4.76/1.98	YES
4.94/2.00	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
4.94/2.00	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.94/2.00	
4.94/2.00	
4.94/2.00	Left Termination of the query pattern
4.94/2.00	
4.94/2.00	ordered(g)
4.94/2.00	
4.94/2.00	w.r.t. the given Prolog program could successfully be proven:
4.94/2.00	
4.94/2.00	(0) Prolog
4.94/2.00	(1) PrologToPiTRSProof [SOUND, 0 ms]
4.94/2.00	(2) PiTRS
4.94/2.00	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
4.94/2.00	(4) PiDP
4.94/2.00	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
4.94/2.00	(6) AND
4.94/2.00	    (7) PiDP
4.94/2.00	        (8) UsableRulesProof [EQUIVALENT, 0 ms]
4.94/2.00	        (9) PiDP
4.94/2.00	        (10) PiDPToQDPProof [EQUIVALENT, 17 ms]
4.94/2.00	        (11) QDP
4.94/2.00	        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
4.94/2.00	        (13) YES
4.94/2.00	    (14) PiDP
4.94/2.00	        (15) UsableRulesProof [EQUIVALENT, 0 ms]
4.94/2.00	        (16) PiDP
4.94/2.00	        (17) PiDPToQDPProof [SOUND, 0 ms]
4.94/2.00	        (18) QDP
4.94/2.00	        (19) UsableRulesReductionPairsProof [EQUIVALENT, 7 ms]
4.94/2.00	        (20) QDP
4.94/2.00	        (21) DependencyGraphProof [EQUIVALENT, 0 ms]
4.94/2.00	        (22) TRUE
4.94/2.00	
4.94/2.00	
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(0)
4.94/2.00	Obligation:
4.94/2.00	Clauses:
4.94/2.00	
4.94/2.00	ordered([]).
4.94/2.00	ordered(.(X, [])).
4.94/2.00	ordered(.(X, .(Y, Xs))) :- ','(le(X, Y), ordered(.(Y, Xs))).
4.94/2.00	le(s(X), s(Y)) :- le(X, Y).
4.94/2.00	le(0, s(0)).
4.94/2.00	le(0, 0).
4.94/2.00	
4.94/2.00	
4.94/2.00	Query: ordered(g)
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(1) PrologToPiTRSProof (SOUND)
4.94/2.00	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
4.94/2.00	
4.94/2.00	ordered_in_1: (b)
4.94/2.00	
4.94/2.00	le_in_2: (b,b)
4.94/2.00	
4.94/2.00	Transforming Prolog into the following Term Rewriting System:
4.94/2.00	
4.94/2.00	Pi-finite rewrite system:
4.94/2.00	The TRS R consists of the following rules:
4.94/2.00	
4.94/2.00	   ordered_in_g([]) -> ordered_out_g([])
4.94/2.00	   ordered_in_g(.(X, [])) -> ordered_out_g(.(X, []))
4.94/2.00	   ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(s(X), s(Y)) -> U3_gg(X, Y, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(0, s(0)) -> le_out_gg(0, s(0))
4.94/2.00	   le_in_gg(0, 0) -> le_out_gg(0, 0)
4.94/2.00	   U3_gg(X, Y, le_out_gg(X, Y)) -> le_out_gg(s(X), s(Y))
4.94/2.00	   U1_g(X, Y, Xs, le_out_gg(X, Y)) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
4.94/2.00	   U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs)))
4.94/2.00	
4.94/2.00	The argument filtering Pi contains the following mapping:
4.94/2.00	ordered_in_g(x1)  =  ordered_in_g(x1)
4.94/2.00	
4.94/2.00	[]  =  []
4.94/2.00	
4.94/2.00	ordered_out_g(x1)  =  ordered_out_g
4.94/2.00	
4.94/2.00	.(x1, x2)  =  .(x1, x2)
4.94/2.00	
4.94/2.00	U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
4.94/2.00	
4.94/2.00	le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
4.94/2.00	
4.94/2.00	s(x1)  =  s(x1)
