3.68/1.77	YES
3.71/1.78	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.71/1.78	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.71/1.78	
3.71/1.78	
3.71/1.78	Left Termination of the query pattern
3.71/1.78	
3.71/1.78	reverse(g,g,a)
3.71/1.78	
3.71/1.78	w.r.t. the given Prolog program could successfully be proven:
3.71/1.78	
3.71/1.78	(0) Prolog
3.71/1.78	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.71/1.78	(2) PiTRS
3.71/1.78	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.71/1.78	(4) PiDP
3.71/1.78	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.71/1.78	(6) PiDP
3.71/1.78	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.71/1.78	(8) PiDP
3.71/1.78	(9) PiDPToQDPProof [SOUND, 0 ms]
3.71/1.78	(10) QDP
3.71/1.78	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.71/1.78	(12) YES
3.71/1.78	
3.71/1.78	
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(0)
3.71/1.78	Obligation:
3.71/1.78	Clauses:
3.71/1.78	
3.71/1.78	reverse([], X, X).
3.71/1.78	reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U)).
3.71/1.78	
3.71/1.78	
3.71/1.78	Query: reverse(g,g,a)
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(1) PrologToPiTRSProof (SOUND)
3.71/1.78	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.71/1.78	
3.71/1.78	reverse_in_3: (b,b,f)
3.71/1.78	
3.71/1.78	Transforming Prolog into the following Term Rewriting System:
3.71/1.78	
3.71/1.78	Pi-finite rewrite system:
3.71/1.78	The TRS R consists of the following rules:
3.71/1.78	
3.71/1.78	   reverse_in_gga([], X, X) -> reverse_out_gga([], X, X)
3.71/1.78	   reverse_in_gga(.(X, Y), Z, U) -> U1_gga(X, Y, Z, U, reverse_in_gga(Y, Z, .(X, U)))
3.71/1.78	   U1_gga(X, Y, Z, U, reverse_out_gga(Y, Z, .(X, U))) -> reverse_out_gga(.(X, Y), Z, U)
3.71/1.78	
3.71/1.78	The argument filtering Pi contains the following mapping:
3.71/1.78	reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
3.71/1.78	
3.71/1.78	[]  =  []
3.71/1.78	
3.71/1.78	reverse_out_gga(x1, x2, x3)  =  reverse_out_gga(x1, x2, x3)
3.71/1.78	
3.71/1.78	.(x1, x2)  =  .(x1, x2)
3.71/1.78	
3.71/1.78	U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x2, x3, x5)
3.71/1.78	
3.71/1.78	
3.71/1.78	
3.71/1.78	
3.71/1.78	
3.71/1.78	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.71/1.78	
3.71/1.78	
3.71/1.78	
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(2)
3.71/1.78	Obligation:
3.71/1.78	Pi-finite rewrite system:
3.71/1.78	The TRS R consists of the following rules:
3.71/1.78	
3.71/1.78	   reverse_in_gga([], X, X) -> reverse_out_gga([], X, X)
3.71/1.78	   reverse_in_gga(.(X, Y), Z, U) -> U1_gga(X, Y, Z, U, reverse_in_gga(Y, Z, .(X, U)))
3.71/1.78	   U1_gga(X, Y, Z, U, reverse_out_gga(Y, Z, .(X, U))) -> reverse_out_gga(.(X, Y), Z, U)
3.71/1.78	
3.71/1.78	The argument filtering Pi contains the following mapping:
3.71/1.78	reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
3.71/1.78	
3.71/1.78	[]  =  []
3.71/1.78	
3.71/1.78	reverse_out_gga(x1, x2, x3)  =  reverse_out_gga(x1, x2, x3)
3.71/1.78	
3.71/1.78	.(x1, x2)  =  .(x1, x2)
3.71/1.78	
3.71/1.78	U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x2, x3, x5)
3.71/1.78	
3.71/1.78	
3.71/1.78	
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(3) DependencyPairsProof (EQUIVALENT)
3.71/1.78	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.71/1.78	Pi DP problem:
3.71/1.78	The TRS P consists of the following rules:
3.71/1.78	
3.71/1.78	   REVERSE_IN_GGA(.(X, Y), Z, U) -> U1_GGA(X, Y, Z, U, reverse_in_gga(Y, Z, .(X, U)))
3.71/1.78	   REVERSE_IN_GGA(.(X, Y), Z, U) -> REVERSE_IN_GGA(Y, Z, .(X, U))
3.71/1.78	
3.71/1.78	The TRS R consists of the following rules:
3.71/1.78	
3.71/1.78	   reverse_in_gga([], X, X) -> reverse_out_gga([], X, X)
3.71/1.78	   reverse_in_gga(.(X, Y), Z, U) -> U1_gga(X, Y, Z, U, reverse_in_gga(Y, Z, .(X, U)))
3.71/1.78	   U1_gga(X, Y, Z, U, reverse_out_gga(Y, Z, .(X, U))) -> reverse_out_gga(.(X, Y), Z, U)
3.71/1.78	
3.71/1.78	The argument filtering Pi contains the following mapping:
3.71/1.78	reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
3.71/1.78	
3.71/1.78	[]  =  []
3.71/1.78	
3.71/1.78	reverse_out_gga(x1, x2, x3)  =  reverse_out_gga(x1, x2, x3)
3.71/1.78	
3.71/1.78	.(x1, x2)  =  .(x1, x2)
3.71/1.78	
3.71/1.78	U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x2, x3, x5)
3.71/1.78	
3.71/1.78	REVERSE_IN_GGA(x1, x2, x3)  =  REVERSE_IN_GGA(x1, x2)
3.71/1.78	
3.71/1.78	U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x1, x2, x3, x5)
3.71/1.78	
3.71/1.78	
3.71/1.78	We have to consider all (P,R,Pi)-chains
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(4)
3.71/1.78	Obligation:
3.71/1.78	Pi DP problem:
3.71/1.78	The TRS P consists of the following rules:
3.71/1.78	
3.71/1.78	   REVERSE_IN_GGA(.(X, Y), Z, U) -> U1_GGA(X, Y, Z, U, reverse_in_gga(Y, Z, .(X, U)))
3.71/1.78	   REVERSE_IN_GGA(.(X, Y), Z, U) -> REVERSE_IN_GGA(Y, Z, .(X, U))
3.71/1.78	
3.71/1.78	The TRS R consists of the following rules:
3.71/1.78	
3.71/1.78	   reverse_in_gga([], X, X) -> reverse_out_gga([], X, X)
3.71/1.78	   reverse_in_gga(.(X, Y), Z, U) -> U1_gga(X, Y, Z, U, reverse_in_gga(Y, Z, .(X, U)))
3.71/1.78	   U1_gga(X, Y, Z, U, reverse_out_gga(Y, Z, .(X, U))) -> reverse_out_gga(.(X, Y), Z, U)
3.71/1.78	
3.71/1.78	The argument filtering Pi contains the following mapping:
3.71/1.78	reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
3.71/1.78	
3.71/1.78	[]  =  []
3.71/1.78	
3.71/1.78	reverse_out_gga(x1, x2, x3)  =  reverse_out_gga(x1, x2, x3)
3.71/1.78	
3.71/1.78	.(x1, x2)  =  .(x1, x2)
3.71/1.78	
3.71/1.78	U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x2, x3, x5)
3.71/1.78	
3.71/1.78	REVERSE_IN_GGA(x1, x2, x3)  =  REVERSE_IN_GGA(x1, x2)
3.71/1.78	
3.71/1.78	U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x1, x2, x3, x5)
3.71/1.78	
3.71/1.78	
3.71/1.78	We have to consider all (P,R,Pi)-chains
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(5) DependencyGraphProof (EQUIVALENT)
3.71/1.78	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(6)
3.71/1.78	Obligation:
3.71/1.78	Pi DP problem:
3.71/1.78	The TRS P consists of the following rules:
3.71/1.78	
3.71/1.78	   REVERSE_IN_GGA(.(X, Y), Z, U) -> REVERSE_IN_GGA(Y, Z, .(X, U))
3.71/1.78	
3.71/1.78	The TRS R consists of the following rules:
3.71/1.78	
3.71/1.78	   reverse_in_gga([], X, X) -> reverse_out_gga([], X, X)
3.71/1.78	   reverse_in_gga(.(X, Y), Z, U) -> U1_gga(X, Y, Z, U, reverse_in_gga(Y, Z, .(X, U)))
3.71/1.78	   U1_gga(X, Y, Z, U, reverse_out_gga(Y, Z, .(X, U))) -> reverse_out_gga(.(X, Y), Z, U)
3.71/1.78	
3.71/1.78	The argument filtering Pi contains the following mapping:
3.71/1.78	reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
3.71/1.78	
3.71/1.78	[]  =  []
3.71/1.78	
3.71/1.78	reverse_out_gga(x1, x2, x3)  =  reverse_out_gga(x1, x2, x3)
3.71/1.78	
3.71/1.78	.(x1, x2)  =  .(x1, x2)
3.71/1.78	
3.71/1.78	U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x2, x3, x5)
3.71/1.78	
3.71/1.78	REVERSE_IN_GGA(x1, x2, x3)  =  REVERSE_IN_GGA(x1, x2)
3.71/1.78	
3.71/1.78	
3.71/1.78	We have to consider all (P,R,Pi)-chains
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(7) UsableRulesProof (EQUIVALENT)
3.71/1.78	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(8)
3.71/1.78	Obligation:
3.71/1.78	Pi DP problem:
3.71/1.78	The TRS P consists of the following rules:
3.71/1.78	
3.71/1.78	   REVERSE_IN_GGA(.(X, Y), Z, U) -> REVERSE_IN_GGA(Y, Z, .(X, U))
3.71/1.78	
3.71/1.78	R is empty.
3.71/1.78	The argument filtering Pi contains the following mapping:
3.71/1.78	.(x1, x2)  =  .(x1, x2)
3.71/1.78	
3.71/1.78	REVERSE_IN_GGA(x1, x2, x3)  =  REVERSE_IN_GGA(x1, x2)
3.71/1.78	
3.71/1.78	
3.71/1.78	We have to consider all (P,R,Pi)-chains
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(9) PiDPToQDPProof (SOUND)
3.71/1.78	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(10)
3.71/1.78	Obligation:
3.71/1.78	Q DP problem:
3.71/1.78	The TRS P consists of the following rules:
3.71/1.78	
3.71/1.78	   REVERSE_IN_GGA(.(X, Y), Z) -> REVERSE_IN_GGA(Y, Z)
3.71/1.78	
3.71/1.78	R is empty.
3.71/1.78	Q is empty.
3.71/1.78	We have to consider all (P,Q,R)-chains.
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(11) QDPSizeChangeProof (EQUIVALENT)
3.71/1.78	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. 
3.71/1.78	
3.71/1.78	From the DPs we obtained the following set of size-change graphs:
3.71/1.78	*REVERSE_IN_GGA(.(X, Y), Z) -> REVERSE_IN_GGA(Y, Z)
3.71/1.78	The graph contains the following edges 1 > 1, 2 >= 2
3.71/1.78	
3.71/1.78	
3.71/1.78	----------------------------------------
3.71/1.78	
3.71/1.78	(12)
3.71/1.78	YES
3.75/1.81	EOF