4.48/2.00	YES
4.48/2.01	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
4.48/2.01	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.48/2.01	
4.48/2.01	
4.48/2.01	Left Termination of the query pattern
4.48/2.01	
4.48/2.01	reverse(g,a,g)
4.48/2.01	
4.48/2.01	w.r.t. the given Prolog program could successfully be proven:
4.48/2.01	
4.48/2.01	(0) Prolog
4.48/2.01	(1) PrologToPiTRSProof [SOUND, 0 ms]
4.48/2.01	(2) PiTRS
4.48/2.01	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
4.48/2.01	(4) PiDP
4.48/2.01	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
4.48/2.01	(6) PiDP
4.48/2.01	(7) UsableRulesProof [EQUIVALENT, 0 ms]
4.48/2.01	(8) PiDP
4.48/2.01	(9) PiDPToQDPProof [SOUND, 1 ms]
4.48/2.01	(10) QDP
4.48/2.01	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
4.48/2.01	(12) YES
4.48/2.01	
4.48/2.01	
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(0)
4.48/2.01	Obligation:
4.48/2.01	Clauses:
4.48/2.01	
4.48/2.01	reverse([], X, X).
4.48/2.01	reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U)).
4.48/2.01	
4.48/2.01	
4.48/2.01	Query: reverse(g,a,g)
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(1) PrologToPiTRSProof (SOUND)
4.48/2.01	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
4.48/2.01	
4.48/2.01	reverse_in_3: (b,f,b)
4.48/2.01	
4.48/2.01	Transforming Prolog into the following Term Rewriting System:
4.48/2.01	
4.48/2.01	Pi-finite rewrite system:
4.48/2.01	The TRS R consists of the following rules:
4.48/2.01	
4.48/2.01	   reverse_in_gag([], X, X) -> reverse_out_gag([], X, X)
4.48/2.01	   reverse_in_gag(.(X, Y), Z, U) -> U1_gag(X, Y, Z, U, reverse_in_gag(Y, Z, .(X, U)))
4.48/2.01	   U1_gag(X, Y, Z, U, reverse_out_gag(Y, Z, .(X, U))) -> reverse_out_gag(.(X, Y), Z, U)
4.48/2.01	
4.48/2.01	The argument filtering Pi contains the following mapping:
4.48/2.01	reverse_in_gag(x1, x2, x3)  =  reverse_in_gag(x1, x3)
4.48/2.01	
4.48/2.01	[]  =  []
4.48/2.01	
4.48/2.01	reverse_out_gag(x1, x2, x3)  =  reverse_out_gag(x2)
4.48/2.01	
4.48/2.01	.(x1, x2)  =  .(x1, x2)
4.48/2.01	
4.48/2.01	U1_gag(x1, x2, x3, x4, x5)  =  U1_gag(x5)
4.48/2.01	
4.48/2.01	
4.48/2.01	
4.48/2.01	
4.48/2.01	
4.48/2.01	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
4.48/2.01	
4.48/2.01	
4.48/2.01	
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(2)
4.48/2.01	Obligation:
4.48/2.01	Pi-finite rewrite system:
4.48/2.01	The TRS R consists of the following rules:
4.48/2.01	
4.48/2.01	   reverse_in_gag([], X, X) -> reverse_out_gag([], X, X)
4.48/2.01	   reverse_in_gag(.(X, Y), Z, U) -> U1_gag(X, Y, Z, U, reverse_in_gag(Y, Z, .(X, U)))
4.48/2.01	   U1_gag(X, Y, Z, U, reverse_out_gag(Y, Z, .(X, U))) -> reverse_out_gag(.(X, Y), Z, U)
4.48/2.01	
4.48/2.01	The argument filtering Pi contains the following mapping:
4.48/2.01	reverse_in_gag(x1, x2, x3)  =  reverse_in_gag(x1, x3)
4.48/2.01	
4.48/2.01	[]  =  []
4.48/2.01	
4.48/2.01	reverse_out_gag(x1, x2, x3)  =  reverse_out_gag(x2)
4.48/2.01	
4.48/2.01	.(x1, x2)  =  .(x1, x2)
4.48/2.01	
4.48/2.01	U1_gag(x1, x2, x3, x4, x5)  =  U1_gag(x5)
4.48/2.01	
4.48/2.01	
4.48/2.01	
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(3) DependencyPairsProof (EQUIVALENT)
4.48/2.01	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
4.48/2.01	Pi DP problem:
4.48/2.01	The TRS P consists of the following rules:
4.48/2.01	
4.48/2.01	   REVERSE_IN_GAG(.(X, Y), Z, U) -> U1_GAG(X, Y, Z, U, reverse_in_gag(Y, Z, .(X, U)))
4.48/2.01	   REVERSE_IN_GAG(.(X, Y), Z, U) -> REVERSE_IN_GAG(Y, Z, .(X, U))
4.48/2.01	
4.48/2.01	The TRS R consists of the following rules:
4.48/2.01	
4.48/2.01	   reverse_in_gag([], X, X) -> reverse_out_gag([], X, X)
4.48/2.01	   reverse_in_gag(.(X, Y), Z, U) -> U1_gag(X, Y, Z, U, reverse_in_gag(Y, Z, .(X, U)))
4.48/2.01	   U1_gag(X, Y, Z, U, reverse_out_gag(Y, Z, .(X, U))) -> reverse_out_gag(.(X, Y), Z, U)
4.48/2.01	
4.48/2.01	The argument filtering Pi contains the following mapping:
4.48/2.01	reverse_in_gag(x1, x2, x3)  =  reverse_in_gag(x1, x3)
4.48/2.01	
4.48/2.01	[]  =  []
4.48/2.01	
4.48/2.01	reverse_out_gag(x1, x2, x3)  =  reverse_out_gag(x2)
4.48/2.01	
4.48/2.01	.(x1, x2)  =  .(x1, x2)
4.48/2.01	
4.48/2.01	U1_gag(x1, x2, x3, x4, x5)  =  U1_gag(x5)
4.48/2.01	
4.48/2.01	REVERSE_IN_GAG(x1, x2, x3)  =  REVERSE_IN_GAG(x1, x3)
4.48/2.01	
4.48/2.01	U1_GAG(x1, x2, x3, x4, x5)  =  U1_GAG(x5)
4.48/2.01	
4.48/2.01	
4.48/2.01	We have to consider all (P,R,Pi)-chains
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(4)
4.48/2.01	Obligation:
4.48/2.01	Pi DP problem:
4.48/2.01	The TRS P consists of the following rules:
4.48/2.01	
4.48/2.01	   REVERSE_IN_GAG(.(X, Y), Z, U) -> U1_GAG(X, Y, Z, U, reverse_in_gag(Y, Z, .(X, U)))
4.48/2.01	   REVERSE_IN_GAG(.(X, Y), Z, U) -> REVERSE_IN_GAG(Y, Z, .(X, U))
4.48/2.01	
4.48/2.01	The TRS R consists of the following rules:
4.48/2.01	
4.48/2.01	   reverse_in_gag([], X, X) -> reverse_out_gag([], X, X)
4.48/2.01	   reverse_in_gag(.(X, Y), Z, U) -> U1_gag(X, Y, Z, U, reverse_in_gag(Y, Z, .(X, U)))
4.48/2.01	   U1_gag(X, Y, Z, U, reverse_out_gag(Y, Z, .(X, U))) -> reverse_out_gag(.(X, Y), Z, U)
4.48/2.01	
4.48/2.01	The argument filtering Pi contains the following mapping:
4.48/2.01	reverse_in_gag(x1, x2, x3)  =  reverse_in_gag(x1, x3)
4.48/2.01	
4.48/2.01	[]  =  []
4.48/2.01	
4.48/2.01	reverse_out_gag(x1, x2, x3)  =  reverse_out_gag(x2)
4.48/2.01	
4.48/2.01	.(x1, x2)  =  .(x1, x2)
4.48/2.01	
4.48/2.01	U1_gag(x1, x2, x3, x4, x5)  =  U1_gag(x5)
4.48/2.01	
4.48/2.01	REVERSE_IN_GAG(x1, x2, x3)  =  REVERSE_IN_GAG(x1, x3)
4.48/2.01	
4.48/2.01	U1_GAG(x1, x2, x3, x4, x5)  =  U1_GAG(x5)
4.48/2.01	
4.48/2.01	
4.48/2.01	We have to consider all (P,R,Pi)-chains
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(5) DependencyGraphProof (EQUIVALENT)
4.48/2.01	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(6)
4.48/2.01	Obligation:
4.48/2.01	Pi DP problem:
4.48/2.01	The TRS P consists of the following rules:
4.48/2.01	
4.48/2.01	   REVERSE_IN_GAG(.(X, Y), Z, U) -> REVERSE_IN_GAG(Y, Z, .(X, U))
4.48/2.01	
4.48/2.01	The TRS R consists of the following rules:
4.48/2.01	
4.48/2.01	   reverse_in_gag([], X, X) -> reverse_out_gag([], X, X)
4.48/2.01	   reverse_in_gag(.(X, Y), Z, U) -> U1_gag(X, Y, Z, U, reverse_in_gag(Y, Z, .(X, U)))
4.48/2.01	   U1_gag(X, Y, Z, U, reverse_out_gag(Y, Z, .(X, U))) -> reverse_out_gag(.(X, Y), Z, U)
4.48/2.01	
4.48/2.01	The argument filtering Pi contains the following mapping:
4.48/2.01	reverse_in_gag(x1, x2, x3)  =  reverse_in_gag(x1, x3)
4.48/2.01	
4.48/2.01	[]  =  []
4.48/2.01	
4.48/2.01	reverse_out_gag(x1, x2, x3)  =  reverse_out_gag(x2)
4.48/2.01	
4.48/2.01	.(x1, x2)  =  .(x1, x2)
4.48/2.01	
4.48/2.01	U1_gag(x1, x2, x3, x4, x5)  =  U1_gag(x5)
4.48/2.01	
4.48/2.01	REVERSE_IN_GAG(x1, x2, x3)  =  REVERSE_IN_GAG(x1, x3)
4.48/2.01	
4.48/2.01	
4.48/2.01	We have to consider all (P,R,Pi)-chains
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(7) UsableRulesProof (EQUIVALENT)
4.48/2.01	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(8)
4.48/2.01	Obligation:
4.48/2.01	Pi DP problem:
4.48/2.01	The TRS P consists of the following rules:
4.48/2.01	
4.48/2.01	   REVERSE_IN_GAG(.(X, Y), Z, U) -> REVERSE_IN_GAG(Y, Z, .(X, U))
4.48/2.01	
4.48/2.01	R is empty.
4.48/2.01	The argument filtering Pi contains the following mapping:
4.48/2.01	.(x1, x2)  =  .(x1, x2)
4.48/2.01	
4.48/2.01	REVERSE_IN_GAG(x1, x2, x3)  =  REVERSE_IN_GAG(x1, x3)
4.48/2.01	
4.48/2.01	
4.48/2.01	We have to consider all (P,R,Pi)-chains
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(9) PiDPToQDPProof (SOUND)
4.48/2.01	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(10)
4.48/2.01	Obligation:
4.48/2.01	Q DP problem:
4.48/2.01	The TRS P consists of the following rules:
4.48/2.01	
4.48/2.01	   REVERSE_IN_GAG(.(X, Y), U) -> REVERSE_IN_GAG(Y, .(X, U))
4.48/2.01	
4.48/2.01	R is empty.
4.48/2.01	Q is empty.
4.48/2.01	We have to consider all (P,Q,R)-chains.
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(11) QDPSizeChangeProof (EQUIVALENT)
4.48/2.01	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. 
4.48/2.01	
4.48/2.01	From the DPs we obtained the following set of size-change graphs:
4.48/2.01	*REVERSE_IN_GAG(.(X, Y), U) -> REVERSE_IN_GAG(Y, .(X, U))
4.48/2.01	The graph contains the following edges 1 > 1
4.48/2.01	
4.48/2.01	
4.48/2.01	----------------------------------------
4.48/2.01	
4.48/2.01	(12)
4.48/2.01	YES
4.48/2.04	EOF