4.94/2.00	
4.94/2.00	U3_gg(x1, x2, x3)  =  U3_gg(x3)
4.94/2.00	
4.94/2.00	0  =  0
4.94/2.00	
4.94/2.00	le_out_gg(x1, x2)  =  le_out_gg
4.94/2.00	
4.94/2.00	U2_g(x1, x2, x3, x4)  =  U2_g(x4)
4.94/2.00	
4.94/2.00	
4.94/2.00	
4.94/2.00	
4.94/2.00	
4.94/2.00	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
4.94/2.00	
4.94/2.00	
4.94/2.00	
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(2)
4.94/2.00	Obligation:
4.94/2.00	Pi-finite rewrite system:
4.94/2.00	The TRS R consists of the following rules:
4.94/2.00	
4.94/2.00	   ordered_in_g([]) -> ordered_out_g([])
4.94/2.00	   ordered_in_g(.(X, [])) -> ordered_out_g(.(X, []))
4.94/2.00	   ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(s(X), s(Y)) -> U3_gg(X, Y, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(0, s(0)) -> le_out_gg(0, s(0))
4.94/2.00	   le_in_gg(0, 0) -> le_out_gg(0, 0)
4.94/2.00	   U3_gg(X, Y, le_out_gg(X, Y)) -> le_out_gg(s(X), s(Y))
4.94/2.00	   U1_g(X, Y, Xs, le_out_gg(X, Y)) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
4.94/2.00	   U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs)))
4.94/2.00	
4.94/2.00	The argument filtering Pi contains the following mapping:
4.94/2.00	ordered_in_g(x1)  =  ordered_in_g(x1)
4.94/2.00	
4.94/2.00	[]  =  []
4.94/2.00	
4.94/2.00	ordered_out_g(x1)  =  ordered_out_g
4.94/2.00	
4.94/2.00	.(x1, x2)  =  .(x1, x2)
4.94/2.00	
4.94/2.00	U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
4.94/2.00	
4.94/2.00	le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
4.94/2.00	
4.94/2.00	s(x1)  =  s(x1)
4.94/2.00	
4.94/2.00	U3_gg(x1, x2, x3)  =  U3_gg(x3)
4.94/2.00	
4.94/2.00	0  =  0
4.94/2.00	
4.94/2.00	le_out_gg(x1, x2)  =  le_out_gg
4.94/2.00	
4.94/2.00	U2_g(x1, x2, x3, x4)  =  U2_g(x4)
4.94/2.00	
4.94/2.00	
4.94/2.00	
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(3) DependencyPairsProof (EQUIVALENT)
4.94/2.00	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
4.94/2.00	Pi DP problem:
4.94/2.00	The TRS P consists of the following rules:
4.94/2.00	
4.94/2.00	   ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(X, Y, Xs, le_in_gg(X, Y))
4.94/2.00	   ORDERED_IN_G(.(X, .(Y, Xs))) -> LE_IN_GG(X, Y)
4.94/2.00	   LE_IN_GG(s(X), s(Y)) -> U3_GG(X, Y, le_in_gg(X, Y))
4.94/2.00	   LE_IN_GG(s(X), s(Y)) -> LE_IN_GG(X, Y)
4.94/2.00	   U1_G(X, Y, Xs, le_out_gg(X, Y)) -> U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
4.94/2.00	   U1_G(X, Y, Xs, le_out_gg(X, Y)) -> ORDERED_IN_G(.(Y, Xs))
4.94/2.00	
4.94/2.00	The TRS R consists of the following rules:
4.94/2.00	
4.94/2.00	   ordered_in_g([]) -> ordered_out_g([])
4.94/2.00	   ordered_in_g(.(X, [])) -> ordered_out_g(.(X, []))
4.94/2.00	   ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(s(X), s(Y)) -> U3_gg(X, Y, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(0, s(0)) -> le_out_gg(0, s(0))
4.94/2.00	   le_in_gg(0, 0) -> le_out_gg(0, 0)
4.94/2.00	   U3_gg(X, Y, le_out_gg(X, Y)) -> le_out_gg(s(X), s(Y))
4.94/2.00	   U1_g(X, Y, Xs, le_out_gg(X, Y)) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
4.94/2.00	   U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs)))
