4.17/1.87	YES
4.43/1.97	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
4.43/1.97	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.43/1.97	
4.43/1.97	
4.43/1.97	Left Termination of the query pattern
4.43/1.97	
4.43/1.97	reverse(g,a)
4.43/1.97	
4.43/1.97	w.r.t. the given Prolog program could successfully be proven:
4.43/1.97	
4.43/1.97	(0) Prolog
4.43/1.97	(1) PrologToPiTRSProof [SOUND, 0 ms]
4.43/1.97	(2) PiTRS
4.43/1.97	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
4.43/1.97	(4) PiDP
4.43/1.97	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
4.43/1.97	(6) AND
4.43/1.97	    (7) PiDP
4.43/1.97	        (8) UsableRulesProof [EQUIVALENT, 0 ms]
4.43/1.97	        (9) PiDP
4.43/1.97	        (10) PiDPToQDPProof [SOUND, 0 ms]
4.43/1.97	        (11) QDP
4.43/1.97	        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
4.43/1.97	        (13) YES
4.43/1.97	    (14) PiDP
4.43/1.97	        (15) UsableRulesProof [EQUIVALENT, 0 ms]
4.43/1.97	        (16) PiDP
4.43/1.97	        (17) PiDPToQDPProof [SOUND, 0 ms]
4.43/1.97	        (18) QDP
4.43/1.97	        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
4.43/1.97	        (20) YES
4.43/1.97	
4.43/1.97	
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(0)
4.43/1.97	Obligation:
4.43/1.97	Clauses:
4.43/1.97	
4.43/1.97	app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
4.43/1.97	app([], Ys, Ys).
4.43/1.97	reverse(.(X, Xs), Ys) :- ','(reverse(Xs, Zs), app(Zs, .(X, []), Ys)).
4.43/1.97	reverse([], []).
4.43/1.97	
4.43/1.97	
4.43/1.97	Query: reverse(g,a)
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(1) PrologToPiTRSProof (SOUND)
4.43/1.97	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
4.43/1.97	
4.43/1.97	reverse_in_2: (b,f)
4.43/1.97	
4.43/1.97	app_in_3: (b,b,f)
4.43/1.97	
4.43/1.97	Transforming Prolog into the following Term Rewriting System:
4.43/1.97	
4.43/1.97	Pi-finite rewrite system:
4.43/1.97	The TRS R consists of the following rules:
4.43/1.97	
4.43/1.97	   reverse_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
4.43/1.97	   reverse_in_ga([], []) -> reverse_out_ga([], [])
4.43/1.97	   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
4.43/1.97	   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
4.43/1.97	   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
4.43/1.97	   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
4.43/1.97	   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)
4.43/1.97	
4.43/1.97	The argument filtering Pi contains the following mapping:
4.43/1.97	reverse_in_ga(x1, x2)  =  reverse_in_ga(x1)
4.43/1.97	
4.43/1.97	.(x1, x2)  =  .(x1, x2)
4.43/1.97	
4.43/1.97	U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
4.43/1.97	
4.43/1.97	[]  =  []
4.43/1.97	
4.43/1.97	reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)
4.43/1.97	
4.43/1.97	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
4.43/1.97	
4.43/1.97	app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
4.43/1.97	
4.43/1.97	U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)
4.43/1.97	
4.43/1.97	app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
4.43/1.97	
4.43/1.97	
4.43/1.97	
4.43/1.97	
4.43/1.97	
4.43/1.97	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
4.43/1.97	
4.43/1.97	
4.43/1.97	
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(2)
4.43/1.97	Obligation:
4.43/1.97	Pi-finite rewrite system:
4.43/1.97	The TRS R consists of the following rules:
4.43/1.97	
4.43/1.97	   reverse_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
4.43/1.97	   reverse_in_ga([], []) -> reverse_out_ga([], [])
4.43/1.97	   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
4.43/1.97	   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
4.43/1.97	   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
4.43/1.97	   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
4.43/1.97	   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)
4.43/1.97	
4.43/1.97	The argument filtering Pi contains the following mapping:
4.43/1.97	reverse_in_ga(x1, x2)  =  reverse_in_ga(x1)
4.43/1.97	
4.43/1.97	.(x1, x2)  =  .(x1, x2)
4.43/1.97	
4.43/1.97	U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
4.43/1.97	
4.43/1.97	[]  =  []
4.43/1.97	
4.43/1.97	reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)
4.43/1.97	
4.43/1.97	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
4.43/1.97	
4.43/1.97	app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
4.43/1.97	
4.43/1.97	U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)
4.43/1.97	
4.43/1.97	app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
4.43/1.97	
4.43/1.97	
4.43/1.97	
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(3) DependencyPairsProof (EQUIVALENT)
4.43/1.97	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
4.43/1.97	Pi DP problem:
4.43/1.97	The TRS P consists of the following rules:
4.43/1.97	
4.43/1.97	   REVERSE_IN_GA(.(X, Xs), Ys) -> U2_GA(X, Xs, Ys, reverse_in_ga(Xs, Zs))
4.43/1.97	   REVERSE_IN_GA(.(X, Xs), Ys) -> REVERSE_IN_GA(Xs, Zs)
4.43/1.97	   U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_GA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
4.43/1.97	   U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> APP_IN_GGA(Zs, .(X, []), Ys)
4.43/1.97	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> U1_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
4.43/1.97	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs)
4.43/1.97	
4.43/1.97	The TRS R consists of the following rules:
4.43/1.97	
4.43/1.97	   reverse_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
4.43/1.97	   reverse_in_ga([], []) -> reverse_out_ga([], [])
4.43/1.97	   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
4.43/1.97	   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
4.43/1.97	   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
4.43/1.97	   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
4.43/1.97	   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)
4.43/1.97	
4.43/1.97	The argument filtering Pi contains the following mapping:
4.43/1.97	reverse_in_ga(x1, x2)  =  reverse_in_ga(x1)
4.43/1.97	
4.43/1.97	.(x1, x2)  =  .(x1, x2)
4.43/1.97	
4.43/1.97	U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
4.43/1.97	
4.43/1.97	[]  =  []
4.43/1.97	
4.43/1.97	reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)
4.43/1.97	
4.43/1.97	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
4.43/1.97	
4.43/1.97	app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
4.43/1.97	
4.43/1.97	U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)
4.43/1.97	
4.43/1.97	app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
4.43/1.97	
4.43/1.97	REVERSE_IN_GA(x1, x2)  =  REVERSE_IN_GA(x1)
4.43/1.97	
4.43/1.97	U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x4)
4.43/1.97	
4.43/1.97	U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)
4.43/1.97	
4.43/1.97	APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
4.43/1.97	
4.43/1.97	U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x1, x5)
4.43/1.97	
4.43/1.97	
4.43/1.97	We have to consider all (P,R,Pi)-chains
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(4)
4.43/1.97	Obligation:
4.43/1.97	Pi DP problem:
4.43/1.97	The TRS P consists of the following rules:
4.43/1.97	
4.43/1.97	   REVERSE_IN_GA(.(X, Xs), Ys) -> U2_GA(X, Xs, Ys, reverse_in_ga(Xs, Zs))
4.43/1.97	   REVERSE_IN_GA(.(X, Xs), Ys) -> REVERSE_IN_GA(Xs, Zs)
4.43/1.97	   U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_GA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
4.43/1.97	   U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> APP_IN_GGA(Zs, .(X, []), Ys)
4.43/1.97	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> U1_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
4.43/1.97	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs)
4.43/1.97	
4.43/1.97	The TRS R consists of the following rules:
4.43/1.97	
4.43/1.97	   reverse_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
4.43/1.97	   reverse_in_ga([], []) -> reverse_out_ga([], [])
4.43/1.97	   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
4.43/1.97	   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
4.43/1.97	   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
4.43/1.97	   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
4.43/1.97	   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)
4.43/1.97	
4.43/1.97	The argument filtering Pi contains the following mapping:
4.43/1.97	reverse_in_ga(x1, x2)  =  reverse_in_ga(x1)
4.43/1.97	
4.43/1.97	.(x1, x2)  =  .(x1, x2)
4.43/1.97	
4.43/1.97	U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
4.43/1.97	
4.43/1.97	[]  =  []
4.43/1.97	
4.43/1.97	reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)
4.43/1.97	
4.43/1.97	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
4.43/1.97	
4.43/1.97	app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
4.43/1.97	
4.43/1.97	U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)
4.43/1.97	
4.43/1.97	app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
4.43/1.97	
4.43/1.97	REVERSE_IN_GA(x1, x2)  =  REVERSE_IN_GA(x1)
4.43/1.97	
4.43/1.97	U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x4)
4.43/1.97	
4.43/1.97	U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)
4.43/1.97	
4.43/1.97	APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
4.43/1.97	
4.43/1.97	U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x1, x5)
4.43/1.97	
4.43/1.97	
4.43/1.97	We have to consider all (P,R,Pi)-chains
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(5) DependencyGraphProof (EQUIVALENT)
4.43/1.97	The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(6)
4.43/1.97	Complex Obligation (AND)
4.43/1.97	
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(7)
4.43/1.97	Obligation:
4.43/1.97	Pi DP problem:
4.43/1.97	The TRS P consists of the following rules:
4.43/1.97	
4.43/1.97	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs)
4.43/1.97	
4.43/1.97	The TRS R consists of the following rules:
4.43/1.97	
4.43/1.97	   reverse_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
4.43/1.97	   reverse_in_ga([], []) -> reverse_out_ga([], [])
4.43/1.97	   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
4.43/1.97	   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
4.43/1.97	   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
4.43/1.97	   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
4.43/1.97	   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)
4.43/1.97	
4.43/1.97	The argument filtering Pi contains the following mapping:
4.43/1.97	reverse_in_ga(x1, x2)  =  reverse_in_ga(x1)
4.43/1.97	
4.43/1.97	.(x1, x2)  =  .(x1, x2)
4.43/1.97	
4.43/1.97	U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
4.43/1.97	
4.43/1.97	[]  =  []
4.43/1.97	
4.43/1.97	reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)
4.43/1.97	
4.43/1.97	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
4.43/1.97	
4.43/1.97	app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
4.43/1.97	
4.43/1.97	U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)
4.43/1.97	
4.43/1.97	app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
4.43/1.97	
4.43/1.97	APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
4.43/1.97	
4.43/1.97	
4.43/1.97	We have to consider all (P,R,Pi)-chains
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(8) UsableRulesProof (EQUIVALENT)
4.43/1.97	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(9)
4.43/1.97	Obligation:
4.43/1.97	Pi DP problem:
4.43/1.97	The TRS P consists of the following rules:
4.43/1.97	
4.43/1.97	   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs)
4.43/1.97	
4.43/1.97	R is empty.
4.43/1.97	The argument filtering Pi contains the following mapping:
4.43/1.97	.(x1, x2)  =  .(x1, x2)
4.43/1.97	
4.43/1.97	APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)
4.43/1.97	
4.43/1.97	
4.43/1.97	We have to consider all (P,R,Pi)-chains
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(10) PiDPToQDPProof (SOUND)
4.43/1.97	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(11)
4.43/1.97	Obligation:
4.43/1.97	Q DP problem:
4.43/1.97	The TRS P consists of the following rules:
4.43/1.97	
4.43/1.97	   APP_IN_GGA(.(X, Xs), Ys) -> APP_IN_GGA(Xs, Ys)
4.43/1.97	
4.43/1.97	R is empty.
4.43/1.97	Q is empty.
4.43/1.97	We have to consider all (P,Q,R)-chains.
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(12) QDPSizeChangeProof (EQUIVALENT)
4.43/1.97	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. 
4.43/1.97	
4.43/1.97	From the DPs we obtained the following set of size-change graphs:
4.43/1.97	*APP_IN_GGA(.(X, Xs), Ys) -> APP_IN_GGA(Xs, Ys)
4.43/1.97	The graph contains the following edges 1 > 1, 2 >= 2
4.43/1.97	
4.43/1.97	
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(13)
4.43/1.97	YES
4.43/1.97	
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(14)
4.43/1.97	Obligation:
4.43/1.97	Pi DP problem:
4.43/1.97	The TRS P consists of the following rules:
4.43/1.97	
4.43/1.97	   REVERSE_IN_GA(.(X, Xs), Ys) -> REVERSE_IN_GA(Xs, Zs)
4.43/1.97	
4.43/1.97	The TRS R consists of the following rules:
4.43/1.97	
4.43/1.97	   reverse_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
4.43/1.97	   reverse_in_ga([], []) -> reverse_out_ga([], [])
4.43/1.97	   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
4.43/1.97	   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
4.43/1.97	   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
4.43/1.97	   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
4.43/1.97	   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)
4.43/1.97	
4.43/1.97	The argument filtering Pi contains the following mapping:
4.43/1.97	reverse_in_ga(x1, x2)  =  reverse_in_ga(x1)
4.43/1.97	
4.43/1.97	.(x1, x2)  =  .(x1, x2)
4.43/1.97	
4.43/1.97	U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)
4.43/1.97	
4.43/1.97	[]  =  []
4.43/1.97	
4.43/1.97	reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)
4.43/1.97	
4.43/1.97	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
4.43/1.97	
4.43/1.97	app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)
4.43/1.97	
4.43/1.97	U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)
4.43/1.97	
4.43/1.97	app_out_gga(x1, x2, x3)  =  app_out_gga(x3)
4.43/1.97	
4.43/1.97	REVERSE_IN_GA(x1, x2)  =  REVERSE_IN_GA(x1)
4.43/1.97	
4.43/1.97	
4.43/1.97	We have to consider all (P,R,Pi)-chains
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(15) UsableRulesProof (EQUIVALENT)
4.43/1.97	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(16)
4.43/1.97	Obligation:
4.43/1.97	Pi DP problem:
4.43/1.97	The TRS P consists of the following rules:
4.43/1.97	
4.43/1.97	   REVERSE_IN_GA(.(X, Xs), Ys) -> REVERSE_IN_GA(Xs, Zs)
4.43/1.97	
4.43/1.97	R is empty.
4.43/1.97	The argument filtering Pi contains the following mapping:
4.43/1.97	.(x1, x2)  =  .(x1, x2)
4.43/1.97	
4.43/1.97	REVERSE_IN_GA(x1, x2)  =  REVERSE_IN_GA(x1)
4.43/1.97	
4.43/1.97	
4.43/1.97	We have to consider all (P,R,Pi)-chains
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(17) PiDPToQDPProof (SOUND)
4.43/1.97	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(18)
4.43/1.97	Obligation:
4.43/1.97	Q DP problem:
4.43/1.97	The TRS P consists of the following rules:
4.43/1.97	
4.43/1.97	   REVERSE_IN_GA(.(X, Xs)) -> REVERSE_IN_GA(Xs)
4.43/1.97	
4.43/1.97	R is empty.
4.43/1.97	Q is empty.
4.43/1.97	We have to consider all (P,Q,R)-chains.
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(19) QDPSizeChangeProof (EQUIVALENT)
4.43/1.97	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. 
4.43/1.97	
4.43/1.97	From the DPs we obtained the following set of size-change graphs:
4.43/1.97	*REVERSE_IN_GA(.(X, Xs)) -> REVERSE_IN_GA(Xs)
4.43/1.97	The graph contains the following edges 1 > 1
4.43/1.97	
4.43/1.97	
4.43/1.97	----------------------------------------
4.43/1.97	
4.43/1.97	(20)
4.43/1.97	YES
4.70/2.01	EOF