5.33/2.17	YES
5.33/2.19	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
5.33/2.19	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
5.33/2.19	
5.33/2.19	
5.33/2.19	Left Termination of the query pattern
5.33/2.19	
5.33/2.19	conf(g)
5.33/2.19	
5.33/2.19	w.r.t. the given Prolog program could successfully be proven:
5.33/2.19	
5.33/2.19	(0) Prolog
5.33/2.19	(1) PrologToPiTRSProof [SOUND, 0 ms]
5.33/2.19	(2) PiTRS
5.33/2.19	(3) DependencyPairsProof [EQUIVALENT, 1 ms]
5.33/2.19	(4) PiDP
5.33/2.19	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
5.33/2.19	(6) AND
5.33/2.19	    (7) PiDP
5.33/2.19	        (8) UsableRulesProof [EQUIVALENT, 0 ms]
5.33/2.19	        (9) PiDP
5.33/2.19	        (10) PiDPToQDPProof [SOUND, 0 ms]
5.33/2.19	        (11) QDP
5.33/2.19	        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
5.33/2.19	        (13) YES
5.33/2.19	    (14) PiDP
5.33/2.19	        (15) UsableRulesProof [EQUIVALENT, 0 ms]
5.33/2.19	        (16) PiDP
5.33/2.19	        (17) PiDPToQDPProof [SOUND, 0 ms]
5.33/2.19	        (18) QDP
5.33/2.19	        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
5.33/2.19	        (20) YES
5.33/2.19	    (21) PiDP
5.33/2.19	        (22) UsableRulesProof [EQUIVALENT, 0 ms]
5.33/2.19	        (23) PiDP
5.33/2.19	        (24) PiDPToQDPProof [SOUND, 0 ms]
5.33/2.19	        (25) QDP
5.33/2.19	        (26) MRRProof [EQUIVALENT, 0 ms]
5.33/2.19	        (27) QDP
5.33/2.19	        (28) PisEmptyProof [EQUIVALENT, 0 ms]
5.33/2.19	        (29) YES
5.33/2.19	
5.33/2.19	
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(0)
5.33/2.19	Obligation:
5.33/2.19	Clauses:
5.33/2.19	
5.33/2.19	conf(X) :- ','(del2(X, Z), ','(del(U, Y, Z), conf(Y))).
5.33/2.19	del2(X, Y) :- ','(del(U, X, Z), del(V, Z, Y)).
5.33/2.19	del(X, cons(X, T), T).
5.33/2.19	del(X, cons(H, T), cons(H, T1)) :- del(X, T, T1).
5.33/2.19	s2l(s(X), cons(Y, Xs)) :- s2l(X, Xs).
5.33/2.19	s2l(0, nil).
5.33/2.19	goal(X) :- ','(s2l(X, XS), conf(XS)).
5.33/2.19	
5.33/2.19	
5.33/2.19	Query: conf(g)
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(1) PrologToPiTRSProof (SOUND)
5.33/2.19	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
5.33/2.19	
5.33/2.19	conf_in_1: (b)
5.33/2.19	
5.33/2.19	del2_in_2: (b,f)
5.33/2.19	
5.33/2.19	del_in_3: (f,b,f) (f,f,b)
5.33/2.19	
5.33/2.19	Transforming Prolog into the following Term Rewriting System:
5.33/2.19	
5.33/2.19	Pi-finite rewrite system:
5.33/2.19	The TRS R consists of the following rules:
5.33/2.19	
5.33/2.19	   conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z))
5.33/2.19	   del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z))
5.33/2.19	   del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T)
5.33/2.19	   del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
5.33/2.19	   U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1))
5.33/2.19	   U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y))
5.33/2.19	   U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y)
5.33/2.19	   U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z))
5.33/2.19	   del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T)
5.33/2.19	   del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
5.33/2.19	   U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1))
5.33/2.19	   U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y))
5.33/2.19	   U3_g(X, conf_out_g(Y)) -> conf_out_g(X)
5.33/2.19	
5.33/2.19	The argument filtering Pi contains the following mapping:
5.33/2.19	conf_in_g(x1)  =  conf_in_g(x1)
5.33/2.19	
5.33/2.19	U1_g(x1, x2)  =  U1_g(x2)
5.33/2.19	
5.33/2.19	del2_in_ga(x1, x2)  =  del2_in_ga(x1)
5.33/2.19	
5.33/2.19	U4_ga(x1, x2, x3)  =  U4_ga(x3)
5.33/2.19	
5.33/2.19	del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
5.33/2.19	
5.33/2.19	cons(x1, x2)  =  cons(x2)
5.33/2.19	
5.33/2.19	del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
5.33/2.19	
5.33/2.19	U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
5.33/2.19	
5.33/2.19	U5_ga(x1, x2, x3)  =  U5_ga(x3)
5.33/2.19	
5.33/2.19	del2_out_ga(x1, x2)  =  del2_out_ga(x2)
5.33/2.19	
5.33/2.19	U2_g(x1, x2)  =  U2_g(x2)
5.33/2.19	
5.33/2.19	del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
5.33/2.19	
5.33/2.19	del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
5.33/2.19	
5.33/2.19	U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
5.33/2.19	
5.33/2.19	U3_g(x1, x2)  =  U3_g(x2)
5.33/2.19	
5.33/2.19	conf_out_g(x1)  =  conf_out_g
