3.82/1.86	YES
3.82/1.87	proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
3.82/1.87	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.82/1.87	
3.82/1.87	
3.82/1.87	Left Termination of the query pattern
3.82/1.87	
3.82/1.87	delmin(g,a,a)
3.82/1.87	
3.82/1.87	w.r.t. the given Prolog program could successfully be proven:
3.82/1.87	
3.82/1.87	(0) Prolog
3.82/1.87	(1) PrologToPiTRSProof [SOUND, 0 ms]
3.82/1.87	(2) PiTRS
3.82/1.87	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
3.82/1.87	(4) PiDP
3.82/1.87	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.82/1.87	(6) PiDP
3.82/1.87	(7) UsableRulesProof [EQUIVALENT, 0 ms]
3.82/1.87	(8) PiDP
3.82/1.87	(9) PiDPToQDPProof [SOUND, 0 ms]
3.82/1.87	(10) QDP
3.82/1.87	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.82/1.87	(12) YES
3.82/1.87	
3.82/1.87	
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(0)
3.82/1.87	Obligation:
3.82/1.87	Clauses:
3.82/1.87	
3.82/1.87	delete(X, tree(X, void, Right), Right).
3.82/1.87	delete(X, tree(X, Left, void), Left).
3.82/1.87	delete(X, tree(X, Left, Right), tree(Y, Left, Right1)) :- delmin(Right, Y, Right1).
3.82/1.87	delete(X, tree(Y, Left, Right), tree(Y, Left1, Right)) :- ','(less(X, Y), delete(X, Left, Left1)).
3.82/1.87	delete(X, tree(Y, Left, Right), tree(Y, Left, Right1)) :- ','(less(Y, X), delete(X, Right, Right1)).
3.82/1.87	delmin(tree(Y, void, Right), Y, Right).
3.82/1.87	delmin(tree(X, Left, X1), Y, tree(X, Left1, X2)) :- delmin(Left, Y, Left1).
3.82/1.87	less(0, s(X3)).
3.82/1.87	less(s(X), s(Y)) :- less(X, Y).
3.82/1.87	
3.82/1.87	
3.82/1.87	Query: delmin(g,a,a)
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(1) PrologToPiTRSProof (SOUND)
3.82/1.87	We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
3.82/1.87	
3.82/1.87	delmin_in_3: (b,f,f)
3.82/1.87	
3.82/1.87	Transforming Prolog into the following Term Rewriting System:
3.82/1.87	
3.82/1.87	Pi-finite rewrite system:
3.82/1.87	The TRS R consists of the following rules:
3.82/1.87	
3.82/1.87	   delmin_in_gaa(tree(Y, void, Right), Y, Right) -> delmin_out_gaa(tree(Y, void, Right), Y, Right)
3.82/1.87	   delmin_in_gaa(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_gaa(X, Left, X1, Y, Left1, X2, delmin_in_gaa(Left, Y, Left1))
3.82/1.87	   U6_gaa(X, Left, X1, Y, Left1, X2, delmin_out_gaa(Left, Y, Left1)) -> delmin_out_gaa(tree(X, Left, X1), Y, tree(X, Left1, X2))
3.82/1.87	
3.82/1.87	The argument filtering Pi contains the following mapping:
3.82/1.87	delmin_in_gaa(x1, x2, x3)  =  delmin_in_gaa(x1)
3.82/1.87	
3.82/1.87	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.82/1.87	
3.82/1.87	void  =  void
3.82/1.87	
3.82/1.87	delmin_out_gaa(x1, x2, x3)  =  delmin_out_gaa(x2)
3.82/1.87	
3.82/1.87	U6_gaa(x1, x2, x3, x4, x5, x6, x7)  =  U6_gaa(x7)
3.82/1.87	
3.82/1.87	
3.82/1.87	
3.82/1.87	
3.82/1.87	
3.82/1.87	Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
3.82/1.87	
3.82/1.87	
3.82/1.87	
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(2)
3.82/1.87	Obligation:
3.82/1.87	Pi-finite rewrite system:
3.82/1.87	The TRS R consists of the following rules:
3.82/1.87	
3.82/1.87	   delmin_in_gaa(tree(Y, void, Right), Y, Right) -> delmin_out_gaa(tree(Y, void, Right), Y, Right)
3.82/1.87	   delmin_in_gaa(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_gaa(X, Left, X1, Y, Left1, X2, delmin_in_gaa(Left, Y, Left1))
