5.15/2.14	YES
5.36/2.15	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
5.36/2.15	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
5.36/2.15	
5.36/2.15	
5.36/2.15	Left Termination of the query pattern
5.36/2.15	
5.36/2.15	perm(g,a)
5.36/2.15	
5.36/2.15	w.r.t. the given Prolog program could successfully be proven:
5.36/2.15	
5.36/2.15	(0) Prolog
5.36/2.15	(1) PrologToPiTRSProof [SOUND, 0 ms]
5.36/2.15	(2) PiTRS
5.36/2.15	(3) DependencyPairsProof [EQUIVALENT, 26 ms]
5.36/2.15	(4) PiDP
5.36/2.15	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
5.36/2.15	(6) AND
5.36/2.15	    (7) PiDP
5.36/2.15	        (8) UsableRulesProof [EQUIVALENT, 0 ms]
5.36/2.15	        (9) PiDP
5.36/2.15	        (10) PiDPToQDPProof [SOUND, 13 ms]
5.36/2.15	        (11) QDP
5.36/2.15	        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
5.36/2.15	        (13) YES
5.36/2.15	    (14) PiDP
5.36/2.15	        (15) UsableRulesProof [EQUIVALENT, 0 ms]
5.36/2.15	        (16) PiDP
5.36/2.15	        (17) PiDPToQDPProof [SOUND, 0 ms]
5.36/2.15	        (18) QDP
5.36/2.15	        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
5.36/2.15	        (20) YES
5.36/2.15	    (21) PiDP
5.36/2.15	        (22) UsableRulesProof [EQUIVALENT, 0 ms]
5.36/2.15	        (23) PiDP
5.36/2.15	        (24) PiDPToQDPProof [SOUND, 0 ms]
5.36/2.15	        (25) QDP
5.36/2.15	        (26) MRRProof [EQUIVALENT, 2 ms]
5.36/2.15	        (27) QDP
5.36/2.15	        (28) PisEmptyProof [EQUIVALENT, 0 ms]
5.36/2.15	        (29) YES
5.36/2.15	
5.36/2.15	
5.36/2.15	----------------------------------------
5.36/2.15	
5.36/2.15	(0)
5.36/2.15	Obligation:
5.36/2.15	Clauses:
5.36/2.15	
5.36/2.15	ap1(nil, X, X).
5.36/2.15	ap1(cons(H, X), Y, cons(H, Z)) :- ap1(X, Y, Z).
5.36/2.15	ap2(nil, X, X).
5.36/2.15	ap2(cons(H, X), Y, cons(H, Z)) :- ap2(X, Y, Z).
5.36/2.15	perm(nil, nil).
5.36/2.15	perm(Xs, cons(X, Ys)) :- ','(ap1(X1s, cons(X, X2s), Xs), ','(ap2(X1s, X2s, Zs), perm(Zs, Ys))).
