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

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

(0) Obligation:

Clauses:

bin_tree(void) :- !.
bin_tree(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).

Queries:

bin_tree(g).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

bin_tree1(tree(T19, T20, T21)) :- bin_tree1(T20).
bin_tree1(tree(T19, T20, T21)) :- ','(bin_treec1(T20), bin_tree1(T21)).

Clauses:

bin_treec1(void).
bin_treec1(tree(T19, T20, T21)) :- ','(bin_treec1(T20), bin_treec1(T21)).

Afs:

bin_tree1(x1)  =  bin_tree1(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
bin_tree1_in: (b)
bin_treec1_in: (b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

BIN_TREE1_IN_G(tree(T19, T20, T21)) → U1_G(T19, T20, T21, bin_tree1_in_g(T20))
BIN_TREE1_IN_G(tree(T19, T20, T21)) → BIN_TREE1_IN_G(T20)
BIN_TREE1_IN_G(tree(T19, T20, T21)) → U2_G(T19, T20, T21, bin_treec1_in_g(T20))
U2_G(T19, T20, T21, bin_treec1_out_g(T20)) → U3_G(T19, T20, T21, bin_tree1_in_g(T21))
U2_G(T19, T20, T21, bin_treec1_out_g(T20)) → BIN_TREE1_IN_G(T21)

The TRS R consists of the following rules:

bin_treec1_in_g(void) → bin_treec1_out_g(void)
bin_treec1_in_g(tree(T19, T20, T21)) → U5_g(T19, T20, T21, bin_treec1_in_g(T20))
U5_g(T19, T20, T21, bin_treec1_out_g(T20)) → U6_g(T19, T20, T21, bin_treec1_in_g(T21))
U6_g(T19, T20, T21, bin_treec1_out_g(T21)) → bin_treec1_out_g(tree(T19, T20, T21))

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_TREE1_IN_G(tree(T19, T20, T21)) → U1_G(T19, T20, T21, bin_tree1_in_g(T20))
BIN_TREE1_IN_G(tree(T19, T20, T21)) → BIN_TREE1_IN_G(T20)
BIN_TREE1_IN_G(tree(T19, T20, T21)) → U2_G(T19, T20, T21, bin_treec1_in_g(T20))
U2_G(T19, T20, T21, bin_treec1_out_g(T20)) → U3_G(T19, T20, T21, bin_tree1_in_g(T21))
U2_G(T19, T20, T21, bin_treec1_out_g(T20)) → BIN_TREE1_IN_G(T21)

The TRS R consists of the following rules:

bin_treec1_in_g(void) → bin_treec1_out_g(void)
bin_treec1_in_g(tree(T19, T20, T21)) → U5_g(T19, T20, T21, bin_treec1_in_g(T20))
U5_g(T19, T20, T21, bin_treec1_out_g(T20)) → U6_g(T19, T20, T21, bin_treec1_in_g(T21))
U6_g(T19, T20, T21, bin_treec1_out_g(T21)) → bin_treec1_out_g(tree(T19, T20, T21))

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_TREE1_IN_G(tree(T19, T20, T21)) → U2_G(T19, T20, T21, bin_treec1_in_g(T20))
U2_G(T19, T20, T21, bin_treec1_out_g(T20)) → BIN_TREE1_IN_G(T21)
BIN_TREE1_IN_G(tree(T19, T20, T21)) → BIN_TREE1_IN_G(T20)

The TRS R consists of the following rules:

bin_treec1_in_g(void) → bin_treec1_out_g(void)
bin_treec1_in_g(tree(T19, T20, T21)) → U5_g(T19, T20, T21, bin_treec1_in_g(T20))
U5_g(T19, T20, T21, bin_treec1_out_g(T20)) → U6_g(T19, T20, T21, bin_treec1_in_g(T21))
U6_g(T19, T20, T21, bin_treec1_out_g(T21)) → bin_treec1_out_g(tree(T19, T20, T21))

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_TREE1_IN_G(tree(T19, T20, T21)) → U2_G(T19, T20, T21, bin_treec1_in_g(T20))
U2_G(T19, T20, T21, bin_treec1_out_g(T20)) → BIN_TREE1_IN_G(T21)
BIN_TREE1_IN_G(tree(T19, T20, T21)) → BIN_TREE1_IN_G(T20)

The TRS R consists of the following rules:

bin_treec1_in_g(void) → bin_treec1_out_g(void)
bin_treec1_in_g(tree(T19, T20, T21)) → U5_g(T19, T20, T21, bin_treec1_in_g(T20))
U5_g(T19, T20, T21, bin_treec1_out_g(T20)) → U6_g(T19, T20, T21, bin_treec1_in_g(T21))
U6_g(T19, T20, T21, bin_treec1_out_g(T21)) → bin_treec1_out_g(tree(T19, T20, T21))

The set Q consists of the following terms:

bin_treec1_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(T19, T20, T21, bin_treec1_out_g(T20)) → BIN_TREE1_IN_G(T21)
    The graph contains the following edges 3 >= 1

  • BIN_TREE1_IN_G(tree(T19, T20, T21)) → BIN_TREE1_IN_G(T20)
    The graph contains the following edges 1 > 1

  • BIN_TREE1_IN_G(tree(T19, T20, T21)) → U2_G(T19, T20, T21, bin_treec1_in_g(T20))
    The graph contains the following edges 1 > 1, 1 > 2, 1 > 3

(10) YES