4.94/2.00	
4.94/2.00	The argument filtering Pi contains the following mapping:
4.94/2.00	ordered_in_g(x1)  =  ordered_in_g(x1)
4.94/2.00	
4.94/2.00	[]  =  []
4.94/2.00	
4.94/2.00	ordered_out_g(x1)  =  ordered_out_g
4.94/2.00	
4.94/2.00	.(x1, x2)  =  .(x1, x2)
4.94/2.00	
4.94/2.00	U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
4.94/2.00	
4.94/2.00	le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
4.94/2.00	
4.94/2.00	s(x1)  =  s(x1)
4.94/2.00	
4.94/2.00	U3_gg(x1, x2, x3)  =  U3_gg(x3)
4.94/2.00	
4.94/2.00	0  =  0
4.94/2.00	
4.94/2.00	le_out_gg(x1, x2)  =  le_out_gg
4.94/2.00	
4.94/2.00	U2_g(x1, x2, x3, x4)  =  U2_g(x4)
4.94/2.00	
4.94/2.00	ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
4.94/2.00	
4.94/2.00	U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)
4.94/2.00	
4.94/2.00	LE_IN_GG(x1, x2)  =  LE_IN_GG(x1, x2)
4.94/2.00	
4.94/2.00	U3_GG(x1, x2, x3)  =  U3_GG(x3)
4.94/2.00	
4.94/2.00	U2_G(x1, x2, x3, x4)  =  U2_G(x4)
4.94/2.00	
4.94/2.00	
4.94/2.00	We have to consider all (P,R,Pi)-chains
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(4)
4.94/2.00	Obligation:
4.94/2.00	Pi DP problem:
4.94/2.00	The TRS P consists of the following rules:
4.94/2.00	
4.94/2.00	   ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(X, Y, Xs, le_in_gg(X, Y))
4.94/2.00	   ORDERED_IN_G(.(X, .(Y, Xs))) -> LE_IN_GG(X, Y)
4.94/2.00	   LE_IN_GG(s(X), s(Y)) -> U3_GG(X, Y, le_in_gg(X, Y))
4.94/2.00	   LE_IN_GG(s(X), s(Y)) -> LE_IN_GG(X, Y)
4.94/2.00	   U1_G(X, Y, Xs, le_out_gg(X, Y)) -> U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
4.94/2.00	   U1_G(X, Y, Xs, le_out_gg(X, Y)) -> ORDERED_IN_G(.(Y, Xs))
4.94/2.00	
4.94/2.00	The TRS R consists of the following rules:
4.94/2.00	
4.94/2.00	   ordered_in_g([]) -> ordered_out_g([])
4.94/2.00	   ordered_in_g(.(X, [])) -> ordered_out_g(.(X, []))
4.94/2.00	   ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(s(X), s(Y)) -> U3_gg(X, Y, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(0, s(0)) -> le_out_gg(0, s(0))
4.94/2.00	   le_in_gg(0, 0) -> le_out_gg(0, 0)
4.94/2.00	   U3_gg(X, Y, le_out_gg(X, Y)) -> le_out_gg(s(X), s(Y))
4.94/2.00	   U1_g(X, Y, Xs, le_out_gg(X, Y)) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
4.94/2.00	   U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs)))
4.94/2.00	
4.94/2.00	The argument filtering Pi contains the following mapping:
4.94/2.00	ordered_in_g(x1)  =  ordered_in_g(x1)
4.94/2.00	
4.94/2.00	[]  =  []
4.94/2.00	
4.94/2.00	ordered_out_g(x1)  =  ordered_out_g
4.94/2.00	
4.94/2.00	.(x1, x2)  =  .(x1, x2)
4.94/2.00	
4.94/2.00	U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
4.94/2.00	
4.94/2.00	le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
4.94/2.00	
4.94/2.00	s(x1)  =  s(x1)
4.94/2.00	
4.94/2.00	U3_gg(x1, x2, x3)  =  U3_gg(x3)
4.94/2.00	
4.94/2.00	0  =  0
4.94/2.00	
4.94/2.00	le_out_gg(x1, x2)  =  le_out_gg
4.94/2.00	
4.94/2.00	U2_g(x1, x2, x3, x4)  =  U2_g(x4)
4.94/2.00	
4.94/2.00	ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
4.94/2.00	
4.94/2.00	U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)
