1.93	app_out_gaa(x1, x2, x3)  =  app_out_gaa
3.72/1.93	
3.72/1.93	.(x1, x2)  =  .(x1, x2)
3.72/1.93	
3.72/1.93	U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
3.72/1.93	
3.72/1.93	APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
3.72/1.93	
3.72/1.93	U1_GAA(x1, x2, x3, x4, x5)  =  U1_GAA(x5)
3.72/1.93	
3.72/1.93	
3.72/1.93	We have to consider all (P,R,Pi)-chains
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(5) DependencyGraphProof (EQUIVALENT)
3.72/1.93	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(6)
3.72/1.93	Obligation:
3.72/1.93	Pi DP problem:
3.72/1.93	The TRS P consists of the following rules:
3.72/1.93	
3.72/1.93	   APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GAA(Xs, Ys, Zs)
3.72/1.93	
3.72/1.93	The TRS R consists of the following rules:
3.72/1.93	
3.72/1.93	   app_in_gaa([], X, X) -> app_out_gaa([], X, X)
3.72/1.93	   app_in_gaa(.(X, Xs), Ys, .(X, Zs)) -> U1_gaa(X, Xs, Ys, Zs, app_in_gaa(Xs, Ys, Zs))
3.72/1.93	   U1_gaa(X, Xs, Ys, Zs, app_out_gaa(Xs, Ys, Zs)) -> app_out_gaa(.(X, Xs), Ys, .(X, Zs))
3.72/1.93	
3.72/1.93	The argument filtering Pi contains the following mapping:
3.72/1.93	app_in_gaa(x1, x2, x3)  =  app_in_gaa(x1)
3.72/1.93	
3.72/1.93	[]  =  []
3.72/1.93	
3.72/1.93	app_out_gaa(x1, x2, x3)  =  app_out_gaa
3.72/1.93	
3.72/1.93	.(x1, x2)  =  .(x1, x2)
3.72/1.93	
3.72/1.93	U1_gaa(x1, x2, x3, x4, x5)  =  U1_gaa(x5)
3.72/1.93	
3.72/1.93	APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
3.72/1.93	
3.72/1.93	
3.72/1.93	We have to consider all (P,R,Pi)-chains
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(7) UsableRulesProof (EQUIVALENT)
3.72/1.93	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(8)
3.72/1.93	Obligation:
3.72/1.93	Pi DP problem:
3.72/1.93	The TRS P consists of the following rules:
3.72/1.93	
3.72/1.93	   APP_IN_GAA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GAA(Xs, Ys, Zs)
3.72/1.93	
3.72/1.93	R is empty.
3.72/1.93	The argument filtering Pi contains the following mapping:
3.72/1.93	.(x1, x2)  =  .(x1, x2)
3.72/1.93	
3.72/1.93	APP_IN_GAA(x1, x2, x3)  =  APP_IN_GAA(x1)
3.72/1.93	
3.72/1.93	
3.72/1.93	We have to consider all (P,R,Pi)-chains
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(9) PiDPToQDPProof (SOUND)
3.72/1.93	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(10)
3.72/1.93	Obligation:
3.72/1.93	Q DP problem:
3.72/1.93	The TRS P consists of the following rules:
3.72/1.93	
3.72/1.93	   APP_IN_GAA(.(X, Xs)) -> APP_IN_GAA(Xs)
3.72/1.93	
3.72/1.93	R is empty.
3.72/1.93	Q is empty.
3.72/1.93	We have to consider all (P,Q,R)-chains.
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(11) QDPSizeChangeProof (EQUIVALENT)
3.72/1.93	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. 
3.72/1.93	
3.72/1.93	From the DPs we obtained the following set of size-change graphs:
3.72/1.93	*APP_IN_GAA(.(X, Xs)) -> APP_IN_GAA(Xs)
3.72/1.93	The graph contains the following edges 1 > 1
3.72/1.93	
3.72/1.93	
3.72/1.93	----------------------------------------
3.72/1.93	
3.72/1.93	(12)
3.72/1.93	YES
4.05/1.96	EOF