5.33/2.19	
5.33/2.19	
5.33/2.19	
5.33/2.19	
5.33/2.19	
5.33/2.19	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
5.33/2.19	
5.33/2.19	
5.33/2.19	
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(2)
5.33/2.19	Obligation:
5.33/2.19	Pi-finite rewrite system:
5.33/2.19	The TRS R consists of the following rules:
5.33/2.19	
5.33/2.19	   conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z))
5.33/2.19	   del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z))
5.33/2.19	   del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T)
5.33/2.19	   del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
5.33/2.19	   U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1))
5.33/2.19	   U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y))
5.33/2.19	   U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y)
5.33/2.19	   U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z))
5.33/2.19	   del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T)
5.33/2.19	   del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
5.33/2.19	   U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1))
5.33/2.19	   U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y))
5.33/2.19	   U3_g(X, conf_out_g(Y)) -> conf_out_g(X)
5.33/2.19	
5.33/2.19	The argument filtering Pi contains the following mapping:
5.33/2.19	conf_in_g(x1)  =  conf_in_g(x1)
5.33/2.19	
5.33/2.19	U1_g(x1, x2)  =  U1_g(x2)
5.33/2.19	
5.33/2.19	del2_in_ga(x1, x2)  =  del2_in_ga(x1)
5.33/2.19	
5.33/2.19	U4_ga(x1, x2, x3)  =  U4_ga(x3)
5.33/2.19	
5.33/2.19	del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
5.33/2.19	
5.33/2.19	cons(x1, x2)  =  cons(x2)
5.33/2.19	
5.33/2.19	del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
5.33/2.19	
5.33/2.19	U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
5.33/2.19	
5.33/2.19	U5_ga(x1, x2, x3)  =  U5_ga(x3)
5.33/2.19	
5.33/2.19	del2_out_ga(x1, x2)  =  del2_out_ga(x2)
5.33/2.19	
5.33/2.19	U2_g(x1, x2)  =  U2_g(x2)
5.33/2.19	
5.33/2.19	del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
5.33/2.19	
5.33/2.19	del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
5.33/2.19	
5.33/2.19	U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
5.33/2.19	
5.33/2.19	U3_g(x1, x2)  =  U3_g(x2)
5.33/2.19	
5.33/2.19	conf_out_g(x1)  =  conf_out_g
5.33/2.19	
5.33/2.19	
5.33/2.19	
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(3) DependencyPairsProof (EQUIVALENT)
5.33/2.19	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
5.33/2.19	Pi DP problem:
5.33/2.19	The TRS P consists of the following rules:
5.33/2.19	
5.33/2.19	   CONF_IN_G(X) -> U1_G(X, del2_in_ga(X, Z))
5.33/2.19	   CONF_IN_G(X) -> DEL2_IN_GA(X, Z)
5.33/2.19	   DEL2_IN_GA(X, Y) -> U4_GA(X, Y, del_in_aga(U, X, Z))
5.33/2.19	   DEL2_IN_GA(X, Y) -> DEL_IN_AGA(U, X, Z)
5.33/2.19	   DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> U6_AGA(X, H, T, T1, del_in_aga(X, T, T1))
5.33/2.19	   DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> DEL_IN_AGA(X, T, T1)
5.33/2.19	   U4_GA(X, Y, del_out_aga(U, X, Z)) -> U5_GA(X, Y, del_in_aga(V, Z, Y))
5.33/2.19	   U4_GA(X, Y, del_out_aga(U, X, Z)) -> DEL_IN_AGA(V, Z, Y)
5.33/2.19	   U1_G(X, del2_out_ga(X, Z)) -> U2_G(X, del_in_aag(U, Y, Z))
5.33/2.19	   U1_G(X, del2_out_ga(X, Z)) -> DEL_IN_AAG(U, Y, Z)
5.33/2.19	   DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> U6_AAG(X, H, T, T1, del_in_aag(X, T, T1))
5.33/2.19	   DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> DEL_IN_AAG(X, T, T1)
5.33/2.19	   U2_G(X, del_out_aag(U, Y, Z)) -> U3_G(X, conf_in_g(Y))
5.33/2.19	   U2_G(X, del_out_aag(U, Y, Z)) -> CONF_IN_G(Y)
5.33/2.19	
5.33/2.19	The TRS R consists of the following rules:
5.33/2.19	
5.33/2.19	   conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z))
5.33/2.19	   del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z))
5.33/2.19	   del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T)
5.33/2.19	   del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
5.33/2.19	   U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1))
5.33/2.19	   U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y))
5.33/2.19	   U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y)
5.33/2.19	   U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z))
5.33/2.19	   del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T)
5.33/2.19	   del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
5.33/2.19	   U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1))
