Left Termination of the query pattern tree_member_in_2(a, g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

tree_member(X, tree(X, X1, X2)).
tree_member(X, tree(X1, Left, X2)) :- tree_member(X, Left).
tree_member(X, tree(X1, X2, Right)) :- tree_member(X, Right).

Queries:

tree_member(a,g).

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:

tree_member_in(X, tree(X1, X2, Right)) → U2(X, X1, X2, Right, tree_member_in(X, Right))
tree_member_in(X, tree(X1, Left, X2)) → U1(X, X1, Left, X2, tree_member_in(X, Left))
tree_member_in(X, tree(X, X1, X2)) → tree_member_out(X, tree(X, X1, X2))
U1(X, X1, Left, X2, tree_member_out(X, Left)) → tree_member_out(X, tree(X1, Left, X2))
U2(X, X1, X2, Right, tree_member_out(X, Right)) → tree_member_out(X, tree(X1, X2, Right))

The argument filtering Pi contains the following mapping:
tree_member_in(x1, x2)  =  tree_member_in(x2)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
tree_member_out(x1, x2)  =  tree_member_out(x1)

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:

tree_member_in(X, tree(X1, X2, Right)) → U2(X, X1, X2, Right, tree_member_in(X, Right))
tree_member_in(X, tree(X1, Left, X2)) → U1(X, X1, Left, X2, tree_member_in(X, Left))
tree_member_in(X, tree(X, X1, X2)) → tree_member_out(X, tree(X, X1, X2))
U1(X, X1, Left, X2, tree_member_out(X, Left)) → tree_member_out(X, tree(X1, Left, X2))
U2(X, X1, X2, Right, tree_member_out(X, Right)) → tree_member_out(X, tree(X1, X2, Right))

The argument filtering Pi contains the following mapping:
tree_member_in(x1, x2)  =  tree_member_in(x2)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
tree_member_out(x1, x2)  =  tree_member_out(x1)


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

TREE_MEMBER_IN(X, tree(X1, X2, Right)) → U21(X, X1, X2, Right, tree_member_in(X, Right))
TREE_MEMBER_IN(X, tree(X1, X2, Right)) → TREE_MEMBER_IN(X, Right)
TREE_MEMBER_IN(X, tree(X1, Left, X2)) → U11(X, X1, Left, X2, tree_member_in(X, Left))
TREE_MEMBER_IN(X, tree(X1, Left, X2)) → TREE_MEMBER_IN(X, Left)

The TRS R consists of the following rules:

tree_member_in(X, tree(X1, X2, Right)) → U2(X, X1, X2, Right, tree_member_in(X, Right))
tree_member_in(X, tree(X1, Left, X2)) → U1(X, X1, Left, X2, tree_member_in(X, Left))
tree_member_in(X, tree(X, X1, X2)) → tree_member_out(X, tree(X, X1, X2))
U1(X, X1, Left, X2, tree_member_out(X, Left)) → tree_member_out(X, tree(X1, Left, X2))
U2(X, X1, X2, Right, tree_member_out(X, Right)) → tree_member_out(X, tree(X1, X2, Right))

The argument filtering Pi contains the following mapping:
tree_member_in(x1, x2)  =  tree_member_in(x2)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
tree_member_out(x1, x2)  =  tree_member_out(x1)
TREE_MEMBER_IN(x1, x2)  =  TREE_MEMBER_IN(x2)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x5)

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:

TREE_MEMBER_IN(X, tree(X1, X2, Right)) → U21(X, X1, X2, Right, tree_member_in(X, Right))
TREE_MEMBER_IN(X, tree(X1, X2, Right)) → TREE_MEMBER_IN(X, Right)
TREE_MEMBER_IN(X, tree(X1, Left, X2)) → U11(X, X1, Left, X2, tree_member_in(X, Left))
TREE_MEMBER_IN(X, tree(X1, Left, X2)) → TREE_MEMBER_IN(X, Left)

The TRS R consists of the following rules:

tree_member_in(X, tree(X1, X2, Right)) → U2(X, X1, X2, Right, tree_member_in(X, Right))
tree_member_in(X, tree(X1, Left, X2)) → U1(X, X1, Left, X2, tree_member_in(X, Left))
tree_member_in(X, tree(X, X1, X2)) → tree_member_out(X, tree(X, X1, X2))
U1(X, X1, Left, X2, tree_member_out(X, Left)) → tree_member_out(X, tree(X1, Left, X2))
U2(X, X1, X2, Right, tree_member_out(X, Right)) → tree_member_out(X, tree(X1, X2, Right))

The argument filtering Pi contains the following mapping:
tree_member_in(x1, x2)  =  tree_member_in(x2)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
tree_member_out(x1, x2)  =  tree_member_out(x1)
TREE_MEMBER_IN(x1, x2)  =  TREE_MEMBER_IN(x2)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

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

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

TREE_MEMBER_IN(X, tree(X1, Left, X2)) → TREE_MEMBER_IN(X, Left)
TREE_MEMBER_IN(X, tree(X1, X2, Right)) → TREE_MEMBER_IN(X, Right)

The TRS R consists of the following rules:

tree_member_in(X, tree(X1, X2, Right)) → U2(X, X1, X2, Right, tree_member_in(X, Right))
tree_member_in(X, tree(X1, Left, X2)) → U1(X, X1, Left, X2, tree_member_in(X, Left))
tree_member_in(X, tree(X, X1, X2)) → tree_member_out(X, tree(X, X1, X2))
U1(X, X1, Left, X2, tree_member_out(X, Left)) → tree_member_out(X, tree(X1, Left, X2))
U2(X, X1, X2, Right, tree_member_out(X, Right)) → tree_member_out(X, tree(X1, X2, Right))

The argument filtering Pi contains the following mapping:
tree_member_in(x1, x2)  =  tree_member_in(x2)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
tree_member_out(x1, x2)  =  tree_member_out(x1)
TREE_MEMBER_IN(x1, x2)  =  TREE_MEMBER_IN(x2)

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:

TREE_MEMBER_IN(X, tree(X1, Left, X2)) → TREE_MEMBER_IN(X, Left)
TREE_MEMBER_IN(X, tree(X1, X2, Right)) → TREE_MEMBER_IN(X, Right)

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

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:

TREE_MEMBER_IN(tree(X1, X2, Right)) → TREE_MEMBER_IN(Right)
TREE_MEMBER_IN(tree(X1, Left, X2)) → TREE_MEMBER_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: