Left Termination of the query pattern delmin_in_3(g, a, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

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(X)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

delmin(g,a,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in(x1, x2, x3)  =  delmin_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in(x1, x2, x3)  =  delmin_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out(x2)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U61(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN(Left, Y, Left1)

The TRS R consists of the following rules:

delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in(x1, x2, x3)  =  delmin_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out(x2)
DELMIN_IN(x1, x2, x3)  =  DELMIN_IN(x1)
U61(x1, x2, x3, x4, x5, x6, x7)  =  U61(x7)

We have to consider all (P,R,Pi)-chains

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U61(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN(Left, Y, Left1)

The TRS R consists of the following rules:

delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in(x1, x2, x3)  =  delmin_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out(x2)
DELMIN_IN(x1, x2, x3)  =  DELMIN_IN(x1)
U61(x1, x2, x3, x4, x5, x6, x7)  =  U61(x7)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN(Left, Y, Left1)

The TRS R consists of the following rules:

delmin_in(tree(X, Left, X1), Y, tree(X, Left1, X2)) → U6(X, Left, X1, Y, Left1, X2, delmin_in(Left, Y, Left1))
delmin_in(tree(Y, void, Right), Y, Right) → delmin_out(tree(Y, void, Right), Y, Right)
U6(X, Left, X1, Y, Left1, X2, delmin_out(Left, Y, Left1)) → delmin_out(tree(X, Left, X1), Y, tree(X, Left1, X2))

The argument filtering Pi contains the following mapping:
delmin_in(x1, x2, x3)  =  delmin_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U6(x1, x2, x3, x4, x5, x6, x7)  =  U6(x7)
void  =  void
delmin_out(x1, x2, x3)  =  delmin_out(x2)
DELMIN_IN(x1, x2, x3)  =  DELMIN_IN(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

DELMIN_IN(tree(X, Left, X1), Y, tree(X, Left1, X2)) → DELMIN_IN(Left, Y, Left1)

R is empty.
The argument filtering Pi contains the following mapping:
tree(x1, x2, x3)  =  tree(x1, x2, x3)
DELMIN_IN(x1, x2, x3)  =  DELMIN_IN(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

Q DP problem:
The TRS P consists of the following rules:

DELMIN_IN(tree(X, Left, X1)) → DELMIN_IN(Left)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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: