3.57/1.82	YES
3.70/1.83	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.70/1.83	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.70/1.83	
3.70/1.83	
3.70/1.83	Left Termination of the query pattern
3.70/1.83	
3.70/1.83	len1(g,a)
3.70/1.83	
3.70/1.83	w.r.t. the given Prolog program could successfully be proven:
3.70/1.83	
3.70/1.83	(0) Prolog
3.70/1.83	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.70/1.83	(2) PiTRS
3.70/1.83	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.70/1.83	(4) PiDP
3.70/1.83	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.70/1.83	(6) PiDP
3.70/1.83	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.70/1.83	(8) PiDP
3.70/1.83	(9) PiDPToQDPProof [SOUND, 8 ms]
3.70/1.83	(10) QDP
3.70/1.83	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.70/1.83	(12) YES
3.70/1.83	
3.70/1.83	
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(0)
3.70/1.83	Obligation:
3.70/1.83	Clauses:
3.70/1.83	
3.70/1.83	len1([], 0).
3.70/1.83	len1(.(X1, Ts), N) :- ','(len1(Ts, M), eq(N, s(M))).
3.70/1.83	eq(X, X).
3.70/1.83	
3.70/1.83	
3.70/1.83	Query: len1(g,a)
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(1) PrologToPiTRSProof (SOUND)
3.70/1.83	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.70/1.83	
3.70/1.83	len1_in_2: (b,f)
3.70/1.83	
3.70/1.83	Transforming Prolog into the following Term Rewriting System:
3.70/1.83	
3.70/1.83	Pi-finite rewrite system:
3.70/1.83	The TRS R consists of the following rules:
3.70/1.83	
3.70/1.83	   len1_in_ga([], 0) -> len1_out_ga([], 0)
3.70/1.83	   len1_in_ga(.(X1, Ts), N) -> U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
3.70/1.83	   U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) -> U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
3.70/1.83	   eq_in_ag(X, X) -> eq_out_ag(X, X)
3.70/1.83	   U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) -> len1_out_ga(.(X1, Ts), N)
3.70/1.83	
3.70/1.83	The argument filtering Pi contains the following mapping:
3.70/1.83	len1_in_ga(x1, x2)  =  len1_in_ga(x1)
3.70/1.83	
3.70/1.83	[]  =  []
3.70/1.83	
3.70/1.83	len1_out_ga(x1, x2)  =  len1_out_ga(x2)
3.70/1.83	
3.70/1.83	.(x1, x2)  =  .(x1, x2)
3.70/1.83	
3.70/1.83	U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
3.70/1.83	
3.70/1.83	U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
3.70/1.83	
3.70/1.83	eq_in_ag(x1, x2)  =  eq_in_ag(x2)
3.70/1.83	
3.70/1.83	eq_out_ag(x1, x2)  =  eq_out_ag(x1)
3.70/1.83	
3.70/1.83	s(x1)  =  s(x1)
3.70/1.83	
3.70/1.83	
3.70/1.83	
3.70/1.83	
3.70/1.83	
3.70/1.83	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.70/1.83	
3.70/1.83	
3.70/1.83	
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(2)
3.70/1.83	Obligation:
3.70/1.83	Pi-finite rewrite system:
3.70/1.83	The TRS R consists of the following rules:
3.70/1.83	
3.70/1.83	   len1_in_ga([], 0) -> len1_out_ga([], 0)
3.70/1.83	   len1_in_ga(.(X1, Ts), N) -> U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
3.70/1.83	   U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) -> U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
3.70/1.83	   eq_in_ag(X, X) -> eq_out_ag(X, X)
3.70/1.83	   U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) -> len1_out_ga(.(X1, Ts), N)
3.70/1.83	
3.70/1.83	The argument filtering Pi contains the following mapping:
3.70/1.83	len1_in_ga(x1, x2)  =  len1_in_ga(x1)
3.70/1.83	
3.70/1.83	[]  =  []
3.70/1.83	
3.70/1.83	len1_out_ga(x1, x2)  =  len1_out_ga(x2)
3.70/1.83	
3.70/1.83	.(x1, x2)  =  .(x1, x2)
3.70/1.83	
3.70/1.83	U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
