3.93/1.77	YES
3.93/1.78	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.93/1.78	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.93/1.78	
3.93/1.78	
3.93/1.78	Left Termination of the query pattern
3.93/1.78	
3.93/1.78	palindrome(g)
3.93/1.78	
3.93/1.78	w.r.t. the given Prolog program could successfully be proven:
3.93/1.78	
3.93/1.78	(0) Prolog
3.93/1.78	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.93/1.78	(2) PiTRS
3.93/1.78	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.93/1.78	(4) PiDP
3.93/1.78	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.93/1.78	(6) PiDP
3.93/1.78	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.93/1.78	(8) PiDP
3.93/1.78	(9) PiDPToQDPProof [EQUIVALENT, 0 ms]
3.93/1.78	(10) QDP
3.93/1.78	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.93/1.78	(12) YES
3.93/1.78	
3.93/1.78	
3.93/1.78	----------------------------------------
3.93/1.78	
3.93/1.78	(0)
3.93/1.78	Obligation:
3.93/1.78	Clauses:
3.93/1.78	
3.93/1.78	palindrome(Xs) :- reverse(Xs, Xs).
3.93/1.78	reverse(X1s, X2s) :- reverse(X1s, [], X2s).
3.93/1.78	reverse([], Xs, Xs).
3.93/1.78	reverse(.(X, X1s), X2s, Ys) :- reverse(X1s, .(X, X2s), Ys).
3.93/1.78	
3.93/1.78	
3.93/1.78	Query: palindrome(g)
3.93/1.78	----------------------------------------
3.93/1.78	
3.93/1.78	(1) PrologToPiTRSProof (SOUND)
3.93/1.78	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.93/1.78	
3.93/1.78	palindrome_in_1: (b)
3.93/1.78	
3.93/1.78	reverse_in_2: (b,b)
3.93/1.78	
3.93/1.78	reverse_in_3: (b,b,b)
3.93/1.78	
3.93/1.78	Transforming Prolog into the following Term Rewriting System:
3.93/1.78	
3.93/1.78	Pi-finite rewrite system:
3.93/1.78	The TRS R consists of the following rules:
3.93/1.78	
3.93/1.78	   palindrome_in_g(Xs) -> U1_g(Xs, reverse_in_gg(Xs, Xs))
3.93/1.78	   reverse_in_gg(X1s, X2s) -> U2_gg(X1s, X2s, reverse_in_ggg(X1s, [], X2s))
3.93/1.78	   reverse_in_ggg([], Xs, Xs) -> reverse_out_ggg([], Xs, Xs)
3.93/1.78	   reverse_in_ggg(.(X, X1s), X2s, Ys) -> U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
3.93/1.78	   U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) -> reverse_out_ggg(.(X, X1s), X2s, Ys)
3.93/1.78	   U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) -> reverse_out_gg(X1s, X2s)
3.93/1.78	   U1_g(Xs, reverse_out_gg(Xs, Xs)) -> palindrome_out_g(Xs)
3.93/1.78	
3.93/1.78	The argument filtering Pi contains the following mapping:
3.93/1.78	palindrome_in_g(x1)  =  palindrome_in_g(x1)
3.93/1.78	
3.93/1.78	U1_g(x1, x2)  =  U1_g(x2)
3.93/1.78	
3.93/1.78	reverse_in_gg(x1, x2)  =  reverse_in_gg(x1, x2)
3.93/1.78	
3.93/1.78	U2_gg(x1, x2, x3)  =  U2_gg(x3)
3.93/1.78	
3.93/1.78	reverse_in_ggg(x1, x2, x3)  =  reverse_in_ggg(x1, x2, x3)
3.93/1.78	
3.93/1.78	[]  =  []
3.93/1.78	
3.93/1.78	reverse_out_ggg(x1, x2, x3)  =  reverse_out_ggg
3.93/1.78	
3.93/1.78	.(x1, x2)  =  .(x1, x2)
3.93/1.78	
3.93/1.78	U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
3.93/1.78	
3.93/1.78	reverse_out_gg(x1, x2)  =  reverse_out_gg
3.93/1.78	
3.93/1.78	palindrome_out_g(x1)  =  palindrome_out_g
3.93/1.78	
3.93/1.78	
3.93/1.78	
3.93/1.78	
3.93/1.78	
3.93/1.78	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.93/1.78	
3.93/1.78	
3.93/1.78	
3.93/1.78	----------------------------------------
3.93/1.78	
3.93/1.78	(2)
3.93/1.78	Obligation:
3.93/1.78	Pi-finite rewrite system:
3.93/1.78	The TRS R consists of the following rules:
3.93/1.78	
3.93/1.78	   palindrome_in_g(Xs) -> U1_g(Xs, reverse_in_gg(Xs, Xs))
3.93/1.78	   reverse_in_gg(X1s, X2s) -> U2_gg(X1s, X2s, reverse_in_ggg(X1s, [], X2s))
3.93/1.78	   reverse_in_ggg([], Xs, Xs) -> reverse_out_ggg([], Xs, Xs)
3.93/1.78	   reverse_in_ggg(.(X, X1s), X2s, Ys) -> U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
3.93/1.78	   U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) -> reverse_out_ggg(.(X, X1s), X2s, Ys)
3.93/1.78	   U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) -> reverse_out_gg(X1s, X2s)
3.93/1.78	   U1_g(Xs, reverse_out_gg(Xs, Xs)) -> palindrome_out_g(Xs)
3.93/1.78	
3.93/1.78	The argument filtering Pi contains the following mapping:
3.93/1.78	palindrome_in_g(x1)  =  palindrome_in_g(x1)
3.93/1.78	
3.93/1.78	U1_g(x1, x2)  =  U1_g(x2)
3.93/1.78	
3.93/1.78	reverse_in_gg(x1, x2)  =  reverse_in_gg(x1, x2)
3.93/1.78	
3.93/1.78	U2_gg(x1, x2, x3)  =  U2_gg(x3)
3.93/1.78	
3.93/1.78	reverse_in_ggg(x1, x2, x3)  =  reverse_in_ggg(x1, x2, x3)
3.93/1.78	
3.93/1.78	[]  =  []
3.93/1.78	
3.93/1.78	reverse_out_ggg(x1, x2, x3)  =  reverse_out_ggg
3.93/1.78	
3.93/1.78	.(x1, x2)  =  .(x1, x2)
3.93/1.78	
3.93/1.78	U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
3.93/1.78	
3.93/1.78	reverse_out_gg(x1, x2)  =  reverse_out_gg
3.93/1.78	
3.93/1.78	palindrome_out_g(x1)  =  palindrome_out_g
3.93/1.78	
3.93/1.78	
3.93/1.78	
3.93/1.78	----------------------------------------
3.93/1.78	
3.93/1.78	(3) DependencyPairsProof (EQUIVALENT)
3.93/1.78	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.93/1.78	Pi DP problem:
3.93/1.78	The TRS P consists of the following rules:
3.93/1.78	
3.93/1.78	   PALINDROME_IN_G(Xs) -> U1_G(Xs, reverse_in_gg(Xs, Xs))
3.93/1.78	   PALINDROME_IN_G(Xs) -> REVERSE_IN_GG(Xs, Xs)
3.93/1.78	   REVERSE_IN_GG(X1s, X2s) -> U2_GG(X1s, X2s, reverse_in_ggg(X1s, [], X2s))
3.93/1.78	   REVERSE_IN_GG(X1s, X2s) -> REVERSE_IN_GGG(X1s, [], X2s)
3.93/1.78	   REVERSE_IN_GGG(.(X, X1s), X2s, Ys) -> U3_GGG(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
