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

0 Prolog
↳1 PrologToDTProblemTransformerProof (⇒, 89 ms)
↳2 TRIPLES
↳3 TriplesToPiDPProof (⇒, 0 ms)
↳4 PiDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 PiDP
↳7 PiDPToQDPProof (⇔, 23 ms)
↳8 QDP
↳9 QDPSizeChangeProof (⇔, 0 ms)
↳10 YES

(0) Obligation:

Clauses:

bin_tree(void).
bin_tree(T) :- ','(no(empty(T)), ','(left(T, L), ','(right(T, R), ','(bin_tree(L), bin_tree(R))))).
left(void, void).
left(tree(X1, L, X2), L).
right(void, void).
right(tree(X3, X4, R), R).
empty(void).
no(X) :- ','(X, ','(!, failure(a))).
no(X5).
failure(b).

Query: bin_tree(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

(2) Obligation:

Triples:

bin_treeA(tree(X1, X2, X3)) :- bin_treeA(X2).
bin_treeA(tree(X1, X2, X3)) :- ','(bin_treecA(X2), bin_treeA(X3)).

Clauses:

bin_treecA(void).
bin_treecA(tree(X1, X2, X3)) :- ','(bin_treecA(X2), bin_treecA(X3)).

Afs:

bin_treeA(x1)  =  bin_treeA(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
bin_treeA_in: (b)
bin_treecA_in: (b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

BIN_TREEA_IN_G(tree(X1, X2, X3)) → U1_G(X1, X2, X3, bin_treeA_in_g(X2))
BIN_TREEA_IN_G(tree(X1, X2, X3)) → BIN_TREEA_IN_G(X2)
BIN_TREEA_IN_G(tree(X1, X2, X3)) → U2_G(X1, X2, X3, bin_treecA_in_g(X2))
U2_G(X1, X2, X3, bin_treecA_out_g(X2)) → U3_G(X1, X2, X3, bin_treeA_in_g(X3))
U2_G(X1, X2, X3, bin_treecA_out_g(X2)) → BIN_TREEA_IN_G(X3)

The TRS R consists of the following rules:

bin_treecA_in_g(void) → bin_treecA_out_g(void)
bin_treecA_in_g(tree(X1, X2, X3)) → U5_g(X1, X2, X3, bin_treecA_in_g(X2))
U5_g(X1, X2, X3, bin_treecA_out_g(X2)) → U6_g(X1, X2, X3, bin_treecA_in_g(X3))
U6_g(X1, X2, X3, bin_treecA_out_g(X3)) → bin_treecA_out_g(tree(X1, X2, X3))

Pi is empty.
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:

BIN_TREEA_IN_G(tree(X1, X2, X3)) → U1_G(X1, X2, X3, bin_treeA_in_g(X2))
BIN_TREEA_IN_G(tree(X1, X2, X3)) → BIN_TREEA_IN_G(X2)
BIN_TREEA_IN_G(tree(X1, X2, X3)) → U2_G(X1, X2, X3, bin_treecA_in_g(X2))
U2_G(X1, X2, X3, bin_treecA_out_g(X2)) → U3_G(X1, X2, X3, bin_treeA_in_g(X3))
U2_G(X1, X2, X3, bin_treecA_out_g(X2)) → BIN_TREEA_IN_G(X3)

The TRS R consists of the following rules:

bin_treecA_in_g(void) → bin_treecA_out_g(void)
bin_treecA_in_g(tree(X1, X2, X3)) → U5_g(X1, X2, X3, bin_treecA_in_g(X2))
U5_g(X1, X2, X3, bin_treecA_out_g(X2)) → U6_g(X1, X2, X3, bin_treecA_in_g(X3))
U6_g(X1, X2, X3, bin_treecA_out_g(X3)) → bin_treecA_out_g(tree(X1, X2, X3))

Pi is empty.
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:

BIN_TREEA_IN_G(tree(X1, X2, X3)) → U2_G(X1, X2, X3, bin_treecA_in_g(X2))
U2_G(X1, X2, X3, bin_treecA_out_g(X2)) → BIN_TREEA_IN_G(X3)
BIN_TREEA_IN_G(tree(X1, X2, X3)) → BIN_TREEA_IN_G(X2)

The TRS R consists of the following rules:

bin_treecA_in_g(void) → bin_treecA_out_g(void)
bin_treecA_in_g(tree(X1, X2, X3)) → U5_g(X1, X2, X3, bin_treecA_in_g(X2))
U5_g(X1, X2, X3, bin_treecA_out_g(X2)) → U6_g(X1, X2, X3, bin_treecA_in_g(X3))
U6_g(X1, X2, X3, bin_treecA_out_g(X3)) → bin_treecA_out_g(tree(X1, X2, X3))

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

(7) PiDPToQDPProof (EQUIVALENT 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:

BIN_TREEA_IN_G(tree(X1, X2, X3)) → U2_G(X1, X2, X3, bin_treecA_in_g(X2))
U2_G(X1, X2, X3, bin_treecA_out_g(X2)) → BIN_TREEA_IN_G(X3)
BIN_TREEA_IN_G(tree(X1, X2, X3)) → BIN_TREEA_IN_G(X2)

The TRS R consists of the following rules:

bin_treecA_in_g(void) → bin_treecA_out_g(void)
bin_treecA_in_g(tree(X1, X2, X3)) → U5_g(X1, X2, X3, bin_treecA_in_g(X2))
U5_g(X1, X2, X3, bin_treecA_out_g(X2)) → U6_g(X1, X2, X3, bin_treecA_in_g(X3))
U6_g(X1, X2, X3, bin_treecA_out_g(X3)) → bin_treecA_out_g(tree(X1, X2, X3))

The set Q consists of the following terms:

bin_treecA_in_g(x0)
U5_g(x0, x1, x2, x3)
U6_g(x0, x1, x2, x3)

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:

  • U2_G(X1, X2, X3, bin_treecA_out_g(X2)) → BIN_TREEA_IN_G(X3)
    The graph contains the following edges 3 >= 1

  • BIN_TREEA_IN_G(tree(X1, X2, X3)) → BIN_TREEA_IN_G(X2)
    The graph contains the following edges 1 > 1

  • BIN_TREEA_IN_G(tree(X1, X2, X3)) → U2_G(X1, X2, X3, bin_treecA_in_g(X2))
    The graph contains the following edges 1 > 1, 1 > 2, 1 > 3

(10) YES