3.70/1.83	
3.70/1.83	U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
3.70/1.83	
3.70/1.83	eq_in_ag(x1, x2)  =  eq_in_ag(x2)
3.70/1.83	
3.70/1.83	eq_out_ag(x1, x2)  =  eq_out_ag(x1)
3.70/1.83	
3.70/1.83	s(x1)  =  s(x1)
3.70/1.83	
3.70/1.83	
3.70/1.83	
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(3) DependencyPairsProof (EQUIVALENT)
3.70/1.83	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.70/1.83	Pi DP problem:
3.70/1.83	The TRS P consists of the following rules:
3.70/1.83	
3.70/1.83	   LEN1_IN_GA(.(X1, Ts), N) -> U1_GA(X1, Ts, N, len1_in_ga(Ts, M))
3.70/1.83	   LEN1_IN_GA(.(X1, Ts), N) -> LEN1_IN_GA(Ts, M)
3.70/1.83	   U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) -> U2_GA(X1, Ts, N, M, eq_in_ag(N, s(M)))
3.70/1.83	   U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) -> EQ_IN_AG(N, s(M))
3.70/1.83	
3.70/1.83	The TRS R consists of the following rules:
3.70/1.83	
3.70/1.83	   len1_in_ga([], 0) -> len1_out_ga([], 0)
3.70/1.83	   len1_in_ga(.(X1, Ts), N) -> U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
3.70/1.83	   U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) -> U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
3.70/1.83	   eq_in_ag(X, X) -> eq_out_ag(X, X)
3.70/1.83	   U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) -> len1_out_ga(.(X1, Ts), N)
3.70/1.83	
3.70/1.83	The argument filtering Pi contains the following mapping:
3.70/1.83	len1_in_ga(x1, x2)  =  len1_in_ga(x1)
3.70/1.83	
3.70/1.83	[]  =  []
3.70/1.83	
3.70/1.83	len1_out_ga(x1, x2)  =  len1_out_ga(x2)
3.70/1.83	
3.70/1.83	.(x1, x2)  =  .(x1, x2)
3.70/1.83	
3.70/1.83	U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
3.70/1.83	
3.70/1.83	U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
3.70/1.83	
3.70/1.83	eq_in_ag(x1, x2)  =  eq_in_ag(x2)
3.70/1.83	
3.70/1.83	eq_out_ag(x1, x2)  =  eq_out_ag(x1)
3.70/1.83	
3.70/1.83	s(x1)  =  s(x1)
3.70/1.83	
3.70/1.83	LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)
3.70/1.83	
3.70/1.83	U1_GA(x1, x2, x3, x4)  =  U1_GA(x4)
3.70/1.83	
3.70/1.83	U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x5)
3.70/1.83	
3.70/1.83	EQ_IN_AG(x1, x2)  =  EQ_IN_AG(x2)
3.70/1.83	
3.70/1.83	
3.70/1.83	We have to consider all (P,R,Pi)-chains
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(4)
3.70/1.83	Obligation:
3.70/1.83	Pi DP problem:
3.70/1.83	The TRS P consists of the following rules:
3.70/1.83	
3.70/1.83	   LEN1_IN_GA(.(X1, Ts), N) -> U1_GA(X1, Ts, N, len1_in_ga(Ts, M))
3.70/1.83	   LEN1_IN_GA(.(X1, Ts), N) -> LEN1_IN_GA(Ts, M)
3.70/1.83	   U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) -> U2_GA(X1, Ts, N, M, eq_in_ag(N, s(M)))
3.70/1.83	   U1_GA(X1, Ts, N, len1_out_ga(Ts, M)) -> EQ_IN_AG(N, s(M))
3.70/1.83	
3.70/1.83	The TRS R consists of the following rules:
3.70/1.83	
3.70/1.83	   len1_in_ga([], 0) -> len1_out_ga([], 0)
3.70/1.83	   len1_in_ga(.(X1, Ts), N) -> U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
3.70/1.83	   U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) -> U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
3.70/1.83	   eq_in_ag(X, X) -> eq_out_ag(X, X)
3.70/1.83	   U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) -> len1_out_ga(.(X1, Ts), N)
3.70/1.83	
3.70/1.83	The argument filtering Pi contains the following mapping:
3.70/1.83	len1_in_ga(x1, x2)  =  len1_in_ga(x1)
3.70/1.83	
3.70/1.83	[]  =  []
3.70/1.83	
3.70/1.83	len1_out_ga(x1, x2)  =  len1_out_ga(x2)
3.70/1.83	
3.70/1.83	.(x1, x2)  =  .(x1, x2)
3.70/1.83	
3.70/1.83	U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
3.70/1.83	
3.70/1.83	U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
3.70/1.83	
3.70/1.83	eq_in_ag(x1, x2)  =  eq_in_ag(x2)
3.70/1.83	
3.70/1.83	eq_out_ag(x1, x2)  =  eq_out_ag(x1)
3.70/1.83	
3.70/1.83	s(x1)  =  s(x1)
3.70/1.83	
3.70/1.83	LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)
3.70/1.83	
3.70/1.83	U1_GA(x1, x2, x3, x4)  =  U1_GA(x4)
3.70/1.83	
3.70/1.83	U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x5)
3.70/1.83	
3.70/1.83	EQ_IN_AG(x1, x2)  =  EQ_IN_AG(x2)
3.70/1.83	
3.70/1.83	
3.70/1.83	We have to consider all (P,R,Pi)-chains
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(5) DependencyGraphProof (EQUIVALENT)
3.70/1.83	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(6)
3.70/1.83	Obligation:
3.70/1.83	Pi DP problem:
3.70/1.83	The TRS P consists of the following rules:
3.70/1.83	
3.70/1.83	   LEN1_IN_GA(.(X1, Ts), N) -> LEN1_IN_GA(Ts, M)
3.70/1.83	
3.70/1.83	The TRS R consists of the following rules:
3.70/1.83	
3.70/1.83	   len1_in_ga([], 0) -> len1_out_ga([], 0)
3.70/1.83	   len1_in_ga(.(X1, Ts), N) -> U1_ga(X1, Ts, N, len1_in_ga(Ts, M))
3.70/1.83	   U1_ga(X1, Ts, N, len1_out_ga(Ts, M)) -> U2_ga(X1, Ts, N, M, eq_in_ag(N, s(M)))
3.70/1.83	   eq_in_ag(X, X) -> eq_out_ag(X, X)
3.70/1.83	   U2_ga(X1, Ts, N, M, eq_out_ag(N, s(M))) -> len1_out_ga(.(X1, Ts), N)
3.70/1.83	
3.70/1.83	The argument filtering Pi contains the following mapping:
3.70/1.83	len1_in_ga(x1, x2)  =  len1_in_ga(x1)
3.70/1.83	
3.70/1.83	[]  =  []
3.70/1.83	
3.70/1.83	len1_out_ga(x1, x2)  =  len1_out_ga(x2)
3.70/1.83	
3.70/1.83	.(x1, x2)  =  .(x1, x2)
3.70/1.83	
3.70/1.83	U1_ga(x1, x2, x3, x4)  =  U1_ga(x4)
3.70/1.83	
3.70/1.83	U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
3.70/1.83	
3.70/1.83	eq_in_ag(x1, x2)  =  eq_in_ag(x2)
3.70/1.83	
3.70/1.83	eq_out_ag(x1, x2)  =  eq_out_ag(x1)
3.70/1.83	
3.70/1.83	s(x1)  =  s(x1)
3.70/1.83	
3.70/1.83	LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)
3.70/1.83	
3.70/1.83	
3.70/1.83	We have to consider all (P,R,Pi)-chains
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(7) UsableRulesProof (EQUIVALENT)
3.70/1.83	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(8)
3.70/1.83	Obligation:
3.70/1.83	Pi DP problem:
3.70/1.83	The TRS P consists of the following rules:
3.70/1.83	
3.70/1.83	   LEN1_IN_GA(.(X1, Ts), N) -> LEN1_IN_GA(Ts, M)
3.70/1.83	
3.70/1.83	R is empty.
3.70/1.83	The argument filtering Pi contains the following mapping:
3.70/1.83	.(x1, x2)  =  .(x1, x2)
3.70/1.83	
3.70/1.83	LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)
3.70/1.83	
3.70/1.83	
3.70/1.83	We have to consider all (P,R,Pi)-chains
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(9) PiDPToQDPProof (SOUND)
3.70/1.83	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(10)
3.70/1.83	Obligation:
3.70/1.83	Q DP problem:
3.70/1.83	The TRS P consists of the following rules:
3.70/1.83	
3.70/1.83	   LEN1_IN_GA(.(X1, Ts)) -> LEN1_IN_GA(Ts)
3.70/1.83	
3.70/1.83	R is empty.
3.70/1.83	Q is empty.
3.70/1.83	We have to consider all (P,Q,R)-chains.
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(11) QDPSizeChangeProof (EQUIVALENT)
3.70/1.83	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. 
3.70/1.83	
3.70/1.83	From the DPs we obtained the following set of size-change graphs:
3.70/1.83	*LEN1_IN_GA(.(X1, Ts)) -> LEN1_IN_GA(Ts)
3.70/1.83	The graph contains the following edges 1 > 1
3.70/1.83	
3.70/1.83	
3.70/1.83	----------------------------------------
3.70/1.83	
3.70/1.83	(12)
3.70/1.83	YES
3.74/1.88	EOF