3.82/1.87	   U6_gaa(X, Left, X1, Y, Left1, X2, delmin_out_gaa(Left, Y, Left1)) -> delmin_out_gaa(tree(X, Left, X1), Y, tree(X, Left1, X2))
3.82/1.87	
3.82/1.87	The argument filtering Pi contains the following mapping:
3.82/1.87	delmin_in_gaa(x1, x2, x3)  =  delmin_in_gaa(x1)
3.82/1.87	
3.82/1.87	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.82/1.87	
3.82/1.87	void  =  void
3.82/1.87	
3.82/1.87	delmin_out_gaa(x1, x2, x3)  =  delmin_out_gaa(x2)
3.82/1.87	
3.82/1.87	U6_gaa(x1, x2, x3, x4, x5, x6, x7)  =  U6_gaa(x7)
3.82/1.87	
3.82/1.87	
3.82/1.87	
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(3) DependencyPairsProof (EQUIVALENT)
3.82/1.87	Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
3.82/1.87	Pi DP problem:
3.82/1.87	The TRS P consists of the following rules:
3.82/1.87	
3.82/1.87	   DELMIN_IN_GAA(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_GAA(X, Left, X1, Y, Left1, X2, delmin_in_gaa(Left, Y, Left1))
3.82/1.87	   DELMIN_IN_GAA(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> DELMIN_IN_GAA(Left, Y, Left1)
3.82/1.87	
3.82/1.87	The TRS R consists of the following rules:
3.82/1.87	
3.82/1.87	   delmin_in_gaa(tree(Y, void, Right), Y, Right) -> delmin_out_gaa(tree(Y, void, Right), Y, Right)
3.82/1.87	   delmin_in_gaa(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_gaa(X, Left, X1, Y, Left1, X2, delmin_in_gaa(Left, Y, Left1))
3.82/1.87	   U6_gaa(X, Left, X1, Y, Left1, X2, delmin_out_gaa(Left, Y, Left1)) -> delmin_out_gaa(tree(X, Left, X1), Y, tree(X, Left1, X2))
3.82/1.87	
3.82/1.87	The argument filtering Pi contains the following mapping:
3.82/1.87	delmin_in_gaa(x1, x2, x3)  =  delmin_in_gaa(x1)
3.82/1.87	
3.82/1.87	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.82/1.87	
3.82/1.87	void  =  void
3.82/1.87	
3.82/1.87	delmin_out_gaa(x1, x2, x3)  =  delmin_out_gaa(x2)
3.82/1.87	
3.82/1.87	U6_gaa(x1, x2, x3, x4, x5, x6, x7)  =  U6_gaa(x7)
3.82/1.87	
3.82/1.87	DELMIN_IN_GAA(x1, x2, x3)  =  DELMIN_IN_GAA(x1)
3.82/1.87	
3.82/1.87	U6_GAA(x1, x2, x3, x4, x5, x6, x7)  =  U6_GAA(x7)
3.82/1.87	
3.82/1.87	
3.82/1.87	We have to consider all (P,R,Pi)-chains
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(4)
3.82/1.87	Obligation:
3.82/1.87	Pi DP problem:
3.82/1.87	The TRS P consists of the following rules:
3.82/1.87	
3.82/1.87	   DELMIN_IN_GAA(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_GAA(X, Left, X1, Y, Left1, X2, delmin_in_gaa(Left, Y, Left1))
3.82/1.87	   DELMIN_IN_GAA(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> DELMIN_IN_GAA(Left, Y, Left1)
3.82/1.87	
3.82/1.87	The TRS R consists of the following rules:
3.82/1.87	
3.82/1.87	   delmin_in_gaa(tree(Y, void, Right), Y, Right) -> delmin_out_gaa(tree(Y, void, Right), Y, Right)
3.82/1.87	   delmin_in_gaa(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_gaa(X, Left, X1, Y, Left1, X2, delmin_in_gaa(Left, Y, Left1))
3.82/1.87	   U6_gaa(X, Left, X1, Y, Left1, X2, delmin_out_gaa(Left, Y, Left1)) -> delmin_out_gaa(tree(X, Left, X1), Y, tree(X, Left1, X2))
3.82/1.87	
3.82/1.87	The argument filtering Pi contains the following mapping:
3.82/1.87	delmin_in_gaa(x1, x2, x3)  =  delmin_in_gaa(x1)
3.82/1.87	
3.82/1.87	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.82/1.87	
3.82/1.87	void  =  void
3.82/1.87	
3.82/1.87	delmin_out_gaa(x1, x2, x3)  =  delmin_out_gaa(x2)
3.82/1.87	
3.82/1.87	U6_gaa(x1, x2, x3, x4, x5, x6, x7)  =  U6_gaa(x7)
3.82/1.87	
3.82/1.87	DELMIN_IN_GAA(x1, x2, x3)  =  DELMIN_IN_GAA(x1)
3.82/1.87	
3.82/1.87	U6_GAA(x1, x2, x3, x4, x5, x6, x7)  =  U6_GAA(x7)
3.82/1.87	
3.82/1.87	
3.82/1.87	We have to consider all (P,R,Pi)-chains
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(5) DependencyGraphProof (EQUIVALENT)
3.82/1.87	The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(6)
3.82/1.87	Obligation:
3.82/1.87	Pi DP problem:
3.82/1.87	The TRS P consists of the following rules:
3.82/1.87	
3.82/1.87	   DELMIN_IN_GAA(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> DELMIN_IN_GAA(Left, Y, Left1)
3.82/1.87	
3.82/1.87	The TRS R consists of the following rules:
3.82/1.87	
3.82/1.87	   delmin_in_gaa(tree(Y, void, Right), Y, Right) -> delmin_out_gaa(tree(Y, void, Right), Y, Right)
3.82/1.87	   delmin_in_gaa(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> U6_gaa(X, Left, X1, Y, Left1, X2, delmin_in_gaa(Left, Y, Left1))
3.82/1.87	   U6_gaa(X, Left, X1, Y, Left1, X2, delmin_out_gaa(Left, Y, Left1)) -> delmin_out_gaa(tree(X, Left, X1), Y, tree(X, Left1, X2))
3.82/1.87	
3.82/1.87	The argument filtering Pi contains the following mapping:
3.82/1.87	delmin_in_gaa(x1, x2, x3)  =  delmin_in_gaa(x1)
3.82/1.87	
3.82/1.87	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.82/1.87	
3.82/1.87	void  =  void
3.82/1.87	
3.82/1.87	delmin_out_gaa(x1, x2, x3)  =  delmin_out_gaa(x2)
3.82/1.87	
3.82/1.87	U6_gaa(x1, x2, x3, x4, x5, x6, x7)  =  U6_gaa(x7)
3.82/1.87	
3.82/1.87	DELMIN_IN_GAA(x1, x2, x3)  =  DELMIN_IN_GAA(x1)
3.82/1.87	
3.82/1.87	
3.82/1.87	We have to consider all (P,R,Pi)-chains
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(7) UsableRulesProof (EQUIVALENT)
3.82/1.87	For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(8)
3.82/1.87	Obligation:
3.82/1.87	Pi DP problem:
3.82/1.87	The TRS P consists of the following rules:
3.82/1.87	
3.82/1.87	   DELMIN_IN_GAA(tree(X, Left, X1), Y, tree(X, Left1, X2)) -> DELMIN_IN_GAA(Left, Y, Left1)
3.82/1.87	
3.82/1.87	R is empty.
3.82/1.87	The argument filtering Pi contains the following mapping:
3.82/1.87	tree(x1, x2, x3)  =  tree(x1, x2, x3)
3.82/1.87	
3.82/1.87	DELMIN_IN_GAA(x1, x2, x3)  =  DELMIN_IN_GAA(x1)
3.82/1.87	
3.82/1.87	
3.82/1.87	We have to consider all (P,R,Pi)-chains
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(9) PiDPToQDPProof (SOUND)
3.82/1.87	Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(10)
3.82/1.87	Obligation:
3.82/1.87	Q DP problem:
3.82/1.87	The TRS P consists of the following rules:
3.82/1.87	
3.82/1.87	   DELMIN_IN_GAA(tree(X, Left, X1)) -> DELMIN_IN_GAA(Left)
3.82/1.87	
3.82/1.87	R is empty.
3.82/1.87	Q is empty.
3.82/1.87	We have to consider all (P,Q,R)-chains.
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(11) QDPSizeChangeProof (EQUIVALENT)
3.82/1.87	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. 
3.82/1.87	
3.82/1.87	From the DPs we obtained the following set of size-change graphs:
3.82/1.87	*DELMIN_IN_GAA(tree(X, Left, X1)) -> DELMIN_IN_GAA(Left)
3.82/1.87	The graph contains the following edges 1 > 1
3.82/1.87	
3.82/1.87	
3.82/1.87	----------------------------------------
3.82/1.87	
3.82/1.87	(12)
3.82/1.87	YES
3.82/1.90	EOF