5.33/2.19	   U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y))
5.33/2.19	   U3_g(X, conf_out_g(Y)) -> conf_out_g(X)
5.33/2.19	
5.33/2.19	The argument filtering Pi contains the following mapping:
5.33/2.19	conf_in_g(x1)  =  conf_in_g(x1)
5.33/2.19	
5.33/2.19	U1_g(x1, x2)  =  U1_g(x2)
5.33/2.19	
5.33/2.19	del2_in_ga(x1, x2)  =  del2_in_ga(x1)
5.33/2.19	
5.33/2.19	U4_ga(x1, x2, x3)  =  U4_ga(x3)
5.33/2.19	
5.33/2.19	del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
5.33/2.19	
5.33/2.19	cons(x1, x2)  =  cons(x2)
5.33/2.19	
5.33/2.19	del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
5.33/2.19	
5.33/2.19	U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
5.33/2.19	
5.33/2.19	U5_ga(x1, x2, x3)  =  U5_ga(x3)
5.33/2.19	
5.33/2.19	del2_out_ga(x1, x2)  =  del2_out_ga(x2)
5.33/2.19	
5.33/2.19	U2_g(x1, x2)  =  U2_g(x2)
5.33/2.19	
5.33/2.19	del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
5.33/2.19	
5.33/2.19	del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
5.33/2.19	
5.33/2.19	U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
5.33/2.19	
5.33/2.19	U3_g(x1, x2)  =  U3_g(x2)
5.33/2.19	
5.33/2.19	conf_out_g(x1)  =  conf_out_g
5.33/2.19	
5.33/2.19	CONF_IN_G(x1)  =  CONF_IN_G(x1)
5.33/2.19	
5.33/2.19	U1_G(x1, x2)  =  U1_G(x2)
5.33/2.19	
5.33/2.19	DEL2_IN_GA(x1, x2)  =  DEL2_IN_GA(x1)
5.33/2.19	
5.33/2.19	U4_GA(x1, x2, x3)  =  U4_GA(x3)
5.33/2.19	
5.33/2.19	DEL_IN_AGA(x1, x2, x3)  =  DEL_IN_AGA(x2)
5.33/2.19	
5.33/2.19	U6_AGA(x1, x2, x3, x4, x5)  =  U6_AGA(x5)
5.33/2.19	
5.33/2.19	U5_GA(x1, x2, x3)  =  U5_GA(x3)
5.33/2.19	
5.33/2.19	U2_G(x1, x2)  =  U2_G(x2)
5.33/2.19	
5.33/2.19	DEL_IN_AAG(x1, x2, x3)  =  DEL_IN_AAG(x3)
5.33/2.19	
5.33/2.19	U6_AAG(x1, x2, x3, x4, x5)  =  U6_AAG(x5)
5.33/2.19	
5.33/2.19	U3_G(x1, x2)  =  U3_G(x2)
5.33/2.19	
5.33/2.19	
5.33/2.19	We have to consider all (P,R,Pi)-chains
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(4)
5.33/2.19	Obligation:
5.33/2.19	Pi DP problem:
5.33/2.19	The TRS P consists of the following rules:
5.33/2.19	
5.33/2.19	   CONF_IN_G(X) -> U1_G(X, del2_in_ga(X, Z))
5.33/2.19	   CONF_IN_G(X) -> DEL2_IN_GA(X, Z)
5.33/2.19	   DEL2_IN_GA(X, Y) -> U4_GA(X, Y, del_in_aga(U, X, Z))
5.33/2.19	   DEL2_IN_GA(X, Y) -> DEL_IN_AGA(U, X, Z)
5.33/2.19	   DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> U6_AGA(X, H, T, T1, del_in_aga(X, T, T1))
5.33/2.19	   DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> DEL_IN_AGA(X, T, T1)
5.33/2.19	   U4_GA(X, Y, del_out_aga(U, X, Z)) -> U5_GA(X, Y, del_in_aga(V, Z, Y))
5.33/2.19	   U4_GA(X, Y, del_out_aga(U, X, Z)) -> DEL_IN_AGA(V, Z, Y)
5.33/2.19	   U1_G(X, del2_out_ga(X, Z)) -> U2_G(X, del_in_aag(U, Y, Z))
5.33/2.19	   U1_G(X, del2_out_ga(X, Z)) -> DEL_IN_AAG(U, Y, Z)
5.33/2.19	   DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> U6_AAG(X, H, T, T1, del_in_aag(X, T, T1))
5.33/2.19	   DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> DEL_IN_AAG(X, T, T1)
5.33/2.19	   U2_G(X, del_out_aag(U, Y, Z)) -> U3_G(X, conf_in_g(Y))
5.33/2.19	   U2_G(X, del_out_aag(U, Y, Z)) -> CONF_IN_G(Y)
5.33/2.19	
5.33/2.19	The TRS R consists of the following rules:
5.33/2.19	
5.33/2.19	   conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z))
5.33/2.19	   del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z))
5.33/2.19	   del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T)
5.33/2.19	   del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
5.33/2.19	   U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1))
5.33/2.19	   U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y))
5.33/2.19	   U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y)
5.33/2.19	   U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z))
5.33/2.19	   del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T)
5.33/2.19	   del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
5.33/2.19	   U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1))
5.33/2.19	   U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y))
5.33/2.19	   U3_g(X, conf_out_g(Y)) -> conf_out_g(X)
5.33/2.19	
5.33/2.19	The argument filtering Pi contains the following mapping:
5.33/2.19	conf_in_g(x1)  =  conf_in_g(x1)
5.33/2.19	
5.33/2.19	U1_g(x1, x2)  =  U1_g(x2)
5.33/2.19	
5.33/2.19	del2_in_ga(x1, x2)  =  del2_in_ga(x1)
5.33/2.19	
5.33/2.19	U4_ga(x1, x2, x3)  =  U4_ga(x3)
5.33/2.19	
5.33/2.19	del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
5.33/2.19	
5.33/2.19	cons(x1, x2)  =  cons(x2)
5.33/2.19	
5.33/2.19	del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
5.33/2.19	
5.33/2.19	U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
5.33/2.19	
5.33/2.19	U5_ga(x1, x2, x3)  =  U5_ga(x3)
5.33/2.19	
5.33/2.19	del2_out_ga(x1, x2)  =  del2_out_ga(x2)
5.33/2.19	
5.33/2.19	U2_g(x1, x2)  =  U2_g(x2)
5.33/2.19	
5.33/2.19	del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
5.33/2.19	
5.33/2.19	del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
5.33/2.19	
5.33/2.19	U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
5.33/2.19	
5.33/2.19	U3_g(x1, x2)  =  U3_g(x2)
5.33/2.19	
5.33/2.19	conf_out_g(x1)  =  conf_out_g
5.33/2.19	
5.33/2.19	CONF_IN_G(x1)  =  CONF_IN_G(x1)
5.33/2.19	
5.33/2.19	U1_G(x1, x2)  =  U1_G(x2)
5.33/2.19	
5.33/2.19	DEL2_IN_GA(x1, x2)  =  DEL2_IN_GA(x1)
5.33/2.19	
5.33/2.19	U4_GA(x1, x2, x3)  =  U4_GA(x3)
5.33/2.19	
5.33/2.19	DEL_IN_AGA(x1, x2, x3)  =  DEL_IN_AGA(x2)
5.33/2.19	
5.33/2.19	U6_AGA(x1, x2, x3, x4, x5)  =  U6_AGA(x5)
5.33/2.19	
5.33/2.19	U5_GA(x1, x2, x3)  =  U5_GA(x3)
5.33/2.19	
5.33/2.19	U2_G(x1, x2)  =  U2_G(x2)
5.33/2.19	
5.33/2.19	DEL_IN_AAG(x1, x2, x3)  =  DEL_IN_AAG(x3)
5.33/2.19	
5.33/2.19	U6_AAG(x1, x2, x3, x4, x5)  =  U6_AAG(x5)
5.33/2.19	
5.33/2.19	U3_G(x1, x2)  =  U3_G(x2)
5.33/2.19	
5.33/2.19	
5.33/2.19	We have to consider all (P,R,Pi)-chains
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(5) DependencyGraphProof (EQUIVALENT)
5.33/2.19	The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 9 less nodes.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(6)
5.33/2.19	Complex Obligation (AND)
5.33/2.19	
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(7)
5.33/2.19	Obligation:
5.33/2.19	Pi DP problem:
5.33/2.19	The TRS P consists of the following rules:
5.33/2.19	
5.33/2.19	   DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> DEL_IN_AAG(X, T, T1)
5.33/2.19	
5.33/2.19	The TRS R consists of the following rules:
5.33/2.19	
5.33/2.19	   conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z))
5.33/2.19	   del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z))
5.33/2.19	   del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T)
5.33/2.19	   del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
5.33/2.19	   U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1))
5.33/2.19	   U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y))
5.33/2.19	   U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y)
5.33/2.19	   U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z))
5.33/2.19	   del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T)
5.33/2.19	   del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
5.33/2.19	   U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1))
5.33/2.19	   U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y))
5.33/2.19	   U3_g(X, conf_out_g(Y)) -> conf_out_g(X)
5.33/2.19	
5.33/2.19	The argument filtering Pi contains the following mapping:
5.33/2.19	conf_in_g(x1)  =  conf_in_g(x1)
5.33/2.19	
5.33/2.19	U1_g(x1, x2)  =  U1_g(x2)
5.33/2.19	
5.33/2.19	del2_in_ga(x1, x2)  =  del2_in_ga(x1)
5.33/2.19	
5.33/2.19	U4_ga(x1, x2, x3)  =  U4_ga(x3)
5.33/2.19	
5.33/2.19	del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
5.33/2.19	
5.33/2.19	cons(x1, x2)  =  cons(x2)
5.33/2.19	
5.33/2.19	del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
5.33/2.19	
5.33/2.19	U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
5.33/2.19	
5.33/2.19	U5_ga(x1, x2, x3)  =  U5_ga(x3)
5.33/2.19	
5.33/2.19	del2_out_ga(x1, x2)  =  del2_out_ga(x2)
5.33/2.19	
5.33/2.19	U2_g(x1, x2)  =  U2_g(x2)
5.33/2.19	
5.33/2.19	del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
5.33/2.19	
5.33/2.19	del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
5.33/2.19	
5.33/2.19	U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
5.33/2.19	
5.33/2.19	U3_g(x1, x2)  =  U3_g(x2)
5.33/2.19	
5.33/2.19	conf_out_g(x1)  =  conf_out_g
5.33/2.19	
5.33/2.19	DEL_IN_AAG(x1, x2, x3)  =  DEL_IN_AAG(x3)
5.33/2.19	
5.33/2.19	
5.33/2.19	We have to consider all (P,R,Pi)-chains
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(8) UsableRulesProof (EQUIVALENT)
5.33/2.19	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(9)
5.33/2.19	Obligation:
5.33/2.19	Pi DP problem:
5.33/2.19	The TRS P consists of the following rules:
5.33/2.19	
5.33/2.19	   DEL_IN_AAG(X, cons(H, T), cons(H, T1)) -> DEL_IN_AAG(X, T, T1)
5.33/2.19	
5.33/2.19	R is empty.
5.33/2.19	The argument filtering Pi contains the following mapping:
5.33/2.19	cons(x1, x2)  =  cons(x2)
5.33/2.19	
5.33/2.19	DEL_IN_AAG(x1, x2, x3)  =  DEL_IN_AAG(x3)
5.33/2.19	
5.33/2.19	
5.33/2.19	We have to consider all (P,R,Pi)-chains
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(10) PiDPToQDPProof (SOUND)
5.33/2.19	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(11)
5.33/2.19	Obligation:
5.33/2.19	Q DP problem:
5.33/2.19	The TRS P consists of the following rules:
5.33/2.19	
5.33/2.19	   DEL_IN_AAG(cons(T1)) -> DEL_IN_AAG(T1)
5.33/2.19	
5.33/2.19	R is empty.
5.33/2.19	Q is empty.
5.33/2.19	We have to consider all (P,Q,R)-chains.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(12) QDPSizeChangeProof (EQUIVALENT)
5.33/2.19	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. 
5.33/2.19	
5.33/2.19	From the DPs we obtained the following set of size-change graphs:
5.33/2.19	*DEL_IN_AAG(cons(T1)) -> DEL_IN_AAG(T1)
5.33/2.19	The graph contains the following edges 1 > 1
5.33/2.19	
5.33/2.19	
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(13)
5.33/2.19	YES
5.33/2.19	
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(14)
5.33/2.19	Obligation:
5.33/2.19	Pi DP problem:
5.33/2.19	The TRS P consists of the following rules:
5.33/2.19	
5.33/2.19	   DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> DEL_IN_AGA(X, T, T1)
5.33/2.19	
5.33/2.19	The TRS R consists of the following rules:
5.33/2.19	
5.33/2.19	   conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z))
5.33/2.19	   del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z))
5.33/2.19	   del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T)
5.33/2.19	   del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
5.33/2.19	   U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1))
5.33/2.19	   U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y))
5.33/2.19	   U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y)
5.33/2.19	   U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z))
5.33/2.19	   del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T)
5.33/2.19	   del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
5.33/2.19	   U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1))
5.33/2.19	   U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y))
5.33/2.19	   U3_g(X, conf_out_g(Y)) -> conf_out_g(X)
5.33/2.19	
5.33/2.19	The argument filtering Pi contains the following mapping:
5.33/2.19	conf_in_g(x1)  =  conf_in_g(x1)
5.33/2.19	
5.33/2.19	U1_g(x1, x2)  =  U1_g(x2)
5.33/2.19	
5.33/2.19	del2_in_ga(x1, x2)  =  del2_in_ga(x1)
5.33/2.19	
5.33/2.19	U4_ga(x1, x2, x3)  =  U4_ga(x3)
5.33/2.19	
5.33/2.19	del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
5.33/2.19	
5.33/2.19	cons(x1, x2)  =  cons(x2)
5.33/2.19	
5.33/2.19	del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
5.33/2.19	
5.33/2.19	U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
5.33/2.19	
5.33/2.19	U5_ga(x1, x2, x3)  =  U5_ga(x3)
5.33/2.19	
5.33/2.19	del2_out_ga(x1, x2)  =  del2_out_ga(x2)
5.33/2.19	
5.33/2.19	U2_g(x1, x2)  =  U2_g(x2)
5.33/2.19	
5.33/2.19	del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
5.33/2.19	
5.33/2.19	del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
5.33/2.19	
5.33/2.19	U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
5.33/2.19	
5.33/2.19	U3_g(x1, x2)  =  U3_g(x2)
5.33/2.19	
5.33/2.19	conf_out_g(x1)  =  conf_out_g
5.33/2.19	
5.33/2.19	DEL_IN_AGA(x1, x2, x3)  =  DEL_IN_AGA(x2)
5.33/2.19	
5.33/2.19	
5.33/2.19	We have to consider all (P,R,Pi)-chains
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(15) UsableRulesProof (EQUIVALENT)
5.33/2.19	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(16)
5.33/2.19	Obligation:
5.33/2.19	Pi DP problem:
5.33/2.19	The TRS P consists of the following rules:
5.33/2.19	
5.33/2.19	   DEL_IN_AGA(X, cons(H, T), cons(H, T1)) -> DEL_IN_AGA(X, T, T1)
5.33/2.19	
5.33/2.19	R is empty.