4.94/2.00	
4.94/2.00	LE_IN_GG(x1, x2)  =  LE_IN_GG(x1, x2)
4.94/2.00	
4.94/2.00	U3_GG(x1, x2, x3)  =  U3_GG(x3)
4.94/2.00	
4.94/2.00	U2_G(x1, x2, x3, x4)  =  U2_G(x4)
4.94/2.00	
4.94/2.00	
4.94/2.00	We have to consider all (P,R,Pi)-chains
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(5) DependencyGraphProof (EQUIVALENT)
4.94/2.00	The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(6)
4.94/2.00	Complex Obligation (AND)
4.94/2.00	
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(7)
4.94/2.00	Obligation:
4.94/2.00	Pi DP problem:
4.94/2.00	The TRS P consists of the following rules:
4.94/2.00	
4.94/2.00	   LE_IN_GG(s(X), s(Y)) -> LE_IN_GG(X, Y)
4.94/2.00	
4.94/2.00	The TRS R consists of the following rules:
4.94/2.00	
4.94/2.00	   ordered_in_g([]) -> ordered_out_g([])
4.94/2.00	   ordered_in_g(.(X, [])) -> ordered_out_g(.(X, []))
4.94/2.00	   ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(s(X), s(Y)) -> U3_gg(X, Y, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(0, s(0)) -> le_out_gg(0, s(0))
4.94/2.00	   le_in_gg(0, 0) -> le_out_gg(0, 0)
4.94/2.00	   U3_gg(X, Y, le_out_gg(X, Y)) -> le_out_gg(s(X), s(Y))
4.94/2.00	   U1_g(X, Y, Xs, le_out_gg(X, Y)) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
4.94/2.00	   U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs)))
4.94/2.00	
4.94/2.00	The argument filtering Pi contains the following mapping:
4.94/2.00	ordered_in_g(x1)  =  ordered_in_g(x1)
4.94/2.00	
4.94/2.00	[]  =  []
4.94/2.00	
4.94/2.00	ordered_out_g(x1)  =  ordered_out_g
4.94/2.00	
4.94/2.00	.(x1, x2)  =  .(x1, x2)
4.94/2.00	
4.94/2.00	U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
4.94/2.00	
4.94/2.00	le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
4.94/2.00	
4.94/2.00	s(x1)  =  s(x1)
4.94/2.00	
4.94/2.00	U3_gg(x1, x2, x3)  =  U3_gg(x3)
4.94/2.00	
4.94/2.00	0  =  0
4.94/2.00	
4.94/2.00	le_out_gg(x1, x2)  =  le_out_gg
4.94/2.00	
4.94/2.00	U2_g(x1, x2, x3, x4)  =  U2_g(x4)
4.94/2.00	
4.94/2.00	LE_IN_GG(x1, x2)  =  LE_IN_GG(x1, x2)
4.94/2.00	
4.94/2.00	
4.94/2.00	We have to consider all (P,R,Pi)-chains
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(8) UsableRulesProof (EQUIVALENT)
4.94/2.00	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(9)
4.94/2.00	Obligation:
4.94/2.00	Pi DP problem:
4.94/2.00	The TRS P consists of the following rules:
4.94/2.00	
4.94/2.00	   LE_IN_GG(s(X), s(Y)) -> LE_IN_GG(X, Y)
4.94/2.00	
4.94/2.00	R is empty.
4.94/2.00	Pi is empty.
4.94/2.00	We have to consider all (P,R,Pi)-chains
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(10) PiDPToQDPProof (EQUIVALENT)
4.94/2.00	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(11)
4.94/2.00	Obligation:
4.94/2.00	Q DP problem:
4.94/2.00	The TRS P consists of the following rules:
4.94/2.00	
4.94/2.00	   LE_IN_GG(s(X), s(Y)) -> LE_IN_GG(X, Y)
4.94/2.00	
4.94/2.00	R is empty.
4.94/2.00	Q is empty.
4.94/2.00	We have to consider all (P,Q,R)-chains.
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(12) QDPSizeChangeProof (EQUIVALENT)
4.94/2.00	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. 