3.93/1.78	   REVERSE_IN_GGG(.(X, X1s), X2s, Ys) -> REVERSE_IN_GGG(X1s, .(X, X2s), Ys)
3.93/1.78	
3.93/1.78	The TRS R consists of the following rules:
3.93/1.78	
3.93/1.78	   palindrome_in_g(Xs) -> U1_g(Xs, reverse_in_gg(Xs, Xs))
3.93/1.78	   reverse_in_gg(X1s, X2s) -> U2_gg(X1s, X2s, reverse_in_ggg(X1s, [], X2s))
3.93/1.78	   reverse_in_ggg([], Xs, Xs) -> reverse_out_ggg([], Xs, Xs)
3.93/1.78	   reverse_in_ggg(.(X, X1s), X2s, Ys) -> U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
3.93/1.78	   U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) -> reverse_out_ggg(.(X, X1s), X2s, Ys)
3.93/1.78	   U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) -> reverse_out_gg(X1s, X2s)
3.93/1.78	   U1_g(Xs, reverse_out_gg(Xs, Xs)) -> palindrome_out_g(Xs)
3.93/1.78	
3.93/1.78	The argument filtering Pi contains the following mapping:
3.93/1.78	palindrome_in_g(x1)  =  palindrome_in_g(x1)
3.93/1.78	
3.93/1.78	U1_g(x1, x2)  =  U1_g(x2)
3.93/1.78	
3.93/1.78	reverse_in_gg(x1, x2)  =  reverse_in_gg(x1, x2)
3.93/1.78	
3.93/1.78	U2_gg(x1, x2, x3)  =  U2_gg(x3)
3.93/1.78	
3.93/1.78	reverse_in_ggg(x1, x2, x3)  =  reverse_in_ggg(x1, x2, x3)
3.93/1.78	
3.93/1.78	[]  =  []
3.93/1.78	
3.93/1.78	reverse_out_ggg(x1, x2, x3)  =  reverse_out_ggg
3.93/1.78	
3.93/1.78	.(x1, x2)  =  .(x1, x2)
3.93/1.78	
3.93/1.78	U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
3.93/1.78	
3.93/1.78	reverse_out_gg(x1, x2)  =  reverse_out_gg
3.93/1.78	
3.93/1.78	palindrome_out_g(x1)  =  palindrome_out_g
3.93/1.78	
3.93/1.78	PALINDROME_IN_G(x1)  =  PALINDROME_IN_G(x1)
3.93/1.78	
3.93/1.78	U1_G(x1, x2)  =  U1_G(x2)
3.93/1.78	
3.93/1.78	REVERSE_IN_GG(x1, x2)  =  REVERSE_IN_GG(x1, x2)
3.93/1.78	
3.93/1.78	U2_GG(x1, x2, x3)  =  U2_GG(x3)
3.93/1.78	
3.93/1.78	REVERSE_IN_GGG(x1, x2, x3)  =  REVERSE_IN_GGG(x1, x2, x3)
3.93/1.78	
3.93/1.78	U3_GGG(x1, x2, x3, x4, x5)  =  U3_GGG(x5)
3.93/1.78	
3.93/1.78	
3.93/1.78	We have to consider all (P,R,Pi)-chains
3.93/1.78	----------------------------------------
3.93/1.78	
3.93/1.78	(4)
3.93/1.78	Obligation:
3.93/1.78	Pi DP problem:
3.93/1.78	The TRS P consists of the following rules:
3.93/1.78	
3.93/1.78	   PALINDROME_IN_G(Xs) -> U1_G(Xs, reverse_in_gg(Xs, Xs))
3.93/1.78	   PALINDROME_IN_G(Xs) -> REVERSE_IN_GG(Xs, Xs)
3.93/1.78	   REVERSE_IN_GG(X1s, X2s) -> U2_GG(X1s, X2s, reverse_in_ggg(X1s, [], X2s))
3.93/1.78	   REVERSE_IN_GG(X1s, X2s) -> REVERSE_IN_GGG(X1s, [], X2s)
3.93/1.78	   REVERSE_IN_GGG(.(X, X1s), X2s, Ys) -> U3_GGG(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
3.93/1.78	   REVERSE_IN_GGG(.(X, X1s), X2s, Ys) -> REVERSE_IN_GGG(X1s, .(X, X2s), Ys)
3.93/1.78	
3.93/1.78	The TRS R consists of the following rules:
3.93/1.78	
3.93/1.78	   palindrome_in_g(Xs) -> U1_g(Xs, reverse_in_gg(Xs, Xs))
3.93/1.78	   reverse_in_gg(X1s, X2s) -> U2_gg(X1s, X2s, reverse_in_ggg(X1s, [], X2s))
3.93/1.78	   reverse_in_ggg([], Xs, Xs) -> reverse_out_ggg([], Xs, Xs)
3.93/1.78	   reverse_in_ggg(.(X, X1s), X2s, Ys) -> U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
3.93/1.78	   U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) -> reverse_out_ggg(.(X, X1s), X2s, Ys)
3.93/1.78	   U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) -> reverse_out_gg(X1s, X2s)
3.93/1.78	   U1_g(Xs, reverse_out_gg(Xs, Xs)) -> palindrome_out_g(Xs)
3.93/1.78	
3.93/1.78	The argument filtering Pi contains the following mapping:
3.93/1.78	palindrome_in_g(x1)  =  palindrome_in_g(x1)
3.93/1.78	
3.93/1.78	U1_g(x1, x2)  =  U1_g(x2)
3.93/1.78	
3.93/1.78	reverse_in_gg(x1, x2)  =  reverse_in_gg(x1, x2)
3.93/1.78	
3.93/1.78	U2_gg(x1, x2, x3)  =  U2_gg(x3)
3.93/1.78	
3.93/1.78	reverse_in_ggg(x1, x2, x3)  =  reverse_in_ggg(x1, x2, x3)
3.93/1.78	
3.93/1.78	[]  =  []
3.93/1.78	
3.93/1.78	reverse_out_ggg(x1, x2, x3)  =  reverse_out_ggg
3.93/1.78	
3.93/1.78	.(x1, x2)  =  .(x1, x2)
3.93/1.78	
3.93/1.78	U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
3.93/1.78	
3.93/1.78	reverse_out_gg(x1, x2)  =  reverse_out_gg
3.93/1.78	
3.93/1.78	palindrome_out_g(x1)  =  palindrome_out_g
3.93/1.78	
3.93/1.78	PALINDROME_IN_G(x1)  =  PALINDROME_IN_G(x1)
3.93/1.78	
3.93/1.78	U1_G(x1, x2)  =  U1_G(x2)
3.93/1.78	
3.93/1.78	REVERSE_IN_GG(x1, x2)  =  REVERSE_IN_GG(x1, x2)
3.93/1.78	
3.93/1.78	U2_GG(x1, x2, x3)  =  U2_GG(x3)
3.93/1.78	
3.93/1.78	REVERSE_IN_GGG(x1, x2, x3)  =  REVERSE_IN_GGG(x1, x2, x3)
3.93/1.78	
3.93/1.78	U3_GGG(x1, x2, x3, x4, x5)  =  U3_GGG(x5)
3.93/1.78	
3.93/1.78	
3.93/1.78	We have to consider all (P,R,Pi)-chains
3.93/1.78	----------------------------------------
3.93/1.78	
3.93/1.78	(5) DependencyGraphProof (EQUIVALENT)
3.93/1.78	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.
3.93/1.78	----------------------------------------
3.93/1.78	
3.93/1.78	(6)
3.93/1.78	Obligation:
3.93/1.78	Pi DP problem:
3.93/1.78	The TRS P consists of the following rules:
3.93/1.78	
3.93/1.78	   REVERSE_IN_GGG(.(X, X1s), X2s, Ys) -> REVERSE_IN_GGG(X1s, .(X, X2s), Ys)
3.93/1.78	
3.93/1.78	The TRS R consists of the following rules:
3.93/1.78	
3.93/1.78	   palindrome_in_g(Xs) -> U1_g(Xs, reverse_in_gg(Xs, Xs))
3.93/1.78	   reverse_in_gg(X1s, X2s) -> U2_gg(X1s, X2s, reverse_in_ggg(X1s, [], X2s))
3.93/1.78	   reverse_in_ggg([], Xs, Xs) -> reverse_out_ggg([], Xs, Xs)
3.93/1.78	   reverse_in_ggg(.(X, X1s), X2s, Ys) -> U3_ggg(X, X1s, X2s, Ys, reverse_in_ggg(X1s, .(X, X2s), Ys))
3.93/1.78	   U3_ggg(X, X1s, X2s, Ys, reverse_out_ggg(X1s, .(X, X2s), Ys)) -> reverse_out_ggg(.(X, X1s), X2s, Ys)