5.33/2.19	The argument filtering Pi contains the following mapping:
5.33/2.19	cons(x1, x2)  =  cons(x2)
5.33/2.19	
5.33/2.19	DEL_IN_AGA(x1, x2, x3)  =  DEL_IN_AGA(x2)
5.33/2.19	
5.33/2.19	
5.33/2.19	We have to consider all (P,R,Pi)-chains
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(17) PiDPToQDPProof (SOUND)
5.33/2.19	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(18)
5.33/2.19	Obligation:
5.33/2.19	Q DP problem:
5.33/2.19	The TRS P consists of the following rules:
5.33/2.19	
5.33/2.19	   DEL_IN_AGA(cons(T)) -> DEL_IN_AGA(T)
5.33/2.19	
5.33/2.19	R is empty.
5.33/2.19	Q is empty.
5.33/2.19	We have to consider all (P,Q,R)-chains.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(19) QDPSizeChangeProof (EQUIVALENT)
5.33/2.19	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. 
5.33/2.19	
5.33/2.19	From the DPs we obtained the following set of size-change graphs:
5.33/2.19	*DEL_IN_AGA(cons(T)) -> DEL_IN_AGA(T)
5.33/2.19	The graph contains the following edges 1 > 1
5.33/2.19	
5.33/2.19	
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(20)
5.33/2.19	YES
5.33/2.19	
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(21)
5.33/2.19	Obligation:
5.33/2.19	Pi DP problem:
5.33/2.19	The TRS P consists of the following rules:
5.33/2.19	
5.33/2.19	   U1_G(X, del2_out_ga(X, Z)) -> U2_G(X, del_in_aag(U, Y, Z))
5.33/2.19	   U2_G(X, del_out_aag(U, Y, Z)) -> CONF_IN_G(Y)
5.33/2.19	   CONF_IN_G(X) -> U1_G(X, del2_in_ga(X, Z))
5.33/2.19	
5.33/2.19	The TRS R consists of the following rules:
5.33/2.19	
5.33/2.19	   conf_in_g(X) -> U1_g(X, del2_in_ga(X, Z))
5.33/2.19	   del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z))
5.33/2.19	   del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T)
5.33/2.19	   del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
5.33/2.19	   U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1))
5.33/2.19	   U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y))
5.33/2.19	   U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y)
5.33/2.19	   U1_g(X, del2_out_ga(X, Z)) -> U2_g(X, del_in_aag(U, Y, Z))
5.33/2.19	   del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T)
5.33/2.19	   del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
5.33/2.19	   U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1))
5.33/2.19	   U2_g(X, del_out_aag(U, Y, Z)) -> U3_g(X, conf_in_g(Y))
5.33/2.19	   U3_g(X, conf_out_g(Y)) -> conf_out_g(X)
5.33/2.19	
5.33/2.19	The argument filtering Pi contains the following mapping:
5.33/2.19	conf_in_g(x1)  =  conf_in_g(x1)
5.33/2.19	
5.33/2.19	U1_g(x1, x2)  =  U1_g(x2)
5.33/2.19	
5.33/2.19	del2_in_ga(x1, x2)  =  del2_in_ga(x1)
5.33/2.19	
5.33/2.19	U4_ga(x1, x2, x3)  =  U4_ga(x3)
5.33/2.19	
5.33/2.19	del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
5.33/2.19	
5.33/2.19	cons(x1, x2)  =  cons(x2)
5.33/2.19	
5.33/2.19	del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
5.33/2.19	
5.33/2.19	U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
5.33/2.19	
5.33/2.19	U5_ga(x1, x2, x3)  =  U5_ga(x3)
5.33/2.19	
5.33/2.19	del2_out_ga(x1, x2)  =  del2_out_ga(x2)
5.33/2.19	
5.33/2.19	U2_g(x1, x2)  =  U2_g(x2)
5.33/2.19	
5.33/2.19	del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