4.94/2.00	
4.94/2.00	From the DPs we obtained the following set of size-change graphs:
4.94/2.00	*LE_IN_GG(s(X), s(Y)) -> LE_IN_GG(X, Y)
4.94/2.00	The graph contains the following edges 1 > 1, 2 > 2
4.94/2.00	
4.94/2.00	
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(13)
4.94/2.00	YES
4.94/2.00	
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(14)
4.94/2.00	Obligation:
4.94/2.00	Pi DP problem:
4.94/2.00	The TRS P consists of the following rules:
4.94/2.00	
4.94/2.00	   U1_G(X, Y, Xs, le_out_gg(X, Y)) -> ORDERED_IN_G(.(Y, Xs))
4.94/2.00	   ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(X, Y, Xs, le_in_gg(X, Y))
4.94/2.00	
4.94/2.00	The TRS R consists of the following rules:
4.94/2.00	
4.94/2.00	   ordered_in_g([]) -> ordered_out_g([])
4.94/2.00	   ordered_in_g(.(X, [])) -> ordered_out_g(.(X, []))
4.94/2.00	   ordered_in_g(.(X, .(Y, Xs))) -> U1_g(X, Y, Xs, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(s(X), s(Y)) -> U3_gg(X, Y, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(0, s(0)) -> le_out_gg(0, s(0))
4.94/2.00	   le_in_gg(0, 0) -> le_out_gg(0, 0)
4.94/2.00	   U3_gg(X, Y, le_out_gg(X, Y)) -> le_out_gg(s(X), s(Y))
4.94/2.00	   U1_g(X, Y, Xs, le_out_gg(X, Y)) -> U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
4.94/2.00	   U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) -> ordered_out_g(.(X, .(Y, Xs)))
4.94/2.00	
4.94/2.00	The argument filtering Pi contains the following mapping:
4.94/2.00	ordered_in_g(x1)  =  ordered_in_g(x1)
4.94/2.00	
4.94/2.00	[]  =  []
4.94/2.00	
4.94/2.00	ordered_out_g(x1)  =  ordered_out_g
4.94/2.00	
4.94/2.00	.(x1, x2)  =  .(x1, x2)
4.94/2.00	
4.94/2.00	U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
4.94/2.00	
4.94/2.00	le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
4.94/2.00	
4.94/2.00	s(x1)  =  s(x1)
4.94/2.00	
4.94/2.00	U3_gg(x1, x2, x3)  =  U3_gg(x3)
4.94/2.00	
4.94/2.00	0  =  0
4.94/2.00	
4.94/2.00	le_out_gg(x1, x2)  =  le_out_gg
4.94/2.00	
4.94/2.00	U2_g(x1, x2, x3, x4)  =  U2_g(x4)
4.94/2.00	
4.94/2.00	ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
4.94/2.00	
4.94/2.00	U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)
4.94/2.00	
4.94/2.00	
4.94/2.00	We have to consider all (P,R,Pi)-chains
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(15) UsableRulesProof (EQUIVALENT)
4.94/2.00	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(16)
4.94/2.00	Obligation:
4.94/2.00	Pi DP problem:
4.94/2.00	The TRS P consists of the following rules:
4.94/2.00	
4.94/2.00	   U1_G(X, Y, Xs, le_out_gg(X, Y)) -> ORDERED_IN_G(.(Y, Xs))
4.94/2.00	   ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(X, Y, Xs, le_in_gg(X, Y))
4.94/2.00	
4.94/2.00	The TRS R consists of the following rules:
4.94/2.00	
4.94/2.00	   le_in_gg(s(X), s(Y)) -> U3_gg(X, Y, le_in_gg(X, Y))
4.94/2.00	   le_in_gg(0, s(0)) -> le_out_gg(0, s(0))
4.94/2.00	   le_in_gg(0, 0) -> le_out_gg(0, 0)
4.94/2.00	   U3_gg(X, Y, le_out_gg(X, Y)) -> le_out_gg(s(X), s(Y))
4.94/2.00	
4.94/2.00	The argument filtering Pi contains the following mapping:
4.94/2.00	.(x1, x2)  =  .(x1, x2)
