3.67/1.71	YES
3.67/1.72	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.67/1.72	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.67/1.72	
3.67/1.72	
3.67/1.72	Left Termination of the query pattern
3.67/1.72	
3.67/1.72	append3(a,a,g)
3.67/1.72	
3.67/1.72	w.r.t. the given Prolog program could successfully be proven:
3.67/1.72	
3.67/1.72	(0) Prolog
3.67/1.72	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.67/1.72	(2) PiTRS
3.67/1.72	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.67/1.72	(4) PiDP
3.67/1.72	(5) DependencyGraphProof [EQUIVALENT, 1 ms]
3.67/1.72	(6) PiDP
3.67/1.72	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.67/1.72	(8) PiDP
3.67/1.72	(9) PiDPToQDPProof [SOUND, 0 ms]
3.67/1.72	(10) QDP
3.67/1.72	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.67/1.72	(12) YES
3.67/1.72	
3.67/1.72	
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(0)
3.67/1.72	Obligation:
3.67/1.72	Clauses:
3.67/1.72	
3.67/1.72	append1([], X, X).
3.67/1.72	append1(.(X, Y), U, .(X, Z)) :- append1(Y, U, Z).
3.67/1.72	append2([], X, X).
3.67/1.72	append2(.(X, Y), U, .(X, Z)) :- append2(Y, U, Z).
3.67/1.72	append3([], X, X).
3.67/1.72	append3(.(X, Y), U, .(X, Z)) :- append3(Y, U, Z).
3.67/1.72	
3.67/1.72	
3.67/1.72	Query: append3(a,a,g)
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(1) PrologToPiTRSProof (SOUND)
3.67/1.72	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.67/1.72	
3.67/1.72	append3_in_3: (f,f,b)
3.67/1.72	
3.67/1.72	Transforming Prolog into the following Term Rewriting System:
3.67/1.72	
3.67/1.72	Pi-finite rewrite system:
3.67/1.72	The TRS R consists of the following rules:
3.67/1.72	
3.67/1.72	   append3_in_aag([], X, X) -> append3_out_aag([], X, X)
3.67/1.72	   append3_in_aag(.(X, Y), U, .(X, Z)) -> U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z))
3.67/1.72	   U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) -> append3_out_aag(.(X, Y), U, .(X, Z))
3.67/1.72	
3.67/1.72	The argument filtering Pi contains the following mapping:
3.67/1.72	append3_in_aag(x1, x2, x3)  =  append3_in_aag(x3)
3.67/1.72	
3.67/1.72	append3_out_aag(x1, x2, x3)  =  append3_out_aag(x1, x2)
3.67/1.72	
3.67/1.72	.(x1, x2)  =  .(x1, x2)
3.67/1.72	
3.67/1.72	U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
3.67/1.72	
3.67/1.72	
3.67/1.72	
3.67/1.72	
3.67/1.72	
3.67/1.72	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.67/1.72	
3.67/1.72	
3.67/1.72	
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(2)
3.67/1.72	Obligation:
3.67/1.72	Pi-finite rewrite system:
3.67/1.72	The TRS R consists of the following rules:
3.67/1.72	
3.67/1.72	   append3_in_aag([], X, X) -> append3_out_aag([], X, X)
3.67/1.72	   append3_in_aag(.(X, Y), U, .(X, Z)) -> U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z))
3.67/1.72	   U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) -> append3_out_aag(.(X, Y), U, .(X, Z))
3.67/1.72	
3.67/1.72	The argument filtering Pi contains the following mapping:
3.67/1.72	append3_in_aag(x1, x2, x3)  =  append3_in_aag(x3)
3.67/1.72	
3.67/1.72	append3_out_aag(x1, x2, x3)  =  append3_out_aag(x1, x2)
3.67/1.72	
3.67/1.72	.(x1, x2)  =  .(x1, x2)
3.67/1.72	
3.67/1.72	U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
3.67/1.72	
3.67/1.72	
3.67/1.72	
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(3) DependencyPairsProof (EQUIVALENT)
3.67/1.72	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.67/1.72	Pi DP problem:
3.67/1.72	The TRS P consists of the following rules:
3.67/1.72	
3.67/1.72	   APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> U3_AAG(X, Y, U, Z, append3_in_aag(Y, U, Z))
3.67/1.72	   APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> APPEND3_IN_AAG(Y, U, Z)
3.67/1.72	
3.67/1.72	The TRS R consists of the following rules:
3.67/1.72	
3.67/1.72	   append3_in_aag([], X, X) -> append3_out_aag([], X, X)
3.67/1.72	   append3_in_aag(.(X, Y), U, .(X, Z)) -> U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z))
3.67/1.72	   U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) -> append3_out_aag(.(X, Y), U, .(X, Z))
3.67/1.72	
3.67/1.72	The argument filtering Pi contains the following mapping:
3.67/1.72	append3_in_aag(x1, x2, x3)  =  append3_in_aag(x3)
3.67/1.72	
3.67/1.72	append3_out_aag(x1, x2, x3)  =  append3_out_aag(x1, x2)
3.67/1.72	
