3.52/1.69	YES
3.52/1.69	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.52/1.69	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.52/1.69	
3.52/1.69	
3.52/1.69	Left Termination of the query pattern
3.52/1.69	
3.52/1.69	max(a,a,g)
3.52/1.69	
3.52/1.69	w.r.t. the given Prolog program could successfully be proven:
3.52/1.69	
3.52/1.69	(0) Prolog
3.52/1.69	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.52/1.69	(2) PiTRS
3.52/1.69	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.52/1.69	(4) PiDP
3.52/1.69	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.52/1.69	(6) PiDP
3.52/1.69	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.52/1.69	(8) PiDP
3.52/1.69	(9) PiDPToQDPProof [SOUND, 0 ms]
3.52/1.69	(10) QDP
3.52/1.69	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.52/1.69	(12) YES
3.52/1.69	
3.52/1.69	
3.52/1.69	----------------------------------------
3.52/1.69	
3.52/1.69	(0)
3.52/1.69	Obligation:
3.52/1.69	Clauses:
3.52/1.69	
3.52/1.69	max(X, Y, X) :- less(Y, X).
3.52/1.69	max(X, Y, Y) :- less(X, s(Y)).
3.52/1.69	less(0, s(X1)).
3.52/1.69	less(s(X), s(Y)) :- less(X, Y).
3.52/1.69	
3.52/1.69	
3.52/1.69	Query: max(a,a,g)
3.52/1.69	----------------------------------------
3.52/1.69	
3.52/1.69	(1) PrologToPiTRSProof (SOUND)
3.52/1.69	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.52/1.69	
3.52/1.69	max_in_3: (f,f,b)
3.52/1.69	
3.52/1.69	less_in_2: (f,b)
3.52/1.69	
3.52/1.69	Transforming Prolog into the following Term Rewriting System:
3.52/1.69	
3.52/1.69	Pi-finite rewrite system:
3.52/1.69	The TRS R consists of the following rules:
3.52/1.69	
3.52/1.69	   max_in_aag(X, Y, X) -> U1_aag(X, Y, less_in_ag(Y, X))
3.52/1.69	   less_in_ag(0, s(X1)) -> less_out_ag(0, s(X1))
3.52/1.69	   less_in_ag(s(X), s(Y)) -> U3_ag(X, Y, less_in_ag(X, Y))
3.52/1.69	   U3_ag(X, Y, less_out_ag(X, Y)) -> less_out_ag(s(X), s(Y))
3.52/1.69	   U1_aag(X, Y, less_out_ag(Y, X)) -> max_out_aag(X, Y, X)
3.52/1.69	   max_in_aag(X, Y, Y) -> U2_aag(X, Y, less_in_ag(X, s(Y)))
3.52/1.69	   U2_aag(X, Y, less_out_ag(X, s(Y))) -> max_out_aag(X, Y, Y)
3.52/1.69	
3.52/1.69	The argument filtering Pi contains the following mapping:
3.52/1.69	max_in_aag(x1, x2, x3)  =  max_in_aag(x3)
3.52/1.69	
3.52/1.69	U1_aag(x1, x2, x3)  =  U1_aag(x1, x3)
3.52/1.69	
3.52/1.69	less_in_ag(x1, x2)  =  less_in_ag(x2)
3.52/1.69	
3.52/1.69	s(x1)  =  s(x1)
3.52/1.69	
3.52/1.69	less_out_ag(x1, x2)  =  less_out_ag(x1)
3.52/1.69	
3.52/1.69	U3_ag(x1, x2, x3)  =  U3_ag(x3)
3.52/1.69	
3.52/1.69	max_out_aag(x1, x2, x3)  =  max_out_aag(x1, x2)
3.52/1.69	
3.52/1.69	U2_aag(x1, x2, x3)  =  U2_aag(x2, x3)
3.52/1.69	
3.52/1.69	
3.52/1.69	
3.52/1.69	
3.52/1.69	
3.52/1.69	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.52/1.69	
3.52/1.69	
3.52/1.69	
3.52/1.69	----------------------------------------
3.52/1.69	
3.52/1.69	(2)
3.52/1.69	Obligation:
3.52/1.69	Pi-finite rewrite system:
3.52/1.69	The TRS R consists of the following rules:
3.52/1.69	
3.52/1.69	   max_in_aag(X, Y, X) -> U1_aag(X, Y, less_in_ag(Y, X))
3.52/1.69	   less_in_ag(0, s(X1)) -> less_out_ag(0, s(X1))
3.52/1.69	   less_in_ag(s(X), s(Y)) -> U3_ag(X, Y, less_in_ag(X, Y))
3.52/1.69	   U3_ag(X, Y, less_out_ag(X, Y)) -> less_out_ag(s(X), s(Y))
3.52/1.69	   U1_aag(X, Y, less_out_ag(Y, X)) -> max_out_aag(X, Y, X)
3.52/1.69	   max_in_aag(X, Y, Y) -> U2_aag(X, Y, less_in_ag(X, s(Y)))
3.52/1.69	   U2_aag(X, Y, less_out_ag(X, s(Y))) -> max_out_aag(X, Y, Y)
3.52/1.69	
3.52/1.69	The argument filtering Pi contains the following mapping:
3.52/1.69	max_in_aag(x1, x2, x3)  =  max_in_aag(x3)
3.52/1.69	
3.52/1.69	U1_aag(x1, x2, x3)  =  U1_aag(x1, x3)
3.52/1.69	
3.52/1.69	less_in_ag(x1, x2)  =  less_in_ag(x2)
3.52/1.69	
3.52/1.69	s(x1)  =  s(x1)
3.52/1.69	
3.52/1.69	less_out_ag(x1, x2)  =  less_out_ag(x1)
3.52/1.69	
3.52/1.69	U3_ag(x1, x2, x3)  =  U3_ag(x3)
3.52/1.69	
3.52/1.69	max_out_aag(x1, x2, x3)  =  max_out_aag(x1, x2)
3.52/1.69	
3.52/1.69	U2_aag(x1, x2, x3)  =  U2_aag(x2, x3)
3.52/1.69	
3.52/1.69	
3.52/1.69	
3.52/1.69	----------------------------------------
3.52/1.69	
3.52/1.69	(3) DependencyPairsProof (EQUIVALENT)
3.52/1.69	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.52/1.69	Pi DP problem:
3.52/1.69	The TRS P consists of the following rules:
3.52/1.69	
3.52/1.69	   MAX_IN_AAG(X, Y, X) -> U1_AAG(X, Y, less_in_ag(Y, X))
3.52/1.69	   MAX_IN_AAG(X, Y, X) -> LESS_IN_AG(Y, X)
3.52/1.69	   LESS_IN_AG(s(X), s(Y)) -> U3_AG(X, Y, less_in_ag(X, Y))
3.52/1.69	   LESS_IN_AG(s(X), s(Y)) -> LESS_IN_AG(X, Y)
3.52/1.69	   MAX_IN_AAG(X, Y, Y) -> U2_AAG(X, Y, less_in_ag(X, s(Y)))
3.52/1.69	   MAX_IN_AAG(X, Y, Y) -> LESS_IN_AG(X, s(Y))
3.52/1.69	
3.52/1.69	The TRS R consists of the following rules:
3.52/1.69	
3.52/1.69	   max_in_aag(X, Y, X) -> U1_aag(X, Y, less_in_ag(Y, X))
3.52/1.69	   less_in_ag(0, s(X1)) -> less_out_ag(0, s(X1))
3.52/1.69	   less_in_ag(s(X), s(Y)) -> U3_ag(X, Y, less_in_ag(X, Y))
3.52/1.69	   U3_ag(X, Y, less_out_ag(X, Y)) -> less_out_ag(s(X), s(Y))
3.52/1.69	   U1_aag(X, Y, less_out_ag(Y, X)) -> max_out_aag(X, Y, X)
3.52/1.69	   max_in_aag(X, Y, Y) -> U2_aag(X, Y, less_in_ag(X, s(Y)))
3.52/1.69	   U2_aag(X, Y, less_out_ag(X, s(Y))) -> max_out_aag(X, Y, Y)
3.52/1.69	
3.52/1.69	The argument filtering Pi contains the following mapping:
3.52/1.69	max_in_aag(x1, x2, x3)  =  max_in_aag(x3)
3.52/1.69	
3.52/1.69	U1_aag(x1, x2, x3)  =  U1_aag(x1, x3)
3.52/1.69	
3.52/1.69	less_in_ag(x1, x2)  =  less_in_ag(x2)
3.52/1.69	
3.52/1.69	s(x1)  =  s(x1)
3.52/1.69	
3.52/1.69	less_out_ag(x1, x2)  =  less_out_ag(x1)
