(0) Obligation:
Clauses:
delete(X, tree(X, void, Right), Right).
delete(X, tree(X, Left, void), Left).
delete(X, tree(X, Left, Right), tree(Y, Left, Right1)) :- delmin(Right, Y, Right1).
delete(X, tree(Y, Left, Right), tree(Y, Left1, Right)) :- ','(less(X, Y), delete(X, Left, Left1)).
delete(X, tree(Y, Left, Right), tree(Y, Left, Right1)) :- ','(less(Y, X), delete(X, Right, Right1)).
delmin(tree(Y, void, Right), Y, Right).
delmin(tree(X, Left, X1), Y, tree(X, Left1, X2)) :- delmin(Left, Y, Left1).
less(0, s(X3)).
less(s(X), s(Y)) :- less(X, Y).
Query: delmin(g,a,a)
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
delminA(tree(X1, tree(X2, X3, X4), X5), X6, tree(X1, tree(X2, X7, X8), X9)) :- delminA(X3, X6, X7).
Clauses:
delmincA(tree(X1, void, X2), X1, X2).
delmincA(tree(X1, tree(X2, void, X3), X4), X2, tree(X1, X3, X5)).
delmincA(tree(X1, tree(X2, X3, X4), X5), X6, tree(X1, tree(X2, X7, X8), X9)) :- delmincA(X3, X6, X7).
Afs:
delminA(x1, x2, x3) = delminA(x1)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
delminA_in: (b,f,f)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
DELMINA_IN_GAA(tree(X1, tree(X2, X3, X4), X5), X6, tree(X1, tree(X2, X7, X8), X9)) → U1_GAA(X1, X2, X3, X4, X5, X6, X7, X8, X9, delminA_in_gaa(X3, X6, X7))
DELMINA_IN_GAA(tree(X1, tree(X2, X3, X4), X5), X6, tree(X1, tree(X2, X7, X8), X9)) → DELMINA_IN_GAA(X3, X6, X7)
R is empty.
The argument filtering Pi contains the following mapping:
delminA_in_gaa(
x1,
x2,
x3) =
delminA_in_gaa(
x1)
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
DELMINA_IN_GAA(
x1,
x2,
x3) =
DELMINA_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x10) =
U1_GAA(
x1,
x2,
x3,
x4,
x5,
x10)
We have to consider all (P,R,Pi)-chains
Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
DELMINA_IN_GAA(tree(X1, tree(X2, X3, X4), X5), X6, tree(X1, tree(X2, X7, X8), X9)) → U1_GAA(X1, X2, X3, X4, X5, X6, X7, X8, X9, delminA_in_gaa(X3, X6, X7))
DELMINA_IN_GAA(tree(X1, tree(X2, X3, X4), X5), X6, tree(X1, tree(X2, X7, X8), X9)) → DELMINA_IN_GAA(X3, X6, X7)
R is empty.
The argument filtering Pi contains the following mapping:
delminA_in_gaa(
x1,
x2,
x3) =
delminA_in_gaa(
x1)
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
DELMINA_IN_GAA(
x1,
x2,
x3) =
DELMINA_IN_GAA(
x1)
U1_GAA(
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8,
x9,
x10) =
U1_GAA(
x1,
x2,
x3,
x4,
x5,
x10)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
DELMINA_IN_GAA(tree(X1, tree(X2, X3, X4), X5), X6, tree(X1, tree(X2, X7, X8), X9)) → DELMINA_IN_GAA(X3, X6, X7)
R is empty.
The argument filtering Pi contains the following mapping:
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
DELMINA_IN_GAA(
x1,
x2,
x3) =
DELMINA_IN_GAA(
x1)
We have to consider all (P,R,Pi)-chains
(7) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DELMINA_IN_GAA(tree(X1, tree(X2, X3, X4), X5)) → DELMINA_IN_GAA(X3)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) QDPSizeChangeProof (EQUIVALENT transformation)
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.
From the DPs we obtained the following set of size-change graphs:
- DELMINA_IN_GAA(tree(X1, tree(X2, X3, X4), X5)) → DELMINA_IN_GAA(X3)
The graph contains the following edges 1 > 1
(10) YES