5.36/2.15	
5.36/2.15	
5.36/2.15	Query: perm(g,a)
5.36/2.15	----------------------------------------
5.36/2.15	
5.36/2.15	(1) PrologToPiTRSProof (SOUND)
5.36/2.15	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
5.36/2.15	
5.36/2.15	perm_in_2: (b,f)
5.36/2.15	
5.36/2.15	ap1_in_3: (f,f,b)
5.36/2.15	
5.36/2.15	ap2_in_3: (b,b,f)
5.36/2.15	
5.36/2.15	Transforming Prolog into the following Term Rewriting System:
5.36/2.15	
5.36/2.15	Pi-finite rewrite system:
5.36/2.15	The TRS R consists of the following rules:
5.36/2.15	
5.36/2.15	   perm_in_ga(nil, nil) -> perm_out_ga(nil, nil)
5.36/2.15	   perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
5.36/2.15	   ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X)
5.36/2.15	   ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
5.36/2.15	   U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z))
5.36/2.15	   U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
5.36/2.15	   ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X)
5.36/2.15	   ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
5.36/2.15	   U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z))
5.36/2.15	   U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
5.36/2.15	   U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys))
5.36/2.15	
5.36/2.15	The argument filtering Pi contains the following mapping:
5.36/2.15	perm_in_ga(x1, x2)  =  perm_in_ga(x1)
5.36/2.15	
5.36/2.15	nil  =  nil
5.36/2.15	
5.36/2.15	perm_out_ga(x1, x2)  =  perm_out_ga(x2)
5.36/2.15	
5.36/2.15	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
5.36/2.15	
5.36/2.15	ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
5.36/2.15	
5.36/2.15	cons(x1, x2)  =  cons(x2)
5.36/2.15	
5.36/2.15	ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
5.36/2.15	
5.36/2.15	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
5.36/2.15	
5.36/2.15	U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
5.36/2.15	
5.36/2.15	ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
5.36/2.15	
5.36/2.15	ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
5.36/2.15	
5.36/2.15	U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
5.36/2.15	
5.36/2.15	U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
5.36/2.15	
5.36/2.15	
5.36/2.15	
5.36/2.15	
5.36/2.15	
5.36/2.15	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
5.36/2.15	
5.36/2.15	
5.36/2.15	
5.36/2.15	----------------------------------------
5.36/2.15	
5.36/2.15	(2)
5.36/2.15	Obligation:
5.36/2.15	Pi-finite rewrite system:
5.36/2.15	The TRS R consists of the following rules:
5.36/2.15	
5.36/2.15	   perm_in_ga(nil, nil) -> perm_out_ga(nil, nil)
5.36/2.15	   perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
5.36/2.15	   ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X)
5.36/2.15	   ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
5.36/2.15	   U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z))
5.36/2.15	   U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
5.36/2.15	   ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X)
5.36/2.15	   ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
5.36/2.15	   U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z))
5.36/2.15	   U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
5.36/2.15	   U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys))
5.36/2.15	
5.36/2.15	The argument filtering Pi contains the following mapping:
5.36/2.15	perm_in_ga(x1, x2)  =  perm_in_ga(x1)
5.36/2.15	
5.36/2.15	nil  =  nil
5.36/2.15	
5.36/2.15	perm_out_ga(x1, x2)  =  perm_out_ga(x2)
5.36/2.15	
5.36/2.15	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
5.36/2.15	
5.36/2.15	ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
5.36/2.15	
5.36/2.15	cons(x1, x2)  =  cons(x2)
5.36/2.15	
5.36/2.15	ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
5.36/2.15	
5.36/2.15	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
5.36/2.15	
5.36/2.15	U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
5.36/2.15	
5.36/2.15	ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
5.36/2.15	
5.36/2.15	ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
5.36/2.15	
5.36/2.15	U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
5.36/2.15	
5.36/2.15	U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
5.36/2.15	
5.36/2.15	
5.36/2.15	
5.36/2.15	----------------------------------------
5.36/2.15	
5.36/2.15	(3) DependencyPairsProof (EQUIVALENT)
5.36/2.15	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
5.36/2.15	Pi DP problem:
5.36/2.15	The TRS P consists of the following rules:
5.36/2.15	
5.36/2.15	   PERM_IN_GA(Xs, cons(X, Ys)) -> U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
5.36/2.15	   PERM_IN_GA(Xs, cons(X, Ys)) -> AP1_IN_AAG(X1s, cons(X, X2s), Xs)
5.36/2.15	   AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> U1_AAG(H, X, Y, Z, ap1_in_aag(X, Y, Z))
5.36/2.15	   AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> AP1_IN_AAG(X, Y, Z)
5.36/2.15	   U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
5.36/2.15	   U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> AP2_IN_GGA(X1s, X2s, Zs)
5.36/2.15	   AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> U2_GGA(H, X, Y, Z, ap2_in_gga(X, Y, Z))
5.36/2.15	   AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> AP2_IN_GGA(X, Y, Z)
5.36/2.15	   U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_GA(Xs, X, Ys, perm_in_ga(Zs, Ys))
5.36/2.15	   U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> PERM_IN_GA(Zs, Ys)
5.36/2.15	
5.36/2.15	The TRS R consists of the following rules:
5.36/2.15	
5.36/2.15	   perm_in_ga(nil, nil) -> perm_out_ga(nil, nil)
5.36/2.15	   perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
5.36/2.15	   ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X)
5.36/2.15	   ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
5.36/2.15	   U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z))
5.36/2.15	   U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
5.36/2.15	   ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X)
5.36/2.15	   ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
5.36/2.15	   U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z))
5.36/2.15	   U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
5.36/2.15	   U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys))
5.36/2.15	
5.36/2.15	The argument filtering Pi contains the following mapping:
5.36/2.15	perm_in_ga(x1, x2)  =  perm_in_ga(x1)
5.36/2.15	
5.36/2.15	nil  =  nil
5.36/2.15	
5.36/2.15	perm_out_ga(x1, x2)  =  perm_out_ga(x2)
5.36/2.15	
5.36/2.15	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
5.36/2.15	
5.36/2.15	ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
5.36/2.15	
5.36/2.15	cons(x1, x2)  =  cons(x2)
5.36/2.15	
5.36/2.15	ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
5.36/2.15	
5.36/2.15	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
5.36/2.15	
5.36/2.15	U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
5.36/2.15	
5.36/2.15	ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
5.36/2.15	
5.36/2.15	ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
5.36/2.15	
5.36/2.15	U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
5.36/2.15	
5.36/2.15	U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
5.36/2.15	
5.36/2.15	PERM_IN_GA(x1, x2)  =  PERM_IN_GA(x1)
5.36/2.15	
5.36/2.15	U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)
5.36/2.15	
5.36/2.15	AP1_IN_AAG(x1, x2, x3)  =  AP1_IN_AAG(x3)
5.36/2.15	
5.36/2.15	U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x5)
5.36/2.15	
5.36/2.15	U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x6)
5.36/2.15	
5.36/2.15	AP2_IN_GGA(x1, x2, x3)  =  AP2_IN_GGA(x1, x2)
5.36/2.15	
5.36/2.15	U2_GGA(x1, x2, x3, x4, x5)  =  U2_GGA(x5)
5.36/2.15	
5.36/2.15	U5_GA(x1, x2, x3, x4)  =  U5_GA(x4)
5.36/2.15	
5.36/2.15	
5.36/2.15	We have to consider all (P,R,Pi)-chains
5.36/2.15	----------------------------------------
5.36/2.15	
5.36/2.15	(4)
5.36/2.15	Obligation:
5.36/2.15	Pi DP problem:
5.36/2.15	The TRS P consists of the following rules:
5.36/2.15	
5.36/2.15	   PERM_IN_GA(Xs, cons(X, Ys)) -> U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
5.36/2.15	   PERM_IN_GA(Xs, cons(X, Ys)) -> AP1_IN_AAG(X1s, cons(X, X2s), Xs)
5.36/2.15	   AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> U1_AAG(H, X, Y, Z, ap1_in_aag(X, Y, Z))
5.36/2.15	   AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> AP1_IN_AAG(X, Y, Z)
5.36/2.16	   U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
5.36/2.16	   U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> AP2_IN_GGA(X1s, X2s, Zs)
5.36/2.16	   AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> U2_GGA(H, X, Y, Z, ap2_in_gga(X, Y, Z))
5.36/2.16	   AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> AP2_IN_GGA(X, Y, Z)
5.36/2.16	   U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_GA(Xs, X, Ys, perm_in_ga(Zs, Ys))
5.36/2.16	   U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> PERM_IN_GA(Zs, Ys)
5.36/2.16	
5.36/2.16	The TRS R consists of the following rules:
5.36/2.16	
5.36/2.16	   perm_in_ga(nil, nil) -> perm_out_ga(nil, nil)
5.36/2.16	   perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
5.36/2.16	   ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X)
5.36/2.16	   ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
5.36/2.16	   U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z))
5.36/2.16	   U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
5.36/2.16	   ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X)
5.36/2.16	   ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
5.36/2.16	   U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z))
5.36/2.16	   U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
5.36/2.16	   U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys))
5.36/2.16	
5.36/2.16	The argument filtering Pi contains the following mapping:
5.36/2.16	perm_in_ga(x1, x2)  =  perm_in_ga(x1)
5.36/2.16	
5.36/2.16	nil  =  nil
5.36/2.16	
5.36/2.16	perm_out_ga(x1, x2)  =  perm_out_ga(x2)
5.36/2.16	
5.36/2.16	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
5.36/2.16	
5.36/2.16	ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
5.36/2.16	
5.36/2.16	cons(x1, x2)  =  cons(x2)
5.36/2.16	
5.36/2.16	ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
5.36/2.16	
5.36/2.16	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
5.36/2.16	
5.36/2.16	U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
5.36/2.16	
5.36/2.16	ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
5.36/2.16	
5.36/2.16	ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
5.36/2.16	
5.36/2.16	U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
5.36/2.16	
5.36/2.16	U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
5.36/2.16	
5.36/2.16	PERM_IN_GA(x1, x2)  =  PERM_IN_GA(x1)
5.43/2.16	
5.43/2.16	U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)
5.43/2.16	
5.43/2.16	AP1_IN_AAG(x1, x2, x3)  =  AP1_IN_AAG(x3)
5.43/2.16	
5.43/2.16	U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x5)
5.43/2.16	
5.43/2.16	U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x6)
5.43/2.16	
5.43/2.16	AP2_IN_GGA(x1, x2, x3)  =  AP2_IN_GGA(x1, x2)
5.43/2.16	
5.43/2.16	U2_GGA(x1, x2, x3, x4, x5)  =  U2_GGA(x5)
5.43/2.16	
5.43/2.16	U5_GA(x1, x2, x3, x4)  =  U5_GA(x4)
5.43/2.16	
5.43/2.16	
5.43/2.16	We have to consider all (P,R,Pi)-chains
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(5) DependencyGraphProof (EQUIVALENT)
5.43/2.16	The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 5 less nodes.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(6)
5.43/2.16	Complex Obligation (AND)
5.43/2.16	
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(7)
5.43/2.16	Obligation:
5.43/2.16	Pi DP problem:
5.43/2.16	The TRS P consists of the following rules:
5.43/2.16	
5.43/2.16	   AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> AP2_IN_GGA(X, Y, Z)
5.43/2.16	
5.43/2.16	The TRS R consists of the following rules:
5.43/2.16	
5.43/2.16	   perm_in_ga(nil, nil) -> perm_out_ga(nil, nil)
5.43/2.16	   perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
5.43/2.16	   ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X)
5.43/2.16	   ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
5.43/2.16	   U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z))
5.43/2.16	   U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
5.43/2.16	   ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X)
5.43/2.16	   ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
5.43/2.16	   U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z))
5.43/2.16	   U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
5.43/2.16	   U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys))
5.43/2.16	
5.43/2.16	The argument filtering Pi contains the following mapping:
5.43/2.16	perm_in_ga(x1, x2)  =  perm_in_ga(x1)
5.43/2.16	
5.43/2.16	nil  =  nil
5.43/2.16	
5.43/2.16	perm_out_ga(x1, x2)  =  perm_out_ga(x2)
5.43/2.16	
5.43/2.16	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
5.43/2.16	
5.43/2.16	ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
5.43/2.16	
5.43/2.16	cons(x1, x2)  =  cons(x2)
5.43/2.16	
5.43/2.16	ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
5.43/2.16	
5.43/2.16	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
5.43/2.16	
5.43/2.16	U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
5.43/2.16	
5.43/2.16	ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
5.43/2.16	
5.43/2.16	ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
5.43/2.16	
5.43/2.16	U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
5.43/2.16	
5.43/2.16	U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
5.43/2.16	
5.43/2.16	AP2_IN_GGA(x1, x2, x3)  =  AP2_IN_GGA(x1, x2)
5.43/2.16	
5.43/2.16	
5.43/2.16	We have to consider all (P,R,Pi)-chains
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(8) UsableRulesProof (EQUIVALENT)
5.43/2.16	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(9)
5.43/2.16	Obligation:
5.43/2.16	Pi DP problem:
5.43/2.16	The TRS P consists of the following rules:
5.43/2.16	
5.43/2.16	   AP2_IN_GGA(cons(H, X), Y, cons(H, Z)) -> AP2_IN_GGA(X, Y, Z)
5.43/2.16	
5.43/2.16	R is empty.
5.43/2.16	The argument filtering Pi contains the following mapping:
5.43/2.16	cons(x1, x2)  =  cons(x2)
5.43/2.16	
5.43/2.16	AP2_IN_GGA(x1, x2, x3)  =  AP2_IN_GGA(x1, x2)
5.43/2.16	
5.43/2.16	
5.43/2.16	We have to consider all (P,R,Pi)-chains
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(10) PiDPToQDPProof (SOUND)
5.43/2.16	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(11)
5.43/2.16	Obligation:
5.43/2.16	Q DP problem:
5.43/2.16	The TRS P consists of the following rules:
5.43/2.16	
5.43/2.16	   AP2_IN_GGA(cons(X), Y) -> AP2_IN_GGA(X, Y)
5.43/2.16	
5.43/2.16	R is empty.
5.43/2.16	Q is empty.
5.43/2.16	We have to consider all (P,Q,R)-chains.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(12) QDPSizeChangeProof (EQUIVALENT)
5.43/2.16	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. 
5.43/2.16	
5.43/2.16	From the DPs we obtained the following set of size-change graphs:
5.43/2.16	*AP2_IN_GGA(cons(X), Y) -> AP2_IN_GGA(X, Y)
5.43/2.16	The graph contains the following edges 1 > 1, 2 >= 2
5.43/2.16	
5.43/2.16	
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(13)
5.43/2.16	YES
5.43/2.16	
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(14)
5.43/2.16	Obligation:
5.43/2.16	Pi DP problem:
5.43/2.16	The TRS P consists of the following rules:
5.43/2.16	
5.43/2.16	   AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> AP1_IN_AAG(X, Y, Z)
5.43/2.16	
5.43/2.16	The TRS R consists of the following rules:
5.43/2.16	
5.43/2.16	   perm_in_ga(nil, nil) -> perm_out_ga(nil, nil)
5.43/2.16	   perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
5.43/2.16	   ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X)
5.43/2.16	   ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
5.43/2.16	   U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z))
5.43/2.16	   U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
5.43/2.16	   ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X)
5.43/2.16	   ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
5.43/2.16	   U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z))
5.43/2.16	   U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
5.43/2.16	   U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys))
5.43/2.16	
5.43/2.16	The argument filtering Pi contains the following mapping:
5.43/2.16	perm_in_ga(x1, x2)  =  perm_in_ga(x1)
5.43/2.16	
5.43/2.16	nil  =  nil
5.43/2.16	
5.43/2.16	perm_out_ga(x1, x2)  =  perm_out_ga(x2)
5.43/2.16	
5.43/2.16	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
5.43/2.16	
5.43/2.16	ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
5.43/2.16	
5.43/2.16	cons(x1, x2)  =  cons(x2)
5.43/2.16	
5.43/2.16	ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
5.43/2.16	
5.43/2.16	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
5.43/2.16	
5.43/2.16	U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
5.43/2.16	
5.43/2.16	ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
5.43/2.16	
5.43/2.16	ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
5.43/2.16	
5.43/2.16	U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
5.43/2.16	
5.43/2.16	U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
5.43/2.16	
5.43/2.16	AP1_IN_AAG(x1, x2, x3)  =  AP1_IN_AAG(x3)
5.43/2.16	
5.43/2.16	
5.43/2.16	We have to consider all (P,R,Pi)-chains
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(15) UsableRulesProof (EQUIVALENT)
5.43/2.16	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(16)
5.43/2.16	Obligation:
5.43/2.16	Pi DP problem:
5.43/2.16	The TRS P consists of the following rules:
5.43/2.16	
5.43/2.16	   AP1_IN_AAG(cons(H, X), Y, cons(H, Z)) -> AP1_IN_AAG(X, Y, Z)
5.43/2.16	
5.43/2.16	R is empty.
