Left Termination of the query pattern bin_tree_in_1(g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

bin_tree(void).
bin_tree(tree(X, Left, Right)) :- ','(bin_tree(Left), bin_tree(Right)).

Queries:

bin_tree(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:

bin_tree_in(tree(X, Left, Right)) → U1(X, Left, Right, bin_tree_in(Left))
bin_tree_in(void) → bin_tree_out(void)
U1(X, Left, Right, bin_tree_out(Left)) → U2(X, Left, Right, bin_tree_in(Right))
U2(X, Left, Right, bin_tree_out(Right)) → bin_tree_out(tree(X, Left, Right))

The argument filtering Pi contains the following mapping:
bin_tree_in(x1)  =  bin_tree_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
void  =  void
bin_tree_out(x1)  =  bin_tree_out
U2(x1, x2, x3, x4)  =  U2(x4)

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:

bin_tree_in(tree(X, Left, Right)) → U1(X, Left, Right, bin_tree_in(Left))
bin_tree_in(void) → bin_tree_out(void)
U1(X, Left, Right, bin_tree_out(Left)) → U2(X, Left, Right, bin_tree_in(Right))
U2(X, Left, Right, bin_tree_out(Right)) → bin_tree_out(tree(X, Left, Right))

The argument filtering Pi contains the following mapping:
bin_tree_in(x1)  =  bin_tree_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
void  =  void
bin_tree_out(x1)  =  bin_tree_out
U2(x1, x2, x3, x4)  =  U2(x4)


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

BIN_TREE_IN(tree(X, Left, Right)) → U11(X, Left, Right, bin_tree_in(Left))
BIN_TREE_IN(tree(X, Left, Right)) → BIN_TREE_IN(Left)
U11(X, Left, Right, bin_tree_out(Left)) → U21(X, Left, Right, bin_tree_in(Right))
U11(X, Left, Right, bin_tree_out(Left)) → BIN_TREE_IN(Right)

The TRS R consists of the following rules:

bin_tree_in(tree(X, Left, Right)) → U1(X, Left, Right, bin_tree_in(Left))
bin_tree_in(void) → bin_tree_out(void)
U1(X, Left, Right, bin_tree_out(Left)) → U2(X, Left, Right, bin_tree_in(Right))
U2(X, Left, Right, bin_tree_out(Right)) → bin_tree_out(tree(X, Left, Right))

The argument filtering Pi contains the following mapping:
bin_tree_in(x1)  =  bin_tree_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
void  =  void
bin_tree_out(x1)  =  bin_tree_out
U2(x1, x2, x3, x4)  =  U2(x4)
BIN_TREE_IN(x1)  =  BIN_TREE_IN(x1)
U21(x1, x2, x3, x4)  =  U21(x4)
U11(x1, x2, x3, x4)  =  U11(x3, x4)

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:

BIN_TREE_IN(tree(X, Left, Right)) → U11(X, Left, Right, bin_tree_in(Left))
BIN_TREE_IN(tree(X, Left, Right)) → BIN_TREE_IN(Left)
U11(X, Left, Right, bin_tree_out(Left)) → U21(X, Left, Right, bin_tree_in(Right))
U11(X, Left, Right, bin_tree_out(Left)) → BIN_TREE_IN(Right)

The TRS R consists of the following rules:

bin_tree_in(tree(X, Left, Right)) → U1(X, Left, Right, bin_tree_in(Left))
bin_tree_in(void) → bin_tree_out(void)
U1(X, Left, Right, bin_tree_out(Left)) → U2(X, Left, Right, bin_tree_in(Right))
U2(X, Left, Right, bin_tree_out(Right)) → bin_tree_out(tree(X, Left, Right))

The argument filtering Pi contains the following mapping:
bin_tree_in(x1)  =  bin_tree_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
void  =  void
bin_tree_out(x1)  =  bin_tree_out
U2(x1, x2, x3, x4)  =  U2(x4)
BIN_TREE_IN(x1)  =  BIN_TREE_IN(x1)
U21(x1, x2, x3, x4)  =  U21(x4)
U11(x1, x2, x3, x4)  =  U11(x3, x4)

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
              ↳ PiDPToQDPProof

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

U11(X, Left, Right, bin_tree_out(Left)) → BIN_TREE_IN(Right)
BIN_TREE_IN(tree(X, Left, Right)) → U11(X, Left, Right, bin_tree_in(Left))
BIN_TREE_IN(tree(X, Left, Right)) → BIN_TREE_IN(Left)

The TRS R consists of the following rules:

bin_tree_in(tree(X, Left, Right)) → U1(X, Left, Right, bin_tree_in(Left))
bin_tree_in(void) → bin_tree_out(void)
U1(X, Left, Right, bin_tree_out(Left)) → U2(X, Left, Right, bin_tree_in(Right))
U2(X, Left, Right, bin_tree_out(Right)) → bin_tree_out(tree(X, Left, Right))

The argument filtering Pi contains the following mapping:
bin_tree_in(x1)  =  bin_tree_in(x1)
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1(x1, x2, x3, x4)  =  U1(x3, x4)
void  =  void
bin_tree_out(x1)  =  bin_tree_out
U2(x1, x2, x3, x4)  =  U2(x4)
BIN_TREE_IN(x1)  =  BIN_TREE_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x3, x4)

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
              ↳ PiDPToQDPProof
                ↳ QDP
                  ↳ QDPSizeChangeProof

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

BIN_TREE_IN(tree(X, Left, Right)) → U11(Right, bin_tree_in(Left))
BIN_TREE_IN(tree(X, Left, Right)) → BIN_TREE_IN(Left)
U11(Right, bin_tree_out) → BIN_TREE_IN(Right)

The TRS R consists of the following rules:

bin_tree_in(tree(X, Left, Right)) → U1(Right, bin_tree_in(Left))
bin_tree_in(void) → bin_tree_out
U1(Right, bin_tree_out) → U2(bin_tree_in(Right))
U2(bin_tree_out) → bin_tree_out

The set Q consists of the following terms:

bin_tree_in(x0)
U1(x0, x1)
U2(x0)

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: