3.93/1.77	YES
3.93/1.79	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.93/1.79	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.93/1.79	
3.93/1.79	
3.93/1.79	Left Termination of the query pattern
3.93/1.79	
3.93/1.79	app3_a(g,g,g,a)
3.93/1.79	
3.93/1.79	w.r.t. the given Prolog program could successfully be proven:
3.93/1.79	
3.93/1.79	(0) Prolog
3.93/1.79	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.93/1.79	(2) PiTRS
3.93/1.79	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.93/1.79	(4) PiDP
3.93/1.79	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.93/1.79	(6) PiDP
3.93/1.79	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.93/1.79	(8) PiDP
3.93/1.79	(9) PiDPToQDPProof [SOUND, 0 ms]
3.93/1.79	(10) QDP
3.93/1.79	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.93/1.79	(12) YES
3.93/1.79	
3.93/1.79	
3.93/1.79	----------------------------------------
3.93/1.79	
3.93/1.79	(0)
3.93/1.79	Obligation:
3.93/1.79	Clauses:
3.93/1.79	
3.93/1.79	app3_a(Xs, Ys, Zs, Us) :- ','(app(Xs, Ys, Vs), app(Vs, Zs, Us)).
3.93/1.79	app3_b(Xs, Ys, Zs, Us) :- ','(app(Ys, Zs, Vs), app(Xs, Vs, Us)).
3.93/1.79	app([], Ys, Ys).
3.93/1.79	app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
3.93/1.79	
3.93/1.79	
3.93/1.79	Query: app3_a(g,g,g,a)
3.93/1.79	----------------------------------------
3.93/1.79	
3.93/1.79	(1) PrologToPiTRSProof (SOUND)
3.93/1.79	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.93/1.79	
3.93/1.79	app3_a_in_4: (b,b,b,f)
3.93/1.79	
3.93/1.79	app_in_3: (b,b,f)
3.93/1.79	
3.93/1.79	Transforming Prolog into the following Term Rewriting System:
3.93/1.79	
3.93/1.79	Pi-finite rewrite system:
3.93/1.79	The TRS R consists of the following rules:
3.93/1.79	
3.93/1.79	   app3_a_in_ggga(Xs, Ys, Zs, Us) -> U1_ggga(Xs, Ys, Zs, Us, app_in_gga(Xs, Ys, Vs))
3.93/1.79	   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
3.93/1.79	   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U5_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
3.93/1.79	   U5_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
3.93/1.79	   U1_ggga(Xs, Ys, Zs, Us, app_out_gga(Xs, Ys, Vs)) -> U2_ggga(Xs, Ys, Zs, Us, app_in_gga(Vs, Zs, Us))
3.93/1.79	   U2_ggga(Xs, Ys, Zs, Us, app_out_gga(Vs, Zs, Us)) -> app3_a_out_ggga(Xs, Ys, Zs, Us)
3.93/1.79	
3.93/1.79	The argument filtering Pi contains the following mapping:
3.93/1.79	app3_a_in_ggga(x1, x2, x3, x4)  =  app3_a_in_ggga(x1, x2, x3)
3.93/1.79	
3.93/1.79	U1_ggga(x1, x2, x3, x4, x5)  =  U1_ggga(x3, x5)
3.93/1.79	
3.93/1.79	app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
3.93/1.79	
3.93/1.79	[]  =  []
3.93/1.79	
3.93/1.79	app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
3.93/1.79	
3.93/1.79	.(x1, x2)  =  .(x1, x2)
3.93/1.79	
3.93/1.79	U5_gga(x1, x2, x3, x4, x5)  =  U5_gga(x1, x5)
3.93/1.79	
3.93/1.79	U2_ggga(x1, x2, x3, x4, x5)  =  U2_ggga(x5)
3.93/1.79	
3.93/1.79	app3_a_out_ggga(x1, x2, x3, x4)  =  app3_a_out_ggga(x4)
3.93/1.79	
3.93/1.79	
3.93/1.79	
3.93/1.79	
3.93/1.79	
3.93/1.79	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.93/1.79	
3.93/1.79	
3.93/1.79	
3.93/1.79	----------------------------------------
3.93/1.79	
3.93/1.79	(2)
3.93/1.79	Obligation:
3.93/1.79	Pi-finite rewrite system:
3.93/1.79	The TRS R consists of the following rules:
3.93/1.79	
3.93/1.79	   app3_a_in_ggga(Xs, Ys, Zs, Us) -> U1_ggga(Xs, Ys, Zs, Us, app_in_gga(Xs, Ys, Vs))
3.93/1.79	   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
3.93/1.79	   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U5_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
3.93/1.79	   U5_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
3.93/1.79	   U1_ggga(Xs, Ys, Zs, Us, app_out_gga(Xs, Ys, Vs)) -> U2_ggga(Xs, Ys, Zs, Us, app_in_gga(Vs, Zs, Us))
3.93/1.79	   U2_ggga(Xs, Ys, Zs, Us, app_out_gga(Vs, Zs, Us)) -> app3_a_out_ggga(Xs, Ys, Zs, Us)
3.93/1.79	
3.93/1.79	The argument filtering Pi contains the following mapping:
3.93/1.79	app3_a_in_ggga(x1, x2, x3, x4)  =  app3_a_in_ggga(x1, x2, x3)
3.93/1.79	
3.93/1.79	U1_ggga(x1, x2, x3, x4, x5)  =  U1_ggga(x3, x5)
3.93/1.79	
3.93/1.79	app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
3.93/1.79	
3.93/1.79	[]  =  []
3.93/1.79	
3.93/1.79	app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
3.93/1.79	
3.93/1.79	.(x1, x2)  =  .(x1, x2)
3.93/1.79	
3.93/1.79	U5_gga(x1, x2, x3, x4, x5)  =  U5_gga(x1, x5)
3.93/1.79	
3.93/1.79	U2_ggga(x1, x2, x3, x4, x5)  =  U2_ggga(x5)
3.93/1.79	
3.93/1.79	app3_a_out_ggga(x1, x2, x3, x4)  =  app3_a_out_ggga(x4)
3.93/1.79	
3.93/1.79	
3.93/1.79	
3.93/1.79	----------------------------------------
3.93/1.79	
3.93/1.79	(3) DependencyPairsProof (EQUIVALENT)
3.93/1.79	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.93/1.79	Pi DP problem:
3.93/1.79	The TRS P consists of the following rules:
3.93/1.79	
3.93/1.79	   APP3_A_IN_GGGA(Xs, Ys, Zs, Us) -> U1_GGGA(Xs, Ys, Zs, Us, app_in_gga(Xs, Ys, Vs))
3.93/1.79	   APP3_A_IN_GGGA(Xs, Ys, Zs, Us) -> APP_IN_GGA(Xs, Ys, Vs)
3.93/1.79	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> U5_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
3.93/1.79	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs)
3.93/1.79	   U1_GGGA(Xs, Ys, Zs, Us, app_out_gga(Xs, Ys, Vs)) -> U2_GGGA(Xs, Ys, Zs, Us, app_in_gga(Vs, Zs, Us))
3.93/1.79	   U1_GGGA(Xs, Ys, Zs, Us, app_out_gga(Xs, Ys, Vs)) -> APP_IN_GGA(Vs, Zs, Us)
3.93/1.79	
3.93/1.79	The TRS R consists of the following rules:
3.93/1.79	
3.93/1.79	   app3_a_in_ggga(Xs, Ys, Zs, Us) -> U1_ggga(Xs, Ys, Zs, Us, app_in_gga(Xs, Ys, Vs))
3.93/1.79	   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
3.93/1.79	   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U5_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
3.93/1.79	   U5_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
3.93/1.79	   U1_ggga(Xs, Ys, Zs, Us, app_out_gga(Xs, Ys, Vs)) -> U2_ggga(Xs, Ys, Zs, Us, app_in_gga(Vs, Zs, Us))
3.93/1.79	   U2_ggga(Xs, Ys, Zs, Us, app_out_gga(Vs, Zs, Us)) -> app3_a_out_ggga(Xs, Ys, Zs, Us)
3.93/1.79	
3.93/1.79	The argument filtering Pi contains the following mapping:
3.93/1.79	app3_a_in_ggga(x1, x2, x3, x4)  =  app3_a_in_ggga(x1, x2, x3)
3.93/1.79	
3.93/1.79	U1_ggga(x1, x2, x3, x4, x5)  =  U1_ggga(x3, x5)
3.93/1.79	
3.93/1.79	app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
3.93/1.79	
3.93/1.79	[]  =  []
3.93/1.79	
3.93/1.79	app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
3.93/1.79	
3.93/1.79	.(x1, x2)  =  .(x1, x2)
3.93/1.79	
3.93/1.79	U5_gga(x1, x2, x3, x4, x5)  =  U5_gga(x1, x5)
3.93/1.79	
3.93/1.79	U2_ggga(x1, x2, x3, x4, x5)  =  U2_ggga(x5)
3.93/1.79	
3.93/1.79	app3_a_out_ggga(x1, x2, x3, x4)  =  app3_a_out_ggga(x4)
3.93/1.79	
3.93/1.79	APP3_A_IN_GGGA(x1, x2, x3, x4)  =  APP3_A_IN_GGGA(x1, x2, x3)
3.93/1.79	
3.93/1.79	U1_GGGA(x1, x2, x3, x4, x5)  =  U1_GGGA(x3, x5)
3.93/1.79	
3.93/1.79	APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
3.93/1.79	
3.93/1.79	U5_GGA(x1, x2, x3, x4, x5)  =  U5_GGA(x1, x5)
3.93/1.79	
3.93/1.79	U2_GGGA(x1, x2, x3, x4, x5)  =  U2_GGGA(x5)
3.93/1.79	
3.93/1.79	
3.93/1.79	We have to consider all (P,R,Pi)-chains
3.93/1.79	----------------------------------------
3.93/1.79	
3.93/1.79	(4)
3.93/1.79	Obligation:
3.93/1.79	Pi DP problem:
3.93/1.79	The TRS P consists of the following rules:
3.93/1.79	
3.93/1.79	   APP3_A_IN_GGGA(Xs, Ys, Zs, Us) -> U1_GGGA(Xs, Ys, Zs, Us, app_in_gga(Xs, Ys, Vs))
3.93/1.79	   APP3_A_IN_GGGA(Xs, Ys, Zs, Us) -> APP_IN_GGA(Xs, Ys, Vs)
3.93/1.79	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> U5_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
3.93/1.79	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs)
3.93/1.79	   U1_GGGA(Xs, Ys, Zs, Us, app_out_gga(Xs, Ys, Vs)) -> U2_GGGA(Xs, Ys, Zs, Us, app_in_gga(Vs, Zs, Us))
3.93/1.79	   U1_GGGA(Xs, Ys, Zs, Us, app_out_gga(Xs, Ys, Vs)) -> APP_IN_GGA(Vs, Zs, Us)
3.93/1.79	
3.93/1.79	The TRS R consists of the following rules:
3.93/1.79	
3.93/1.79	   app3_a_in_ggga(Xs, Ys, Zs, Us) -> U1_ggga(Xs, Ys, Zs, Us, app_in_gga(Xs, Ys, Vs))
3.93/1.79	   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
3.93/1.79	   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U5_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
3.93/1.79	   U5_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
3.93/1.79	   U1_ggga(Xs, Ys, Zs, Us, app_out_gga(Xs, Ys, Vs)) -> U2_ggga(Xs, Ys, Zs, Us, app_in_gga(Vs, Zs, Us))
3.93/1.79	   U2_ggga(Xs, Ys, Zs, Us, app_out_gga(Vs, Zs, Us)) -> app3_a_out_ggga(Xs, Ys, Zs, Us)
3.93/1.79	
3.93/1.79	The argument filtering Pi contains the following mapping:
3.93/1.79	app3_a_in_ggga(x1, x2, x3, x4)  =  app3_a_in_ggga(x1, x2, x3)
3.93/1.79	
3.93/1.79	U1_ggga(x1, x2, x3, x4, x5)  =  U1_ggga(x3, x5)
3.93/1.79	
3.93/1.79	app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
3.93/1.79	
3.93/1.79	[]  =  []
3.93/1.79	
3.93/1.79	app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
3.93/1.79	
3.93/1.79	.(x1, x2)  =  .(x1, x2)
3.93/1.79	
3.93/1.79	U5_gga(x1, x2, x3, x4, x5)  =  U5_gga(x1, x5)
3.93/1.79	
3.93/1.79	U2_ggga(x1, x2, x3, x4, x5)  =  U2_ggga(x5)
3.93/1.79	
3.93/1.79	app3_a_out_ggga(x1, x2, x3, x4)  =  app3_a_out_ggga(x4)
3.93/1.79	
3.93/1.79	APP3_A_IN_GGGA(x1, x2, x3, x4)  =  APP3_A_IN_GGGA(x1, x2, x3)
3.93/1.79	
3.93/1.79	U1_GGGA(x1, x2, x3, x4, x5)  =  U1_GGGA(x3, x5)
3.93/1.79	
3.93/1.79	APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
3.93/1.79	
3.93/1.79	U5_GGA(x1, x2, x3, x4, x5)  =  U5_GGA(x1, x5)
3.93/1.79	
3.93/1.79	U2_GGGA(x1, x2, x3, x4, x5)  =  U2_GGGA(x5)
3.93/1.79	
3.93/1.79	
3.93/1.79	We have to consider all (P,R,Pi)-chains
3.93/1.79	----------------------------------------
3.93/1.79	
3.93/1.79	(5) DependencyGraphProof (EQUIVALENT)
3.93/1.79	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.
3.93/1.79	----------------------------------------
3.93/1.79	
3.93/1.79	(6)
3.93/1.79	Obligation:
3.93/1.79	Pi DP problem:
3.93/1.79	The TRS P consists of the following rules:
3.93/1.79	
3.93/1.79	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs)
3.93/1.79	
3.93/1.79	The TRS R consists of the following rules:
3.93/1.79	
3.93/1.79	   app3_a_in_ggga(Xs, Ys, Zs, Us) -> U1_ggga(Xs, Ys, Zs, Us, app_in_gga(Xs, Ys, Vs))
3.93/1.79	   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
3.93/1.79	   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U5_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
3.93/1.79	   U5_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
3.93/1.79	   U1_ggga(Xs, Ys, Zs, Us, app_out_gga(Xs, Ys, Vs)) -> U2_ggga(Xs, Ys, Zs, Us, app_in_gga(Vs, Zs, Us))
3.93/1.79	   U2_ggga(Xs, Ys, Zs, Us, app_out_gga(Vs, Zs, Us)) -> app3_a_out_ggga(Xs, Ys, Zs, Us)
3.93/1.79	
3.93/1.79	The argument filtering Pi contains the following mapping:
3.93/1.79	app3_a_in_ggga(x1, x2, x3, x4)  =  app3_a_in_ggga(x1, x2, x3)
3.93/1.79	
3.93/1.79	U1_ggga(x1, x2, x3, x4, x5)  =  U1_ggga(x3, x5)
3.93/1.79	
3.93/1.79	app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
3.93/1.79	
3.93/1.79	[]  =  []
3.93/1.79	
3.93/1.79	app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
3.93/1.79	
3.93/1.79	.(x1, x2)  =  .(x1, x2)
3.93/1.79	
3.93/1.79	U5_gga(x1, x2, x3, x4, x5)  =  U5_gga(x1, x5)
3.93/1.79	
3.93/1.79	U2_ggga(x1, x2, x3, x4, x5)  =  U2_ggga(x5)
3.93/1.79	
3.93/1.79	app3_a_out_ggga(x1, x2, x3, x4)  =  app3_a_out_ggga(x4)
3.93/1.79	
3.93/1.79	APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
3.93/1.79	
3.93/1.79	
3.93/1.79	We have to consider all (P,R,Pi)-chains
3.93/1.79	----------------------------------------
3.93/1.79	
3.93/1.79	(7) UsableRulesProof (EQUIVALENT)
3.93/1.79	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.93/1.79	----------------------------------------
3.93/1.79	
3.93/1.79	(8)
3.93/1.79	Obligation:
3.93/1.79	Pi DP problem:
3.93/1.79	The TRS P consists of the following rules:
3.93/1.79	
3.93/1.79	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs)
3.93/1.79	
3.93/1.79	R is empty.
3.93/1.79	The argument filtering Pi contains the following mapping:
3.93/1.79	.(x1, x2)  =  .(x1, x2)
3.93/1.79	
3.93/1.79	APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
3.93/1.79	
3.93/1.79	
3.93/1.79	We have to consider all (P,R,Pi)-chains
3.93/1.79	----------------------------------------
3.93/1.79	
3.93/1.79	(9) PiDPToQDPProof (SOUND)
3.93/1.79	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.93/1.79	----------------------------------------
3.93/1.79	
3.93/1.79	(10)
3.93/1.79	Obligation:
3.93/1.79	Q DP problem:
3.93/1.79	The TRS P consists of the following rules:
3.93/1.79	
3.93/1.79	   APP_IN_GGA(.(X, Xs), Ys) -> APP_IN_GGA(Xs, Ys)
3.93/1.79	
3.93/1.79	R is empty.
3.93/1.79	Q is empty.
3.93/1.79	We have to consider all (P,Q,R)-chains.
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3.93/1.79	
3.93/1.79	(11) QDPSizeChangeProof (EQUIVALENT)
3.93/1.79	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. 
3.93/1.79	
3.93/1.79	From the DPs we obtained the following set of size-change graphs:
3.93/1.79	*APP_IN_GGA(.(X, Xs), Ys) -> APP_IN_GGA(Xs, Ys)
3.93/1.79	The graph contains the following edges 1 > 1, 2 >= 2
3.93/1.79	
3.93/1.79	
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3.93/1.79	
3.93/1.79	(12)
3.93/1.79	YES
3.93/1.82	EOF