Left Termination proof of /tmp/tmp6DZD1k/tree.pl

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]. With regard to the inferred argument filtering the predicates were used in the following modes:
bin_tree_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

bin_tree_in_g(void) → bin_tree_out_g(void)
bin_tree_in_g(tree(X, Left, Right)) → U1_g(X, Left, Right, bin_tree_in_g(Left))
U1_g(X, Left, Right, bin_tree_out_g(Left)) → U2_g(X, Left, Right, bin_tree_in_g(Right))
U2_g(X, Left, Right, bin_tree_out_g(Right)) → bin_tree_out_g(tree(X, Left, Right))

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_g(void) → bin_tree_out_g(void)
bin_tree_in_g(tree(X, Left, Right)) → U1_g(X, Left, Right, bin_tree_in_g(Left))
U1_g(X, Left, Right, bin_tree_out_g(Left)) → U2_g(X, Left, Right, bin_tree_in_g(Right))
U2_g(X, Left, Right, bin_tree_out_g(Right)) → bin_tree_out_g(tree(X, Left, Right))

The argument filtering Pi contains the following mapping:
bin_tree_in_g(x1) = bin_tree_in_g(x1)
void = void
bin_tree_out_g(x1) = bin_tree_out_g
tree(x1, x2, x3) = tree(x1, x2, x3)
U1_g(x1, x2, x3, x4) = U1_g(x3, x4)
U2_g(x1, x2, x3, x4) = U2_g(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_G(tree(X, Left, Right)) → U1_G(X, Left, Right, bin_tree_in_g(Left))
BIN_TREE_IN_G(tree(X, Left, Right)) → BIN_TREE_IN_G(Left)
U1_G(X, Left, Right, bin_tree_out_g(Left)) → U2_G(X, Left, Right, bin_tree_in_g(Right))
U1_G(X, Left, Right, bin_tree_out_g(Left)) → BIN_TREE_IN_G(Right)

The TRS R consists of the following rules:

bin_tree_in_g(void) → bin_tree_out_g(void)
bin_tree_in_g(tree(X, Left, Right)) → U1_g(X, Left, Right, bin_tree_in_g(Left))
U1_g(X, Left, Right, bin_tree_out_g(Left)) → U2_g(X, Left, Right, bin_tree_in_g(Right))
U2_g(X, Left, Right, bin_tree_out_g(Right)) → bin_tree_out_g(tree(X, Left, Right))

The argument filtering Pi contains the following mapping:
bin_tree_in_g(x1) = bin_tree_in_g(x1)
void = void
bin_tree_out_g(x1) = bin_tree_out_g
tree(x1, x2, x3) = tree(x1, x2, x3)
U1_g(x1, x2, x3, x4) = U1_g(x3, x4)
U2_g(x1, x2, x3, x4) = U2_g(x4)
U1_G(x1, x2, x3, x4) = U1_G(x3, x4)
BIN_TREE_IN_G(x1) = BIN_TREE_IN_G(x1)
U2_G(x1, x2, x3, x4) = U2_G(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_G(tree(X, Left, Right)) → U1_G(X, Left, Right, bin_tree_in_g(Left))
BIN_TREE_IN_G(tree(X, Left, Right)) → BIN_TREE_IN_G(Left)
U1_G(X, Left, Right, bin_tree_out_g(Left)) → U2_G(X, Left, Right, bin_tree_in_g(Right))
U1_G(X, Left, Right, bin_tree_out_g(Left)) → BIN_TREE_IN_G(Right)

The TRS R consists of the following rules:

bin_tree_in_g(void) → bin_tree_out_g(void)
bin_tree_in_g(tree(X, Left, Right)) → U1_g(X, Left, Right, bin_tree_in_g(Left))
U1_g(X, Left, Right, bin_tree_out_g(Left)) → U2_g(X, Left, Right, bin_tree_in_g(Right))
U2_g(X, Left, Right, bin_tree_out_g(Right)) → bin_tree_out_g(tree(X, Left, Right))

The argument filtering Pi contains the following mapping:
bin_tree_in_g(x1) = bin_tree_in_g(x1)
void = void
bin_tree_out_g(x1) = bin_tree_out_g
tree(x1, x2, x3) = tree(x1, x2, x3)
U1_g(x1, x2, x3, x4) = U1_g(x3, x4)
U2_g(x1, x2, x3, x4) = U2_g(x4)
U1_G(x1, x2, x3, x4) = U1_G(x3, x4)
BIN_TREE_IN_G(x1) = BIN_TREE_IN_G(x1)
U2_G(x1, x2, x3, x4) = U2_G(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:

U1_G(X, Left, Right, bin_tree_out_g(Left)) → BIN_TREE_IN_G(Right)
BIN_TREE_IN_G(tree(X, Left, Right)) → BIN_TREE_IN_G(Left)
BIN_TREE_IN_G(tree(X, Left, Right)) → U1_G(X, Left, Right, bin_tree_in_g(Left))

The TRS R consists of the following rules:

bin_tree_in_g(void) → bin_tree_out_g(void)
bin_tree_in_g(tree(X, Left, Right)) → U1_g(X, Left, Right, bin_tree_in_g(Left))
U1_g(X, Left, Right, bin_tree_out_g(Left)) → U2_g(X, Left, Right, bin_tree_in_g(Right))
U2_g(X, Left, Right, bin_tree_out_g(Right)) → bin_tree_out_g(tree(X, Left, Right))

The argument filtering Pi contains the following mapping:
bin_tree_in_g(x1) = bin_tree_in_g(x1)
void = void
bin_tree_out_g(x1) = bin_tree_out_g
tree(x1, x2, x3) = tree(x1, x2, x3)
U1_g(x1, x2, x3, x4) = U1_g(x3, x4)
U2_g(x1, x2, x3, x4) = U2_g(x4)
U1_G(x1, x2, x3, x4) = U1_G(x3, x4)
BIN_TREE_IN_G(x1) = BIN_TREE_IN_G(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

              ↳ PiDPToQDPProof

                ↳ QDP

                  ↳ QDPSizeChangeProof

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

U1_G(Right, bin_tree_out_g) → BIN_TREE_IN_G(Right)
BIN_TREE_IN_G(tree(X, Left, Right)) → BIN_TREE_IN_G(Left)
BIN_TREE_IN_G(tree(X, Left, Right)) → U1_G(Right, bin_tree_in_g(Left))

The TRS R consists of the following rules:

bin_tree_in_g(void) → bin_tree_out_g
bin_tree_in_g(tree(X, Left, Right)) → U1_g(Right, bin_tree_in_g(Left))
U1_g(Right, bin_tree_out_g) → U2_g(bin_tree_in_g(Right))
U2_g(bin_tree_out_g) → bin_tree_out_g

The set Q consists of the following terms:

bin_tree_in_g(x0)
U1_g(x0, x1)
U2_g(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:

BIN_TREE_IN_G(tree(X, Left, Right)) → U1_G(Right, bin_tree_in_g(Left))
The graph contains the following edges 1 > 1
BIN_TREE_IN_G(tree(X, Left, Right)) → BIN_TREE_IN_G(Left)
The graph contains the following edges 1 > 1
U1_G(Right, bin_tree_out_g) → BIN_TREE_IN_G(Right)
The graph contains the following edges 1 >= 1