(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).
Queries:
delmin(a,a,g).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:delmin1(tree(T14, tree(T44, T50, T46), T16), T51, tree(T14, tree(T44, T48, T49), T19)) :- delmin1(T50, T51, T48).
delmin1(tree(T62, tree(T92, T98, T94), T64), T99, tree(T62, tree(T92, T96, T97), T67)) :- delmin1(T98, T99, T96).
Clauses:
delminc1(tree(T6, void, T7), T6, T7).
delminc1(tree(T14, tree(T30, void, T31), T16), T30, tree(T14, T31, T19)).
delminc1(tree(T14, tree(T44, T50, T46), T16), T51, tree(T14, tree(T44, T48, T49), T19)) :- delminc1(T50, T51, T48).
delminc1(tree(T62, tree(T78, void, T79), T64), T78, tree(T62, T79, T67)).
delminc1(tree(T62, tree(T92, T98, T94), T64), T99, tree(T62, tree(T92, T96, T97), T67)) :- delminc1(T98, T99, T96).
Afs:
delmin1(x1, x2, x3) = delmin1(x3)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:delmin1_in: (f,f,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
DELMIN1_IN_AAG(tree(T14, tree(T44, T50, T46), T16), T51, tree(T14, tree(T44, T48, T49), T19)) → U1_AAG(T14, T44, T50, T46, T16, T51, T48, T49, T19, delmin1_in_aag(T50, T51, T48))
DELMIN1_IN_AAG(tree(T14, tree(T44, T50, T46), T16), T51, tree(T14, tree(T44, T48, T49), T19)) → DELMIN1_IN_AAG(T50, T51, T48)
DELMIN1_IN_AAG(tree(T62, tree(T92, T98, T94), T64), T99, tree(T62, tree(T92, T96, T97), T67)) → U2_AAG(T62, T92, T98, T94, T64, T99, T96, T97, T67, delmin1_in_aag(T98, T99, T96))
R is empty.
The argument filtering Pi contains the following mapping:
delmin1_in_aag(x1, x2, x3) = delmin1_in_aag(x3)
tree(x1, x2, x3) = tree(x1, x2, x3)
DELMIN1_IN_AAG(x1, x2, x3) = DELMIN1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) = U1_AAG(x1, x2, x7, x8, x9, x10)
U2_AAG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) = U2_AAG(x1, x2, x7, x8, x9, 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:
DELMIN1_IN_AAG(tree(T14, tree(T44, T50, T46), T16), T51, tree(T14, tree(T44, T48, T49), T19)) → U1_AAG(T14, T44, T50, T46, T16, T51, T48, T49, T19, delmin1_in_aag(T50, T51, T48))
DELMIN1_IN_AAG(tree(T14, tree(T44, T50, T46), T16), T51, tree(T14, tree(T44, T48, T49), T19)) → DELMIN1_IN_AAG(T50, T51, T48)
DELMIN1_IN_AAG(tree(T62, tree(T92, T98, T94), T64), T99, tree(T62, tree(T92, T96, T97), T67)) → U2_AAG(T62, T92, T98, T94, T64, T99, T96, T97, T67, delmin1_in_aag(T98, T99, T96))
R is empty.
The argument filtering Pi contains the following mapping:
delmin1_in_aag(x1, x2, x3) = delmin1_in_aag(x3)
tree(x1, x2, x3) = tree(x1, x2, x3)
DELMIN1_IN_AAG(x1, x2, x3) = DELMIN1_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) = U1_AAG(x1, x2, x7, x8, x9, x10)
U2_AAG(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) = U2_AAG(x1, x2, x7, x8, x9, 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 2 less nodes.(6) Obligation:
Pi DP problem:The TRS P consists of the following rules:
DELMIN1_IN_AAG(tree(T14, tree(T44, T50, T46), T16), T51, tree(T14, tree(T44, T48, T49), T19)) → DELMIN1_IN_AAG(T50, T51, T48)
R is empty.
The argument filtering Pi contains the following mapping:
tree(x1, x2, x3) = tree(x1, x2, x3)
DELMIN1_IN_AAG(x1, x2, x3) = DELMIN1_IN_AAG(x3)
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:
DELMIN1_IN_AAG(tree(T14, tree(T44, T48, T49), T19)) → DELMIN1_IN_AAG(T48)
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:
- DELMIN1_IN_AAG(tree(T14, tree(T44, T48, T49), T19)) → DELMIN1_IN_AAG(T48)
The graph contains the following edges 1 > 1