3.77/1.87	YES
3.77/1.89	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.77/1.89	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.77/1.89	
3.77/1.89	
3.77/1.89	Left Termination of the query pattern
3.77/1.89	
3.77/1.89	delmin(a,a,g)
3.77/1.89	
3.77/1.89	w.r.t. the given Prolog program could successfully be proven:
3.77/1.89	
3.77/1.89	(0) Prolog
3.77/1.89	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.77/1.89	(2) PiTRS
3.77/1.89	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.77/1.89	(4) PiDP
3.77/1.89	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.77/1.89	(6) PiDP
3.77/1.89	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.77/1.89	(8) PiDP
3.77/1.89	(9) PiDPToQDPProof [SOUND, 0 ms]
3.77/1.89	(10) QDP
3.77/1.89	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.77/1.89	(12) YES
3.77/1.89	
3.77/1.89	
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(0)
3.77/1.89	Obligation:
3.77/1.89	Clauses:
3.77/1.89	
3.77/1.89	delete(X, tree(X, void, Right), Right).
3.77/1.89	delete(X, tree(X, Left, void), Left).
3.77/1.89	delete(X, tree(X, Left, Right), tree(Y, Left, Right1)) :- delmin(Right, Y, Right1).
3.77/1.89	delete(X, tree(Y, Left, Right), tree(Y, Left1, Right)) :- ','(less(X, Y), delete(X, Left, Left1)).
3.77/1.89	delete(X, tree(Y, Left, Right), tree(Y, Left, Right1)) :- ','(less(Y, X), delete(X, Right, Right1)).
3.77/1.89	delmin(tree(Y, void, Right), Y, Right).
3.77/1.89	delmin(tree(X, Left, X1), Y, tree(X, Left1, X2)) :- delmin(Left, Y, Left1).
3.77/1.89	less(0, s(X3)).
3.77/1.89	less(s(X), s(Y)) :- less(X, Y).
3.77/1.89	
3.77/1.89	
3.77/1.89	Query: delmin(a,a,g)
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(1) PrologToPiTRSProof (SOUND)
3.77/1.89	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.77/1.89	
3.77/1.89	delmin_in_3: (f,f,b)
3.77/1.89	
3.77/1.89	Transforming Prolog into the following Term Rewriting System:
3.77/1.89	
3.77/1.89	Pi-finite rewrite system:
3.77/1.89	The TRS R consists of the following rules:
3.77/1.89	
3.77/1.89	   delmin_in_aag(tree(Y, void, Right), Y, Right) -> delmin_out_aag(tree(Y, void, Right), Y, Right)
3.77/1.89	   delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
3.77/1.89	   U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) -> delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))
3.77/1.89	
3.77/1.89	The argument filtering Pi contains the following mapping:
3.77/1.89	delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
3.77/1.89	
3.77/1.89	delmin_out_aag(x1, x2, x3)  =  delmin_out_aag
3.77/1.89	
3.77/1.89	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.77/1.89	
3.77/1.89	U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x7)
3.77/1.89	
3.77/1.89	
3.77/1.89	
3.77/1.89	
3.77/1.89	
3.77/1.89	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.77/1.89	
3.77/1.89	
3.77/1.89	
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(2)
3.77/1.89	Obligation:
3.77/1.89	Pi-finite rewrite system:
3.77/1.89	The TRS R consists of the following rules:
3.77/1.89	
3.77/1.89	   delmin_in_aag(tree(Y, void, Right), Y, Right) -> delmin_out_aag(tree(Y, void, Right), Y, Right)
3.77/1.89	   delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
3.77/1.89	   U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) -> delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))
3.77/1.89	
3.77/1.89	The argument filtering Pi contains the following mapping:
3.77/1.89	delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
3.77/1.89	
3.77/1.89	delmin_out_aag(x1, x2, x3)  =  delmin_out_aag
3.77/1.89	
3.77/1.89	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.77/1.89	
3.77/1.89	U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x7)
3.77/1.89	
3.77/1.89	
3.77/1.89	
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(3) DependencyPairsProof (EQUIVALENT)
3.77/1.89	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.77/1.89	Pi DP problem:
3.77/1.89	The TRS P consists of the following rules:
3.77/1.89	
3.77/1.89	   DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_AAG(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
3.77/1.89	   DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> DELMIN_IN_AAG(Left, Y, Left1)
3.77/1.89	
3.77/1.89	The TRS R consists of the following rules:
3.77/1.89	
3.77/1.89	   delmin_in_aag(tree(Y, void, Right), Y, Right) -> delmin_out_aag(tree(Y, void, Right), Y, Right)
3.77/1.89	   delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
3.77/1.89	   U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) -> delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))
3.77/1.89	
3.77/1.89	The argument filtering Pi contains the following mapping:
3.77/1.89	delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
3.77/1.89	
3.77/1.89	delmin_out_aag(x1, x2, x3)  =  delmin_out_aag
3.77/1.89	
3.77/1.89	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.77/1.89	
3.77/1.89	U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x7)
3.77/1.89	
3.77/1.89	DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)
3.77/1.89	
3.77/1.89	U6_AAG(x1, x2, x3, x4, x5, x6, x7)  =  U6_AAG(x7)
3.77/1.89	
3.77/1.89	
3.77/1.89	We have to consider all (P,R,Pi)-chains
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(4)
3.77/1.89	Obligation:
3.77/1.89	Pi DP problem:
3.77/1.89	The TRS P consists of the following rules:
3.77/1.89	
3.77/1.89	   DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_AAG(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
3.77/1.89	   DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> DELMIN_IN_AAG(Left, Y, Left1)
3.77/1.89	
3.77/1.89	The TRS R consists of the following rules:
3.77/1.89	
3.77/1.89	   delmin_in_aag(tree(Y, void, Right), Y, Right) -> delmin_out_aag(tree(Y, void, Right), Y, Right)
3.77/1.89	   delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
3.77/1.89	   U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) -> delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))
3.77/1.89	
3.77/1.89	The argument filtering Pi contains the following mapping:
3.77/1.89	delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
3.77/1.89	
3.77/1.89	delmin_out_aag(x1, x2, x3)  =  delmin_out_aag
3.77/1.89	
3.77/1.89	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.77/1.89	
3.77/1.89	U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x7)
3.77/1.89	
3.77/1.89	DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)
3.77/1.89	
3.77/1.89	U6_AAG(x1, x2, x3, x4, x5, x6, x7)  =  U6_AAG(x7)
3.77/1.89	
3.77/1.89	
3.77/1.89	We have to consider all (P,R,Pi)-chains
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(5) DependencyGraphProof (EQUIVALENT)
3.77/1.89	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(6)
3.77/1.89	Obligation:
3.77/1.89	Pi DP problem:
3.77/1.89	The TRS P consists of the following rules:
3.77/1.89	
3.77/1.89	   DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> DELMIN_IN_AAG(Left, Y, Left1)
3.77/1.89	
3.77/1.89	The TRS R consists of the following rules:
3.77/1.89	
3.77/1.89	   delmin_in_aag(tree(Y, void, Right), Y, Right) -> delmin_out_aag(tree(Y, void, Right), Y, Right)
3.77/1.89	   delmin_in_aag(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_aag(X, Left, X1, Y, Left1, X2, delmin_in_aag(Left, Y, Left1))
3.77/1.89	   U6_aag(X, Left, X1, Y, Left1, X2, delmin_out_aag(Left, Y, Left1)) -> delmin_out_aag(tree(X, Left, X1), Y, tree(X, Left1, X2))
3.77/1.89	
3.77/1.89	The argument filtering Pi contains the following mapping:
3.77/1.89	delmin_in_aag(x1, x2, x3)  =  delmin_in_aag(x3)
3.77/1.89	
3.77/1.89	delmin_out_aag(x1, x2, x3)  =  delmin_out_aag
3.77/1.89	
3.77/1.89	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.77/1.89	
3.77/1.89	U6_aag(x1, x2, x3, x4, x5, x6, x7)  =  U6_aag(x7)
3.77/1.89	
3.77/1.89	DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)
3.77/1.89	
3.77/1.89	
3.77/1.89	We have to consider all (P,R,Pi)-chains
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(7) UsableRulesProof (EQUIVALENT)
3.77/1.89	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(8)
3.77/1.89	Obligation:
3.77/1.89	Pi DP problem:
3.77/1.89	The TRS P consists of the following rules:
3.77/1.89	
3.77/1.89	   DELMIN_IN_AAG(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> DELMIN_IN_AAG(Left, Y, Left1)
3.77/1.89	
3.77/1.89	R is empty.
3.77/1.89	The argument filtering Pi contains the following mapping:
3.77/1.89	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.77/1.89	
3.77/1.89	DELMIN_IN_AAG(x1, x2, x3)  =  DELMIN_IN_AAG(x3)
3.77/1.89	
3.77/1.89	
3.77/1.89	We have to consider all (P,R,Pi)-chains
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(9) PiDPToQDPProof (SOUND)
3.77/1.89	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(10)
3.77/1.89	Obligation:
3.77/1.89	Q DP problem:
3.77/1.89	The TRS P consists of the following rules:
3.77/1.89	
3.77/1.89	   DELMIN_IN_AAG(tree(X, Left1, X2)) -> DELMIN_IN_AAG(Left1)
3.77/1.89	
3.77/1.89	R is empty.
3.77/1.89	Q is empty.
3.77/1.89	We have to consider all (P,Q,R)-chains.
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(11) QDPSizeChangeProof (EQUIVALENT)
3.77/1.89	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. 
3.77/1.89	
3.77/1.89	From the DPs we obtained the following set of size-change graphs:
3.77/1.89	*DELMIN_IN_AAG(tree(X, Left1, X2)) -> DELMIN_IN_AAG(Left1)
3.77/1.89	The graph contains the following edges 1 > 1
3.77/1.89	
3.77/1.89	
3.77/1.89	----------------------------------------
3.77/1.89	
3.77/1.89	(12)
3.77/1.89	YES
4.04/2.25	EOF