3.67/1.72	.(x1, x2)  =  .(x1, x2)
3.67/1.72	
3.67/1.72	U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
3.67/1.72	
3.67/1.72	APPEND3_IN_AAG(x1, x2, x3)  =  APPEND3_IN_AAG(x3)
3.67/1.72	
3.67/1.72	U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(x1, x5)
3.67/1.72	
3.67/1.72	
3.67/1.72	We have to consider all (P,R,Pi)-chains
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(4)
3.67/1.72	Obligation:
3.67/1.72	Pi DP problem:
3.67/1.72	The TRS P consists of the following rules:
3.67/1.72	
3.67/1.72	   APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> U3_AAG(X, Y, U, Z, append3_in_aag(Y, U, Z))
3.67/1.72	   APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> APPEND3_IN_AAG(Y, U, Z)
3.67/1.72	
3.67/1.72	The TRS R consists of the following rules:
3.67/1.72	
3.67/1.72	   append3_in_aag([], X, X) -> append3_out_aag([], X, X)
3.67/1.72	   append3_in_aag(.(X, Y), U, .(X, Z)) -> U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z))
3.67/1.72	   U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) -> append3_out_aag(.(X, Y), U, .(X, Z))
3.67/1.72	
3.67/1.72	The argument filtering Pi contains the following mapping:
3.67/1.72	append3_in_aag(x1, x2, x3)  =  append3_in_aag(x3)
3.67/1.72	
3.67/1.72	append3_out_aag(x1, x2, x3)  =  append3_out_aag(x1, x2)
3.67/1.72	
3.67/1.72	.(x1, x2)  =  .(x1, x2)
3.67/1.72	
3.67/1.72	U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
3.67/1.72	
3.67/1.72	APPEND3_IN_AAG(x1, x2, x3)  =  APPEND3_IN_AAG(x3)
3.67/1.72	
3.67/1.72	U3_AAG(x1, x2, x3, x4, x5)  =  U3_AAG(x1, x5)
3.67/1.72	
3.67/1.72	
3.67/1.72	We have to consider all (P,R,Pi)-chains
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(5) DependencyGraphProof (EQUIVALENT)
3.67/1.72	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(6)
3.67/1.72	Obligation:
3.67/1.72	Pi DP problem:
3.67/1.72	The TRS P consists of the following rules:
3.67/1.72	
3.67/1.72	   APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> APPEND3_IN_AAG(Y, U, Z)
3.67/1.72	
3.67/1.72	The TRS R consists of the following rules:
3.67/1.72	
3.67/1.72	   append3_in_aag([], X, X) -> append3_out_aag([], X, X)
3.67/1.72	   append3_in_aag(.(X, Y), U, .(X, Z)) -> U3_aag(X, Y, U, Z, append3_in_aag(Y, U, Z))
3.67/1.72	   U3_aag(X, Y, U, Z, append3_out_aag(Y, U, Z)) -> append3_out_aag(.(X, Y), U, .(X, Z))
3.67/1.72	
3.67/1.72	The argument filtering Pi contains the following mapping:
3.67/1.72	append3_in_aag(x1, x2, x3)  =  append3_in_aag(x3)
3.67/1.72	
3.67/1.72	append3_out_aag(x1, x2, x3)  =  append3_out_aag(x1, x2)
3.67/1.72	
3.67/1.72	.(x1, x2)  =  .(x1, x2)
3.67/1.72	
3.67/1.72	U3_aag(x1, x2, x3, x4, x5)  =  U3_aag(x1, x5)
3.67/1.72	
3.67/1.72	APPEND3_IN_AAG(x1, x2, x3)  =  APPEND3_IN_AAG(x3)
3.67/1.72	
3.67/1.72	
3.67/1.72	We have to consider all (P,R,Pi)-chains
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(7) UsableRulesProof (EQUIVALENT)
3.67/1.72	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(8)
3.67/1.72	Obligation:
3.67/1.72	Pi DP problem:
3.67/1.72	The TRS P consists of the following rules:
3.67/1.72	
3.67/1.72	   APPEND3_IN_AAG(.(X, Y), U, .(X, Z)) -> APPEND3_IN_AAG(Y, U, Z)
3.67/1.72	
3.67/1.72	R is empty.
3.67/1.72	The argument filtering Pi contains the following mapping:
3.67/1.72	.(x1, x2)  =  .(x1, x2)
3.67/1.72	
3.67/1.72	APPEND3_IN_AAG(x1, x2, x3)  =  APPEND3_IN_AAG(x3)
3.67/1.72	
3.67/1.72	
3.67/1.72	We have to consider all (P,R,Pi)-chains
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(9) PiDPToQDPProof (SOUND)
3.67/1.72	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(10)
3.67/1.72	Obligation:
3.67/1.72	Q DP problem:
3.67/1.72	The TRS P consists of the following rules:
3.67/1.72	
3.67/1.72	   APPEND3_IN_AAG(.(X, Z)) -> APPEND3_IN_AAG(Z)
3.67/1.72	
3.67/1.72	R is empty.
3.67/1.72	Q is empty.
3.67/1.72	We have to consider all (P,Q,R)-chains.
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(11) QDPSizeChangeProof (EQUIVALENT)
3.67/1.72	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. 
3.67/1.72	
3.67/1.72	From the DPs we obtained the following set of size-change graphs:
3.67/1.72	*APPEND3_IN_AAG(.(X, Z)) -> APPEND3_IN_AAG(Z)
3.67/1.72	The graph contains the following edges 1 > 1
3.67/1.72	
3.67/1.72	
3.67/1.72	----------------------------------------
3.67/1.72	
3.67/1.72	(12)
3.67/1.72	YES
3.81/1.78	EOF