3.52/1.69	
3.52/1.69	U3_ag(x1, x2, x3)  =  U3_ag(x3)
3.52/1.69	
3.52/1.69	max_out_aag(x1, x2, x3)  =  max_out_aag(x1, x2)
3.52/1.69	
3.52/1.69	U2_aag(x1, x2, x3)  =  U2_aag(x2, x3)
3.52/1.69	
3.52/1.69	MAX_IN_AAG(x1, x2, x3)  =  MAX_IN_AAG(x3)
3.52/1.69	
3.52/1.69	U1_AAG(x1, x2, x3)  =  U1_AAG(x1, x3)
3.52/1.69	
3.52/1.69	LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)
3.52/1.69	
3.52/1.69	U3_AG(x1, x2, x3)  =  U3_AG(x3)
3.52/1.69	
3.52/1.69	U2_AAG(x1, x2, x3)  =  U2_AAG(x2, x3)
3.52/1.69	
3.52/1.69	
3.52/1.69	We have to consider all (P,R,Pi)-chains
3.52/1.69	----------------------------------------
3.52/1.69	
3.52/1.69	(4)
3.52/1.69	Obligation:
3.52/1.69	Pi DP problem:
3.52/1.69	The TRS P consists of the following rules:
3.52/1.69	
3.52/1.69	   MAX_IN_AAG(X, Y, X) -> U1_AAG(X, Y, less_in_ag(Y, X))
3.52/1.69	   MAX_IN_AAG(X, Y, X) -> LESS_IN_AG(Y, X)
3.52/1.69	   LESS_IN_AG(s(X), s(Y)) -> U3_AG(X, Y, less_in_ag(X, Y))
3.52/1.69	   LESS_IN_AG(s(X), s(Y)) -> LESS_IN_AG(X, Y)
3.52/1.69	   MAX_IN_AAG(X, Y, Y) -> U2_AAG(X, Y, less_in_ag(X, s(Y)))
3.52/1.69	   MAX_IN_AAG(X, Y, Y) -> LESS_IN_AG(X, s(Y))
3.52/1.69	
3.52/1.69	The TRS R consists of the following rules:
3.52/1.69	
3.52/1.69	   max_in_aag(X, Y, X) -> U1_aag(X, Y, less_in_ag(Y, X))
3.52/1.69	   less_in_ag(0, s(X1)) -> less_out_ag(0, s(X1))
3.52/1.69	   less_in_ag(s(X), s(Y)) -> U3_ag(X, Y, less_in_ag(X, Y))
3.52/1.69	   U3_ag(X, Y, less_out_ag(X, Y)) -> less_out_ag(s(X), s(Y))
3.52/1.69	   U1_aag(X, Y, less_out_ag(Y, X)) -> max_out_aag(X, Y, X)
3.52/1.69	   max_in_aag(X, Y, Y) -> U2_aag(X, Y, less_in_ag(X, s(Y)))
3.52/1.69	   U2_aag(X, Y, less_out_ag(X, s(Y))) -> max_out_aag(X, Y, Y)
3.52/1.69	
3.52/1.69	The argument filtering Pi contains the following mapping:
3.52/1.69	max_in_aag(x1, x2, x3)  =  max_in_aag(x3)
3.52/1.69	
3.52/1.69	U1_aag(x1, x2, x3)  =  U1_aag(x1, x3)
3.52/1.69	
3.52/1.69	less_in_ag(x1, x2)  =  less_in_ag(x2)
3.52/1.69	
3.52/1.69	s(x1)  =  s(x1)
3.52/1.69	
3.52/1.69	less_out_ag(x1, x2)  =  less_out_ag(x1)
3.52/1.69	
3.52/1.69	U3_ag(x1, x2, x3)  =  U3_ag(x3)
3.52/1.69	
3.52/1.69	max_out_aag(x1, x2, x3)  =  max_out_aag(x1, x2)
3.52/1.69	
3.52/1.69	U2_aag(x1, x2, x3)  =  U2_aag(x2, x3)
3.52/1.69	
3.52/1.69	MAX_IN_AAG(x1, x2, x3)  =  MAX_IN_AAG(x3)
3.52/1.69	
3.52/1.69	U1_AAG(x1, x2, x3)  =  U1_AAG(x1, x3)
3.52/1.69	
3.52/1.69	LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)
3.52/1.69	
3.52/1.69	U3_AG(x1, x2, x3)  =  U3_AG(x3)
3.52/1.69	
3.52/1.69	U2_AAG(x1, x2, x3)  =  U2_AAG(x2, x3)
3.52/1.69	
3.52/1.69	
3.52/1.69	We have to consider all (P,R,Pi)-chains
3.52/1.69	----------------------------------------
3.52/1.69	
3.52/1.69	(5) DependencyGraphProof (EQUIVALENT)
3.52/1.69	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.
3.52/1.69	----------------------------------------
3.52/1.70	
3.52/1.70	(6)
3.52/1.70	Obligation:
3.52/1.70	Pi DP problem:
3.52/1.70	The TRS P consists of the following rules:
3.52/1.70	
3.52/1.70	   LESS_IN_AG(s(X), s(Y)) -> LESS_IN_AG(X, Y)
3.52/1.70	
3.52/1.70	The TRS R consists of the following rules:
3.52/1.70	
3.52/1.70	   max_in_aag(X, Y, X) -> U1_aag(X, Y, less_in_ag(Y, X))
3.52/1.70	   less_in_ag(0, s(X1)) -> less_out_ag(0, s(X1))
3.52/1.70	   less_in_ag(s(X), s(Y)) -> U3_ag(X, Y, less_in_ag(X, Y))
3.52/1.70	   U3_ag(X, Y, less_out_ag(X, Y)) -> less_out_ag(s(X), s(Y))
3.52/1.70	   U1_aag(X, Y, less_out_ag(Y, X)) -> max_out_aag(X, Y, X)
3.52/1.70	   max_in_aag(X, Y, Y) -> U2_aag(X, Y, less_in_ag(X, s(Y)))
3.52/1.70	   U2_aag(X, Y, less_out_ag(X, s(Y))) -> max_out_aag(X, Y, Y)
3.52/1.70	
3.52/1.70	The argument filtering Pi contains the following mapping:
3.52/1.70	max_in_aag(x1, x2, x3)  =  max_in_aag(x3)
3.52/1.70	
3.52/1.70	U1_aag(x1, x2, x3)  =  U1_aag(x1, x3)
3.52/1.70	
3.52/1.70	less_in_ag(x1, x2)  =  less_in_ag(x2)
3.52/1.70	
3.52/1.70	s(x1)  =  s(x1)
3.52/1.70	
3.52/1.70	less_out_ag(x1, x2)  =  less_out_ag(x1)
3.52/1.70	
3.52/1.70	U3_ag(x1, x2, x3)  =  U3_ag(x3)
3.52/1.70	
3.52/1.70	max_out_aag(x1, x2, x3)  =  max_out_aag(x1, x2)
3.52/1.70	
3.52/1.70	U2_aag(x1, x2, x3)  =  U2_aag(x2, x3)
3.52/1.70	
3.52/1.70	LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)
3.52/1.70	
3.52/1.70	
3.52/1.70	We have to consider all (P,R,Pi)-chains
3.52/1.70	----------------------------------------
3.52/1.70	
3.52/1.70	(7) UsableRulesProof (EQUIVALENT)
3.52/1.70	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.52/1.70	----------------------------------------
3.52/1.70	
3.52/1.70	(8)
3.52/1.70	Obligation:
3.52/1.70	Pi DP problem:
3.52/1.70	The TRS P consists of the following rules:
3.52/1.70	
3.52/1.70	   LESS_IN_AG(s(X), s(Y)) -> LESS_IN_AG(X, Y)
3.52/1.70	
3.52/1.70	R is empty.
3.52/1.70	The argument filtering Pi contains the following mapping:
3.52/1.70	s(x1)  =  s(x1)
3.52/1.70	
3.52/1.70	LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)
3.52/1.70	
3.52/1.70	
3.52/1.70	We have to consider all (P,R,Pi)-chains
3.52/1.70	----------------------------------------
3.52/1.70	
3.52/1.70	(9) PiDPToQDPProof (SOUND)
3.52/1.70	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.52/1.70	----------------------------------------
3.52/1.70	
3.52/1.70	(10)
3.52/1.70	Obligation:
3.52/1.70	Q DP problem:
3.52/1.70	The TRS P consists of the following rules:
3.52/1.70	
3.52/1.70	   LESS_IN_AG(s(Y)) -> LESS_IN_AG(Y)
3.52/1.70	
3.52/1.70	R is empty.
3.52/1.70	Q is empty.
3.52/1.70	We have to consider all (P,Q,R)-chains.
3.52/1.70	----------------------------------------
3.52/1.70	
3.52/1.70	(11) QDPSizeChangeProof (EQUIVALENT)
3.52/1.70	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. 
3.52/1.70	
3.52/1.70	From the DPs we obtained the following set of size-change graphs:
3.52/1.70	*LESS_IN_AG(s(Y)) -> LESS_IN_AG(Y)
3.52/1.70	The graph contains the following edges 1 > 1
3.52/1.70	
3.52/1.70	
3.52/1.70	----------------------------------------
3.52/1.70	
3.52/1.70	(12)
3.52/1.70	YES
3.76/1.75	EOF