4.94/2.00	
4.94/2.00	le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
4.94/2.00	
4.94/2.00	s(x1)  =  s(x1)
4.94/2.00	
4.94/2.00	U3_gg(x1, x2, x3)  =  U3_gg(x3)
4.94/2.00	
4.94/2.00	0  =  0
4.94/2.00	
4.94/2.00	le_out_gg(x1, x2)  =  le_out_gg
4.94/2.00	
4.94/2.00	ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
4.94/2.00	
4.94/2.00	U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)
4.94/2.00	
4.94/2.00	
4.94/2.00	We have to consider all (P,R,Pi)-chains
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(17) PiDPToQDPProof (SOUND)
4.94/2.00	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(18)
4.94/2.00	Obligation:
4.94/2.00	Q DP problem:
4.94/2.00	The TRS P consists of the following rules:
4.94/2.00	
4.94/2.00	   U1_G(Y, Xs, le_out_gg) -> ORDERED_IN_G(.(Y, Xs))
4.94/2.00	   ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(Y, Xs, le_in_gg(X, Y))
4.94/2.00	
4.94/2.00	The TRS R consists of the following rules:
4.94/2.00	
4.94/2.00	   le_in_gg(s(X), s(Y)) -> U3_gg(le_in_gg(X, Y))
4.94/2.00	   le_in_gg(0, s(0)) -> le_out_gg
4.94/2.00	   le_in_gg(0, 0) -> le_out_gg
4.94/2.00	   U3_gg(le_out_gg) -> le_out_gg
4.94/2.00	
4.94/2.00	The set Q consists of the following terms:
4.94/2.00	
4.94/2.00	   le_in_gg(x0, x1)
4.94/2.00	   U3_gg(x0)
4.94/2.00	
4.94/2.00	We have to consider all (P,Q,R)-chains.
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(19) UsableRulesReductionPairsProof (EQUIVALENT)
4.94/2.00	By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
4.94/2.00	
4.94/2.00	No dependency pairs are removed.
4.94/2.00	
4.94/2.00	The following rules are removed from R:
4.94/2.00	
4.94/2.00	   le_in_gg(s(X), s(Y)) -> U3_gg(le_in_gg(X, Y))
4.94/2.00	   le_in_gg(0, s(0)) -> le_out_gg
4.94/2.00	   le_in_gg(0, 0) -> le_out_gg
4.94/2.00	Used ordering: POLO with Polynomial interpretation [POLO]:
4.94/2.00	
4.94/2.00	   POL(.(x_1, x_2)) = 2*x_1 + 2*x_2
4.94/2.00	   POL(0) = 0
4.94/2.00	   POL(ORDERED_IN_G(x_1)) = x_1
4.94/2.00	   POL(U1_G(x_1, x_2, x_3)) = 2*x_1 + 2*x_2 + 2*x_3
4.94/2.00	   POL(U3_gg(x_1)) = x_1
4.94/2.00	   POL(le_in_gg(x_1, x_2)) = x_1 + x_2
4.94/2.00	   POL(le_out_gg) = 0
4.94/2.00	   POL(s(x_1)) = x_1
4.94/2.00	
4.94/2.00	
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(20)
4.94/2.00	Obligation:
4.94/2.00	Q DP problem:
4.94/2.00	The TRS P consists of the following rules:
4.94/2.00	
4.94/2.00	   U1_G(Y, Xs, le_out_gg) -> ORDERED_IN_G(.(Y, Xs))
4.94/2.00	   ORDERED_IN_G(.(X, .(Y, Xs))) -> U1_G(Y, Xs, le_in_gg(X, Y))
4.94/2.00	
4.94/2.00	The TRS R consists of the following rules:
4.94/2.00	
4.94/2.00	   U3_gg(le_out_gg) -> le_out_gg
4.94/2.00	
4.94/2.00	The set Q consists of the following terms:
4.94/2.00	
4.94/2.00	   le_in_gg(x0, x1)
4.94/2.00	   U3_gg(x0)
4.94/2.00	
4.94/2.00	We have to consider all (P,Q,R)-chains.
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(21) DependencyGraphProof (EQUIVALENT)
4.94/2.00	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
4.94/2.00	----------------------------------------
4.94/2.00	
4.94/2.00	(22)
4.94/2.00	TRUE
4.94/2.04	EOF