3.68/1.74	YES
3.68/1.75	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.68/1.75	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.68/1.75	
3.68/1.75	
3.68/1.75	Left Termination of the query pattern
3.68/1.75	
3.68/1.75	sublist(a,g)
3.68/1.75	
3.68/1.75	w.r.t. the given Prolog program could successfully be proven:
3.68/1.75	
3.68/1.75	(0) Prolog
3.68/1.75	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.68/1.75	(2) PiTRS
3.68/1.75	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.68/1.75	(4) PiDP
3.68/1.75	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.68/1.75	(6) PiDP
3.68/1.75	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.68/1.75	(8) PiDP
3.68/1.75	(9) PiDPToQDPProof [SOUND, 0 ms]
3.68/1.75	(10) QDP
3.68/1.75	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.68/1.75	(12) YES
3.68/1.75	
3.68/1.75	
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(0)
3.68/1.75	Obligation:
3.68/1.75	Clauses:
3.68/1.75	
3.68/1.75	append([], Ys, Ys).
3.68/1.75	append(.(X, Xs), Ys, .(X, Zs)) :- append(Xs, Ys, Zs).
3.68/1.75	sublist(X, Y) :- ','(append(P, X1, Y), append(X2, X, P)).
3.68/1.75	
3.68/1.75	
3.68/1.75	Query: sublist(a,g)
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(1) PrologToPiTRSProof (SOUND)
3.68/1.75	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.68/1.75	
3.68/1.75	sublist_in_2: (f,b)
3.68/1.75	
3.68/1.75	append_in_3: (f,f,b)
3.68/1.75	
3.68/1.75	Transforming Prolog into the following Term Rewriting System:
3.68/1.75	
3.68/1.75	Pi-finite rewrite system:
3.68/1.75	The TRS R consists of the following rules:
3.68/1.75	
3.68/1.75	   sublist_in_ag(X, Y) -> U2_ag(X, Y, append_in_aag(P, X1, Y))
3.68/1.75	   append_in_aag([], Ys, Ys) -> append_out_aag([], Ys, Ys)
3.68/1.75	   append_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
3.68/1.75	   U1_aag(X, Xs, Ys, Zs, append_out_aag(Xs, Ys, Zs)) -> append_out_aag(.(X, Xs), Ys, .(X, Zs))
3.68/1.75	   U2_ag(X, Y, append_out_aag(P, X1, Y)) -> U3_ag(X, Y, append_in_aag(X2, X, P))
3.68/1.75	   U3_ag(X, Y, append_out_aag(X2, X, P)) -> sublist_out_ag(X, Y)
3.68/1.75	
3.68/1.75	The argument filtering Pi contains the following mapping:
3.68/1.75	sublist_in_ag(x1, x2)  =  sublist_in_ag(x2)
3.68/1.75	
3.68/1.75	U2_ag(x1, x2, x3)  =  U2_ag(x3)
3.68/1.75	
3.68/1.75	append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
3.68/1.75	
3.68/1.75	append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)
3.68/1.75	
3.68/1.75	.(x1, x2)  =  .(x1, x2)
3.68/1.75	
3.68/1.75	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
3.68/1.75	
3.68/1.75	U3_ag(x1, x2, x3)  =  U3_ag(x3)
3.68/1.75	
3.68/1.75	sublist_out_ag(x1, x2)  =  sublist_out_ag(x1)
3.68/1.75	
3.68/1.75	
3.68/1.75	
3.68/1.75	
3.68/1.75	
3.68/1.75	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.68/1.75	
3.68/1.75	
3.68/1.75	
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(2)
3.68/1.75	Obligation:
3.68/1.75	Pi-finite rewrite system:
3.68/1.75	The TRS R consists of the following rules:
3.68/1.75	
3.68/1.75	   sublist_in_ag(X, Y) -> U2_ag(X, Y, append_in_aag(P, X1, Y))
3.68/1.75	   append_in_aag([], Ys, Ys) -> append_out_aag([], Ys, Ys)
3.68/1.75	   append_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
3.68/1.75	   U1_aag(X, Xs, Ys, Zs, append_out_aag(Xs, Ys, Zs)) -> append_out_aag(.(X, Xs), Ys, .(X, Zs))
3.68/1.75	   U2_ag(X, Y, append_out_aag(P, X1, Y)) -> U3_ag(X, Y, append_in_aag(X2, X, P))
3.68/1.75	   U3_ag(X, Y, append_out_aag(X2, X, P)) -> sublist_out_ag(X, Y)
3.68/1.75	
3.68/1.75	The argument filtering Pi contains the following mapping:
3.68/1.75	sublist_in_ag(x1, x2)  =  sublist_in_ag(x2)
3.68/1.75	
3.68/1.75	U2_ag(x1, x2, x3)  =  U2_ag(x3)
3.68/1.75	
3.68/1.75	append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
3.68/1.75	
3.68/1.75	append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)
3.68/1.75	
3.68/1.75	.(x1, x2)  =  .(x1, x2)
3.68/1.75	
3.68/1.75	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
3.68/1.75	
3.68/1.75	U3_ag(x1, x2, x3)  =  U3_ag(x3)
3.68/1.75	
3.68/1.75	sublist_out_ag(x1, x2)  =  sublist_out_ag(x1)
3.68/1.75	
3.68/1.75	
3.68/1.75	
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(3) DependencyPairsProof (EQUIVALENT)
3.68/1.75	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.68/1.75	Pi DP problem:
3.68/1.75	The TRS P consists of the following rules:
3.68/1.75	
3.68/1.75	   SUBLIST_IN_AG(X, Y) -> U2_AG(X, Y, append_in_aag(P, X1, Y))
3.68/1.75	   SUBLIST_IN_AG(X, Y) -> APPEND_IN_AAG(P, X1, Y)
3.68/1.75	   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U1_AAG(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
3.68/1.75	   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AAG(Xs, Ys, Zs)
3.68/1.75	   U2_AG(X, Y, append_out_aag(P, X1, Y)) -> U3_AG(X, Y, append_in_aag(X2, X, P))
3.68/1.75	   U2_AG(X, Y, append_out_aag(P, X1, Y)) -> APPEND_IN_AAG(X2, X, P)
3.68/1.75	
3.68/1.75	The TRS R consists of the following rules:
3.68/1.75	
3.68/1.75	   sublist_in_ag(X, Y) -> U2_ag(X, Y, append_in_aag(P, X1, Y))
3.68/1.75	   append_in_aag([], Ys, Ys) -> append_out_aag([], Ys, Ys)
3.68/1.75	   append_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
3.68/1.75	   U1_aag(X, Xs, Ys, Zs, append_out_aag(Xs, Ys, Zs)) -> append_out_aag(.(X, Xs), Ys, .(X, Zs))
3.68/1.75	   U2_ag(X, Y, append_out_aag(P, X1, Y)) -> U3_ag(X, Y, append_in_aag(X2, X, P))
3.68/1.75	   U3_ag(X, Y, append_out_aag(X2, X, P)) -> sublist_out_ag(X, Y)
3.68/1.75	
3.68/1.75	The argument filtering Pi contains the following mapping:
3.68/1.75	sublist_in_ag(x1, x2)  =  sublist_in_ag(x2)
3.68/1.75	
3.68/1.75	U2_ag(x1, x2, x3)  =  U2_ag(x3)
3.68/1.75	
3.68/1.75	append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
3.68/1.75	
3.68/1.75	append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)
3.68/1.75	
3.68/1.75	.(x1, x2)  =  .(x1, x2)
3.68/1.75	
3.68/1.75	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
3.68/1.75	
3.68/1.75	U3_ag(x1, x2, x3)  =  U3_ag(x3)
3.68/1.75	
3.68/1.75	sublist_out_ag(x1, x2)  =  sublist_out_ag(x1)
3.68/1.75	
3.68/1.75	SUBLIST_IN_AG(x1, x2)  =  SUBLIST_IN_AG(x2)
3.68/1.75	
3.68/1.75	U2_AG(x1, x2, x3)  =  U2_AG'(x3)
3.68/1.75	
3.68/1.75	APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)
3.68/1.75	
3.68/1.75	U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x5)
3.68/1.75	
3.68/1.75	U3_AG(x1, x2, x3)  =  U3_AG(x3)
3.68/1.75	
3.68/1.75	
3.68/1.75	We have to consider all (P,R,Pi)-chains
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(4)
3.68/1.75	Obligation:
3.68/1.75	Pi DP problem:
3.68/1.75	The TRS P consists of the following rules:
3.68/1.75	
3.68/1.75	   SUBLIST_IN_AG(X, Y) -> U2_AG(X, Y, append_in_aag(P, X1, Y))
3.68/1.75	   SUBLIST_IN_AG(X, Y) -> APPEND_IN_AAG(P, X1, Y)
3.68/1.75	   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U1_AAG(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
3.68/1.75	   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AAG(Xs, Ys, Zs)
3.68/1.75	   U2_AG(X, Y, append_out_aag(P, X1, Y)) -> U3_AG(X, Y, append_in_aag(X2, X, P))
3.68/1.75	   U2_AG(X, Y, append_out_aag(P, X1, Y)) -> APPEND_IN_AAG(X2, X, P)
3.68/1.75	
3.68/1.75	The TRS R consists of the following rules:
3.68/1.75	
3.68/1.75	   sublist_in_ag(X, Y) -> U2_ag(X, Y, append_in_aag(P, X1, Y))
3.68/1.75	   append_in_aag([], Ys, Ys) -> append_out_aag([], Ys, Ys)
3.68/1.75	   append_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
3.68/1.75	   U1_aag(X, Xs, Ys, Zs, append_out_aag(Xs, Ys, Zs)) -> append_out_aag(.(X, Xs), Ys, .(X, Zs))
3.68/1.75	   U2_ag(X, Y, append_out_aag(P, X1, Y)) -> U3_ag(X, Y, append_in_aag(X2, X, P))
3.68/1.75	   U3_ag(X, Y, append_out_aag(X2, X, P)) -> sublist_out_ag(X, Y)
3.68/1.75	
3.68/1.75	The argument filtering Pi contains the following mapping:
3.68/1.75	sublist_in_ag(x1, x2)  =  sublist_in_ag(x2)
3.68/1.75	
3.68/1.75	U2_ag(x1, x2, x3)  =  U2_ag(x3)
3.68/1.75	
3.68/1.75	append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
3.68/1.75	
3.68/1.75	append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)
3.68/1.75	
3.68/1.75	.(x1, x2)  =  .(x1, x2)
3.68/1.75	
3.68/1.75	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
3.68/1.75	
3.68/1.75	U3_ag(x1, x2, x3)  =  U3_ag(x3)
3.68/1.75	
3.68/1.75	sublist_out_ag(x1, x2)  =  sublist_out_ag(x1)
3.68/1.75	
3.68/1.75	SUBLIST_IN_AG(x1, x2)  =  SUBLIST_IN_AG(x2)
3.68/1.75	
3.68/1.75	U2_AG(x1, x2, x3)  =  U2_AG(x3)
3.68/1.75	
3.68/1.75	APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)
3.68/1.75	
3.68/1.75	U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x5)
3.68/1.75	
3.68/1.75	U3_AG(x1, x2, x3)  =  U3_AG(x3)
3.68/1.75	
3.68/1.75	
3.68/1.75	We have to consider all (P,R,Pi)-chains
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(5) DependencyGraphProof (EQUIVALENT)
3.68/1.75	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(6)
3.68/1.75	Obligation:
3.68/1.75	Pi DP problem:
3.68/1.75	The TRS P consists of the following rules:
3.68/1.75	
3.68/1.75	   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AAG(Xs, Ys, Zs)
3.68/1.75	
3.68/1.75	The TRS R consists of the following rules:
3.68/1.75	
3.68/1.75	   sublist_in_ag(X, Y) -> U2_ag(X, Y, append_in_aag(P, X1, Y))
3.68/1.75	   append_in_aag([], Ys, Ys) -> append_out_aag([], Ys, Ys)
3.68/1.75	   append_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U1_aag(X, Xs, Ys, Zs, append_in_aag(Xs, Ys, Zs))
3.68/1.75	   U1_aag(X, Xs, Ys, Zs, append_out_aag(Xs, Ys, Zs)) -> append_out_aag(.(X, Xs), Ys, .(X, Zs))
3.68/1.75	   U2_ag(X, Y, append_out_aag(P, X1, Y)) -> U3_ag(X, Y, append_in_aag(X2, X, P))
3.68/1.75	   U3_ag(X, Y, append_out_aag(X2, X, P)) -> sublist_out_ag(X, Y)
3.68/1.75	
3.68/1.75	The argument filtering Pi contains the following mapping:
3.68/1.75	sublist_in_ag(x1, x2)  =  sublist_in_ag(x2)
3.68/1.75	
3.68/1.75	U2_ag(x1, x2, x3)  =  U2_ag(x3)
3.68/1.75	
3.68/1.75	append_in_aag(x1, x2, x3)  =  append_in_aag(x3)
3.68/1.75	
3.68/1.75	append_out_aag(x1, x2, x3)  =  append_out_aag(x1, x2)
3.68/1.75	
3.68/1.75	.(x1, x2)  =  .(x1, x2)
3.68/1.75	
3.68/1.75	U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
3.68/1.75	
3.68/1.75	U3_ag(x1, x2, x3)  =  U3_ag(x3)
3.68/1.75	
3.68/1.75	sublist_out_ag(x1, x2)  =  sublist_out_ag(x1)
3.68/1.75	
3.68/1.75	APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)
3.68/1.75	
3.68/1.75	
3.68/1.75	We have to consider all (P,R,Pi)-chains
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(7) UsableRulesProof (EQUIVALENT)
3.68/1.75	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(8)
3.68/1.75	Obligation:
3.68/1.75	Pi DP problem:
3.68/1.75	The TRS P consists of the following rules:
3.68/1.75	
3.68/1.75	   APPEND_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APPEND_IN_AAG(Xs, Ys, Zs)
3.68/1.75	
3.68/1.75	R is empty.
3.68/1.75	The argument filtering Pi contains the following mapping:
3.68/1.75	.(x1, x2)  =  .(x1, x2)
3.68/1.75	
3.68/1.75	APPEND_IN_AAG(x1, x2, x3)  =  APPEND_IN_AAG(x3)
3.68/1.75	
3.68/1.75	
3.68/1.75	We have to consider all (P,R,Pi)-chains
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(9) PiDPToQDPProof (SOUND)
3.68/1.75	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(10)
3.68/1.75	Obligation:
3.68/1.75	Q DP problem:
3.68/1.75	The TRS P consists of the following rules:
3.68/1.75	
3.68/1.75	   APPEND_IN_AAG(.(X, Zs)) -> APPEND_IN_AAG(Zs)
3.68/1.75	
3.68/1.75	R is empty.
3.68/1.75	Q is empty.
3.68/1.75	We have to consider all (P,Q,R)-chains.
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(11) QDPSizeChangeProof (EQUIVALENT)
3.68/1.75	By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
3.68/1.75	
3.68/1.75	From the DPs we obtained the following set of size-change graphs:
3.68/1.75	*APPEND_IN_AAG(.(X, Zs)) -> APPEND_IN_AAG(Zs)
3.68/1.75	The graph contains the following edges 1 > 1
3.68/1.75	
3.68/1.75	
3.68/1.75	----------------------------------------
3.68/1.75	
3.68/1.75	(12)
3.68/1.75	YES
3.90/1.78	EOF