5.33/2.19	
5.33/2.19	del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
5.33/2.19	
5.33/2.19	U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
5.33/2.19	
5.33/2.19	U3_g(x1, x2)  =  U3_g(x2)
5.33/2.19	
5.33/2.19	conf_out_g(x1)  =  conf_out_g
5.33/2.19	
5.33/2.19	CONF_IN_G(x1)  =  CONF_IN_G(x1)
5.33/2.19	
5.33/2.19	U1_G(x1, x2)  =  U1_G(x2)
5.33/2.19	
5.33/2.19	U2_G(x1, x2)  =  U2_G(x2)
5.33/2.19	
5.33/2.19	
5.33/2.19	We have to consider all (P,R,Pi)-chains
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(22) UsableRulesProof (EQUIVALENT)
5.33/2.19	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(23)
5.33/2.19	Obligation:
5.33/2.19	Pi DP problem:
5.33/2.19	The TRS P consists of the following rules:
5.33/2.19	
5.33/2.19	   U1_G(X, del2_out_ga(X, Z)) -> U2_G(X, del_in_aag(U, Y, Z))
5.33/2.19	   U2_G(X, del_out_aag(U, Y, Z)) -> CONF_IN_G(Y)
5.33/2.19	   CONF_IN_G(X) -> U1_G(X, del2_in_ga(X, Z))
5.33/2.19	
5.33/2.19	The TRS R consists of the following rules:
5.33/2.19	
5.33/2.19	   del_in_aag(X, cons(X, T), T) -> del_out_aag(X, cons(X, T), T)
5.33/2.19	   del_in_aag(X, cons(H, T), cons(H, T1)) -> U6_aag(X, H, T, T1, del_in_aag(X, T, T1))
5.33/2.19	   del2_in_ga(X, Y) -> U4_ga(X, Y, del_in_aga(U, X, Z))
5.33/2.19	   U6_aag(X, H, T, T1, del_out_aag(X, T, T1)) -> del_out_aag(X, cons(H, T), cons(H, T1))
5.33/2.19	   U4_ga(X, Y, del_out_aga(U, X, Z)) -> U5_ga(X, Y, del_in_aga(V, Z, Y))
5.33/2.19	   del_in_aga(X, cons(X, T), T) -> del_out_aga(X, cons(X, T), T)
5.33/2.19	   del_in_aga(X, cons(H, T), cons(H, T1)) -> U6_aga(X, H, T, T1, del_in_aga(X, T, T1))
5.33/2.19	   U5_ga(X, Y, del_out_aga(V, Z, Y)) -> del2_out_ga(X, Y)
5.33/2.19	   U6_aga(X, H, T, T1, del_out_aga(X, T, T1)) -> del_out_aga(X, cons(H, T), cons(H, T1))
5.33/2.19	
5.33/2.19	The argument filtering Pi contains the following mapping:
5.33/2.19	del2_in_ga(x1, x2)  =  del2_in_ga(x1)
5.33/2.19	
5.33/2.19	U4_ga(x1, x2, x3)  =  U4_ga(x3)
5.33/2.19	
5.33/2.19	del_in_aga(x1, x2, x3)  =  del_in_aga(x2)
5.33/2.19	
5.33/2.19	cons(x1, x2)  =  cons(x2)
5.33/2.19	
5.33/2.19	del_out_aga(x1, x2, x3)  =  del_out_aga(x3)
5.33/2.19	
5.33/2.19	U6_aga(x1, x2, x3, x4, x5)  =  U6_aga(x5)
5.33/2.19	
5.33/2.19	U5_ga(x1, x2, x3)  =  U5_ga(x3)
5.33/2.19	
5.33/2.19	del2_out_ga(x1, x2)  =  del2_out_ga(x2)
5.33/2.19	
5.33/2.19	del_in_aag(x1, x2, x3)  =  del_in_aag(x3)
5.33/2.19	
5.33/2.19	del_out_aag(x1, x2, x3)  =  del_out_aag(x2)
5.33/2.19	
5.33/2.19	U6_aag(x1, x2, x3, x4, x5)  =  U6_aag(x5)
5.33/2.19	
5.33/2.19	CONF_IN_G(x1)  =  CONF_IN_G(x1)
5.33/2.19	
5.33/2.19	U1_G(x1, x2)  =  U1_G(x2)
5.33/2.19	
5.33/2.19	U2_G(x1, x2)  =  U2_G(x2)
5.33/2.19	
5.33/2.19	
5.33/2.19	We have to consider all (P,R,Pi)-chains
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(24) PiDPToQDPProof (SOUND)
5.33/2.19	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(25)
5.33/2.19	Obligation:
5.33/2.19	Q DP problem:
5.33/2.19	The TRS P consists of the following rules:
5.33/2.19	
5.33/2.19	   U1_G(del2_out_ga(Z)) -> U2_G(del_in_aag(Z))
5.33/2.19	   U2_G(del_out_aag(Y)) -> CONF_IN_G(Y)
5.33/2.19	   CONF_IN_G(X) -> U1_G(del2_in_ga(X))
5.33/2.19	
5.33/2.19	The TRS R consists of the following rules:
5.33/2.19	
5.33/2.19	   del_in_aag(T) -> del_out_aag(cons(T))
5.33/2.19	   del_in_aag(cons(T1)) -> U6_aag(del_in_aag(T1))
5.33/2.19	   del2_in_ga(X) -> U4_ga(del_in_aga(X))
5.33/2.19	   U6_aag(del_out_aag(T)) -> del_out_aag(cons(T))
5.33/2.19	   U4_ga(del_out_aga(Z)) -> U5_ga(del_in_aga(Z))
5.33/2.19	   del_in_aga(cons(T)) -> del_out_aga(T)
5.33/2.19	   del_in_aga(cons(T)) -> U6_aga(del_in_aga(T))
5.33/2.19	   U5_ga(del_out_aga(Y)) -> del2_out_ga(Y)
5.33/2.19	   U6_aga(del_out_aga(T1)) -> del_out_aga(cons(T1))
5.33/2.19	
5.33/2.19	The set Q consists of the following terms:
5.33/2.19	
5.33/2.19	   del_in_aag(x0)
5.33/2.19	   del2_in_ga(x0)
5.33/2.19	   U6_aag(x0)
5.33/2.19	   U4_ga(x0)
5.33/2.19	   del_in_aga(x0)
5.33/2.19	   U5_ga(x0)
5.33/2.19	   U6_aga(x0)
5.33/2.19	
5.33/2.19	We have to consider all (P,Q,R)-chains.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(26) MRRProof (EQUIVALENT)
5.33/2.19	By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
5.33/2.19	
5.33/2.19	Strictly oriented dependency pairs:
5.33/2.19	
5.33/2.19	   U1_G(del2_out_ga(Z)) -> U2_G(del_in_aag(Z))
5.33/2.19	   U2_G(del_out_aag(Y)) -> CONF_IN_G(Y)
5.33/2.19	   CONF_IN_G(X) -> U1_G(del2_in_ga(X))
5.33/2.19	
5.33/2.19	Strictly oriented rules of the TRS R:
5.33/2.19	
5.33/2.19	   del_in_aag(T) -> del_out_aag(cons(T))
5.33/2.19	   del_in_aag(cons(T1)) -> U6_aag(del_in_aag(T1))
5.33/2.19	   del2_in_ga(X) -> U4_ga(del_in_aga(X))
5.33/2.19	   U6_aag(del_out_aag(T)) -> del_out_aag(cons(T))
5.33/2.19	   U4_ga(del_out_aga(Z)) -> U5_ga(del_in_aga(Z))
5.33/2.19	   del_in_aga(cons(T)) -> del_out_aga(T)
5.33/2.19	   del_in_aga(cons(T)) -> U6_aga(del_in_aga(T))
5.33/2.19	   U5_ga(del_out_aga(Y)) -> del2_out_ga(Y)
5.33/2.19	   U6_aga(del_out_aga(T1)) -> del_out_aga(cons(T1))
5.33/2.19	
5.33/2.19	Used ordering: Knuth-Bendix order [KBO] with precedence:del2_in_ga_1 > U4_ga_1 > U1_G_1 > U2_G_1 > CONF_IN_G_1 > del_in_aag_1 > del_in_aga_1 > U6_aga_1 > del2_out_ga_1 > U5_ga_1 > del_out_aga_1 > U6_aag_1 > del_out_aag_1 > cons_1
5.33/2.19	
5.33/2.19	and weight map:
5.33/2.19	
5.33/2.19	   del_in_aag_1=3
5.33/2.19	   del_out_aag_1=1
5.33/2.19	   cons_1=2
5.33/2.19	   U6_aag_1=2
5.33/2.19	   del2_in_ga_1=4
5.33/2.19	   U4_ga_1=1
5.33/2.19	   del_in_aga_1=3
5.33/2.19	   del_out_aga_1=5
5.33/2.19	   U5_ga_1=3
5.33/2.19	   U6_aga_1=2
5.33/2.19	   del2_out_ga_1=7
5.33/2.19	   U1_G_1=1
5.33/2.19	   U2_G_1=5
5.33/2.19	   CONF_IN_G_1=6
5.33/2.19	
5.33/2.19	The variable weight is 1
5.33/2.19	
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(27)
5.33/2.19	Obligation:
5.33/2.19	Q DP problem:
5.33/2.19	P is empty.
5.33/2.19	R is empty.
5.33/2.19	The set Q consists of the following terms:
5.33/2.19	
5.33/2.19	   del_in_aag(x0)
5.33/2.19	   del2_in_ga(x0)
5.33/2.19	   U6_aag(x0)
5.33/2.19	   U4_ga(x0)
5.33/2.19	   del_in_aga(x0)
5.33/2.19	   U5_ga(x0)
5.33/2.19	   U6_aga(x0)
5.33/2.19	
5.33/2.19	We have to consider all (P,Q,R)-chains.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(28) PisEmptyProof (EQUIVALENT)
5.33/2.19	The TRS P is empty. Hence, there is no (P,Q,R) chain.
5.33/2.19	----------------------------------------
5.33/2.19	
5.33/2.19	(29)
5.33/2.19	YES
5.70/2.23	EOF