3.93/1.78	   U2_gg(X1s, X2s, reverse_out_ggg(X1s, [], X2s)) -> reverse_out_gg(X1s, X2s)
3.93/1.78	   U1_g(Xs, reverse_out_gg(Xs, Xs)) -> palindrome_out_g(Xs)
3.93/1.78	
3.93/1.78	The argument filtering Pi contains the following mapping:
3.93/1.78	palindrome_in_g(x1)  =  palindrome_in_g(x1)
3.93/1.78	
3.93/1.78	U1_g(x1, x2)  =  U1_g(x2)
3.93/1.78	
3.93/1.78	reverse_in_gg(x1, x2)  =  reverse_in_gg(x1, x2)
3.93/1.78	
3.93/1.78	U2_gg(x1, x2, x3)  =  U2_gg(x3)
3.93/1.78	
3.93/1.78	reverse_in_ggg(x1, x2, x3)  =  reverse_in_ggg(x1, x2, x3)
3.93/1.78	
3.93/1.78	[]  =  []
3.93/1.78	
3.93/1.78	reverse_out_ggg(x1, x2, x3)  =  reverse_out_ggg
3.93/1.78	
3.93/1.78	.(x1, x2)  =  .(x1, x2)
3.93/1.78	
3.93/1.78	U3_ggg(x1, x2, x3, x4, x5)  =  U3_ggg(x5)
3.93/1.78	
3.93/1.78	reverse_out_gg(x1, x2)  =  reverse_out_gg
3.93/1.78	
3.93/1.78	palindrome_out_g(x1)  =  palindrome_out_g
3.93/1.78	
3.93/1.78	REVERSE_IN_GGG(x1, x2, x3)  =  REVERSE_IN_GGG(x1, x2, x3)
3.93/1.78	
3.93/1.78	
3.93/1.78	We have to consider all (P,R,Pi)-chains
3.93/1.78	----------------------------------------
3.93/1.78	
3.93/1.78	(7) UsableRulesProof (EQUIVALENT)
3.93/1.78	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.93/1.78	----------------------------------------
3.93/1.78	
3.93/1.78	(8)
3.93/1.78	Obligation:
3.93/1.78	Pi DP problem:
3.93/1.78	The TRS P consists of the following rules:
3.93/1.78	
3.93/1.78	   REVERSE_IN_GGG(.(X, X1s), X2s, Ys) -> REVERSE_IN_GGG(X1s, .(X, X2s), Ys)
3.93/1.78	
3.93/1.78	R is empty.
3.93/1.78	Pi is empty.
3.93/1.78	We have to consider all (P,R,Pi)-chains
3.93/1.78	----------------------------------------
3.93/1.78	
3.93/1.78	(9) PiDPToQDPProof (EQUIVALENT)
3.93/1.78	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.93/1.78	----------------------------------------
3.93/1.78	
3.93/1.78	(10)
3.93/1.78	Obligation:
3.93/1.78	Q DP problem:
3.93/1.78	The TRS P consists of the following rules:
3.93/1.78	
3.93/1.78	   REVERSE_IN_GGG(.(X, X1s), X2s, Ys) -> REVERSE_IN_GGG(X1s, .(X, X2s), Ys)
3.93/1.78	
3.93/1.78	R is empty.
3.93/1.78	Q is empty.
3.93/1.78	We have to consider all (P,Q,R)-chains.
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3.93/1.78	
3.93/1.78	(11) QDPSizeChangeProof (EQUIVALENT)
3.93/1.78	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. 
3.93/1.78	
3.93/1.78	From the DPs we obtained the following set of size-change graphs:
3.93/1.78	*REVERSE_IN_GGG(.(X, X1s), X2s, Ys) -> REVERSE_IN_GGG(X1s, .(X, X2s), Ys)
3.93/1.78	The graph contains the following edges 1 > 1, 3 >= 3
3.93/1.78	
3.93/1.78	
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3.93/1.78	
3.93/1.78	(12)
3.93/1.78	YES
4.01/1.82	EOF