5.43/2.16	The argument filtering Pi contains the following mapping:
5.43/2.16	cons(x1, x2)  =  cons(x2)
5.43/2.16	
5.43/2.16	AP1_IN_AAG(x1, x2, x3)  =  AP1_IN_AAG(x3)
5.43/2.16	
5.43/2.16	
5.43/2.16	We have to consider all (P,R,Pi)-chains
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(17) PiDPToQDPProof (SOUND)
5.43/2.16	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(18)
5.43/2.16	Obligation:
5.43/2.16	Q DP problem:
5.43/2.16	The TRS P consists of the following rules:
5.43/2.16	
5.43/2.16	   AP1_IN_AAG(cons(Z)) -> AP1_IN_AAG(Z)
5.43/2.16	
5.43/2.16	R is empty.
5.43/2.16	Q is empty.
5.43/2.16	We have to consider all (P,Q,R)-chains.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(19) QDPSizeChangeProof (EQUIVALENT)
5.43/2.16	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. 
5.43/2.16	
5.43/2.16	From the DPs we obtained the following set of size-change graphs:
5.43/2.16	*AP1_IN_AAG(cons(Z)) -> AP1_IN_AAG(Z)
5.43/2.16	The graph contains the following edges 1 > 1
5.43/2.16	
5.43/2.16	
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(20)
5.43/2.16	YES
5.43/2.16	
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(21)
5.43/2.16	Obligation:
5.43/2.16	Pi DP problem:
5.43/2.16	The TRS P consists of the following rules:
5.43/2.16	
5.43/2.16	   U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
5.43/2.16	   U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> PERM_IN_GA(Zs, Ys)
5.43/2.16	   PERM_IN_GA(Xs, cons(X, Ys)) -> U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
5.43/2.16	
5.43/2.16	The TRS R consists of the following rules:
5.43/2.16	
5.43/2.16	   perm_in_ga(nil, nil) -> perm_out_ga(nil, nil)
5.43/2.16	   perm_in_ga(Xs, cons(X, Ys)) -> U3_ga(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
5.43/2.16	   ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X)
5.43/2.16	   ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
5.43/2.16	   U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z))
5.43/2.16	   U3_ga(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_ga(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
5.43/2.16	   ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X)
5.43/2.16	   ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
5.43/2.16	   U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z))
5.43/2.16	   U4_ga(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> U5_ga(Xs, X, Ys, perm_in_ga(Zs, Ys))
5.43/2.16	   U5_ga(Xs, X, Ys, perm_out_ga(Zs, Ys)) -> perm_out_ga(Xs, cons(X, Ys))
5.43/2.16	
5.43/2.16	The argument filtering Pi contains the following mapping:
5.43/2.16	perm_in_ga(x1, x2)  =  perm_in_ga(x1)
5.43/2.16	
5.43/2.16	nil  =  nil
5.43/2.16	
5.43/2.16	perm_out_ga(x1, x2)  =  perm_out_ga(x2)
5.43/2.16	
5.43/2.16	U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)
5.43/2.16	
5.43/2.16	ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
5.43/2.16	
5.43/2.16	cons(x1, x2)  =  cons(x2)
5.43/2.16	
5.43/2.16	ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
5.43/2.16	
5.43/2.16	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
5.43/2.16	
5.43/2.16	U4_ga(x1, x2, x3, x4, x5, x6)  =  U4_ga(x6)
5.43/2.16	
5.43/2.16	ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
5.43/2.16	
5.43/2.16	ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
5.43/2.16	
5.43/2.16	U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
5.43/2.16	
5.43/2.16	U5_ga(x1, x2, x3, x4)  =  U5_ga(x4)
5.43/2.16	
5.43/2.16	PERM_IN_GA(x1, x2)  =  PERM_IN_GA(x1)
5.43/2.16	
5.43/2.16	U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)
5.43/2.16	
5.43/2.16	U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x6)
5.43/2.16	
5.43/2.16	
5.43/2.16	We have to consider all (P,R,Pi)-chains
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(22) UsableRulesProof (EQUIVALENT)
5.43/2.16	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(23)
5.43/2.16	Obligation:
5.43/2.16	Pi DP problem:
5.43/2.16	The TRS P consists of the following rules:
5.43/2.16	
5.43/2.16	   U3_GA(Xs, X, Ys, ap1_out_aag(X1s, cons(X, X2s), Xs)) -> U4_GA(Xs, X, Ys, X1s, X2s, ap2_in_gga(X1s, X2s, Zs))
5.43/2.16	   U4_GA(Xs, X, Ys, X1s, X2s, ap2_out_gga(X1s, X2s, Zs)) -> PERM_IN_GA(Zs, Ys)
5.43/2.16	   PERM_IN_GA(Xs, cons(X, Ys)) -> U3_GA(Xs, X, Ys, ap1_in_aag(X1s, cons(X, X2s), Xs))
5.43/2.16	
5.43/2.16	The TRS R consists of the following rules:
5.43/2.16	
5.43/2.16	   ap2_in_gga(nil, X, X) -> ap2_out_gga(nil, X, X)
5.43/2.16	   ap2_in_gga(cons(H, X), Y, cons(H, Z)) -> U2_gga(H, X, Y, Z, ap2_in_gga(X, Y, Z))
5.43/2.16	   ap1_in_aag(nil, X, X) -> ap1_out_aag(nil, X, X)
5.43/2.16	   ap1_in_aag(cons(H, X), Y, cons(H, Z)) -> U1_aag(H, X, Y, Z, ap1_in_aag(X, Y, Z))
5.43/2.16	   U2_gga(H, X, Y, Z, ap2_out_gga(X, Y, Z)) -> ap2_out_gga(cons(H, X), Y, cons(H, Z))
5.43/2.16	   U1_aag(H, X, Y, Z, ap1_out_aag(X, Y, Z)) -> ap1_out_aag(cons(H, X), Y, cons(H, Z))
5.43/2.16	
5.43/2.16	The argument filtering Pi contains the following mapping:
5.43/2.16	nil  =  nil
5.43/2.16	
5.43/2.16	ap1_in_aag(x1, x2, x3)  =  ap1_in_aag(x3)
5.43/2.16	
5.43/2.16	cons(x1, x2)  =  cons(x2)
5.43/2.16	
5.43/2.16	ap1_out_aag(x1, x2, x3)  =  ap1_out_aag(x1, x2)
5.43/2.16	
5.43/2.16	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x5)
5.43/2.16	
5.43/2.16	ap2_in_gga(x1, x2, x3)  =  ap2_in_gga(x1, x2)
5.43/2.16	
5.43/2.16	ap2_out_gga(x1, x2, x3)  =  ap2_out_gga(x3)
5.43/2.16	
5.43/2.16	U2_gga(x1, x2, x3, x4, x5)  =  U2_gga(x5)
5.43/2.16	
5.43/2.16	PERM_IN_GA(x1, x2)  =  PERM_IN_GA(x1)
5.43/2.16	
5.43/2.16	U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)
5.43/2.16	
5.43/2.16	U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x6)
5.43/2.16	
5.43/2.16	
5.43/2.16	We have to consider all (P,R,Pi)-chains
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(24) PiDPToQDPProof (SOUND)
5.43/2.16	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(25)
5.43/2.16	Obligation:
5.43/2.16	Q DP problem:
5.43/2.16	The TRS P consists of the following rules:
5.43/2.16	
5.43/2.16	   U3_GA(ap1_out_aag(X1s, cons(X2s))) -> U4_GA(ap2_in_gga(X1s, X2s))
5.43/2.16	   U4_GA(ap2_out_gga(Zs)) -> PERM_IN_GA(Zs)
5.43/2.16	   PERM_IN_GA(Xs) -> U3_GA(ap1_in_aag(Xs))
5.43/2.16	
5.43/2.16	The TRS R consists of the following rules:
5.43/2.16	
5.43/2.16	   ap2_in_gga(nil, X) -> ap2_out_gga(X)
5.43/2.16	   ap2_in_gga(cons(X), Y) -> U2_gga(ap2_in_gga(X, Y))
5.43/2.16	   ap1_in_aag(X) -> ap1_out_aag(nil, X)
5.43/2.16	   ap1_in_aag(cons(Z)) -> U1_aag(ap1_in_aag(Z))
5.43/2.16	   U2_gga(ap2_out_gga(Z)) -> ap2_out_gga(cons(Z))
5.43/2.16	   U1_aag(ap1_out_aag(X, Y)) -> ap1_out_aag(cons(X), Y)
5.43/2.16	
5.43/2.16	The set Q consists of the following terms:
5.43/2.16	
5.43/2.16	   ap2_in_gga(x0, x1)
5.43/2.16	   ap1_in_aag(x0)
5.43/2.16	   U2_gga(x0)
5.43/2.16	   U1_aag(x0)
5.43/2.16	
5.43/2.16	We have to consider all (P,Q,R)-chains.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(26) MRRProof (EQUIVALENT)
5.43/2.16	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.
5.43/2.16	
5.43/2.16	Strictly oriented dependency pairs:
5.43/2.16	
5.43/2.16	   U3_GA(ap1_out_aag(X1s, cons(X2s))) -> U4_GA(ap2_in_gga(X1s, X2s))
5.43/2.16	   U4_GA(ap2_out_gga(Zs)) -> PERM_IN_GA(Zs)
5.43/2.16	   PERM_IN_GA(Xs) -> U3_GA(ap1_in_aag(Xs))
5.43/2.16	
5.43/2.16	Strictly oriented rules of the TRS R:
5.43/2.16	
5.43/2.16	   ap2_in_gga(nil, X) -> ap2_out_gga(X)
5.43/2.16	   ap2_in_gga(cons(X), Y) -> U2_gga(ap2_in_gga(X, Y))
5.43/2.16	   ap1_in_aag(X) -> ap1_out_aag(nil, X)
5.43/2.16	   ap1_in_aag(cons(Z)) -> U1_aag(ap1_in_aag(Z))
5.43/2.16	   U2_gga(ap2_out_gga(Z)) -> ap2_out_gga(cons(Z))
5.43/2.16	   U1_aag(ap1_out_aag(X, Y)) -> ap1_out_aag(cons(X), Y)
5.43/2.16	
5.43/2.16	Used ordering: Knuth-Bendix order [KBO] with precedence:ap2_in_gga_2 > nil > PERM_IN_GA_1 > cons_1 > U4_GA_1 > ap1_in_aag_1 > U3_GA_1 > U1_aag_1 > ap1_out_aag_2 > U2_gga_1 > ap2_out_gga_1
5.43/2.16	
5.43/2.16	and weight map:
5.43/2.16	
5.43/2.16	   nil=2
5.43/2.16	   ap2_out_gga_1=6
5.43/2.16	   cons_1=7
5.43/2.16	   U2_gga_1=7
5.43/2.16	   ap1_in_aag_1=2
5.43/2.16	   U1_aag_1=7
5.43/2.16	   U3_GA_1=3
5.43/2.16	   U4_GA_1=1
5.43/2.16	   PERM_IN_GA_1=6
5.43/2.16	   ap2_in_gga_2=8
5.43/2.16	   ap1_out_aag_2=0
5.43/2.16	
5.43/2.16	The variable weight is 1
5.43/2.16	
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(27)
5.43/2.16	Obligation:
5.43/2.16	Q DP problem:
5.43/2.16	P is empty.
5.43/2.16	R is empty.
5.43/2.16	The set Q consists of the following terms:
5.43/2.16	
5.43/2.16	   ap2_in_gga(x0, x1)
5.43/2.16	   ap1_in_aag(x0)
5.43/2.16	   U2_gga(x0)
5.43/2.16	   U1_aag(x0)
5.43/2.16	
5.43/2.16	We have to consider all (P,Q,R)-chains.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(28) PisEmptyProof (EQUIVALENT)
5.43/2.16	The TRS P is empty. Hence, there is no (P,Q,R) chain.
5.43/2.16	----------------------------------------
5.43/2.16	
5.43/2.16	(29)
5.43/2.16	YES
5.50/2.22	EOF