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

(0) Obligation:

Clauses:

bin_tree(void).
bin_tree(tree(X1, Left, Right)) :- ','(bin_tree(Left), bin_tree(Right)).

Query: bin_tree(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: bin_tree(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: bin_tree(T1), SCOPE: 1, CLAUSE: 0 | bin_tree(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | bin_tree(void), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: bin_tree(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nbin_tree(T1) !=? bin_tree(void)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: bin_tree(void), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="B: ','(bin_tree(T6), bin_tree(T7))\n\nT6 is ground\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(bin_tree(T6), bin_tree(T7)), SCOPE: 2, CLAUSE: 0 | ','(bin_tree(T6), bin_tree(T7)), SCOPE: 2, CLAUSE: 1\n\nT6 is ground\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(bin_tree(T6), bin_tree(T7)), SCOPE: 2, CLAUSE: 0\n\nT6 is ground\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(bin_tree(T6), bin_tree(T7)), SCOPE: 2, CLAUSE: 1\n\nT6 is ground\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: bin_tree(T7)\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(','(bin_tree(T15), bin_tree(T16)), bin_tree(T7))\n\nT7 is ground\nT15 is ground\nT16 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: bin_tree(T15)\n\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(bin_tree(T16), bin_tree(T7))\n\nT7 is ground\nT16 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 18 [arrowhead = none ];
18 -> 2;

19 [label="EVAL with clause\nbin_tree(void).\nand substitutionT1 -> void", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 19 [arrowhead = none ];
19 -> 3;

20 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 20 [arrowhead = none ];
20 -> 4;

21 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 21 [arrowhead = none ];
21 -> 5;

22 [label="EVAL with clause\nbin_tree(tree(X8, X9, X10)) :- ','(bin_tree(X9), bin_tree(X10)).\nand substitutionX8 -> T5,\nX9 -> T6,\nX10 -> T7,\nT1 -> tree(T5, T6, T7)", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 22 [arrowhead = none ];
22 -> 7;

23 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 23 [arrowhead = none ];
23 -> 8;

24 [label="BACKTRACK\nfor clause: bin_tree(tree(X1, Left, Right)) :- ','(bin_tree(Left), bin_tree(Right))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 24 [arrowhead = none ];
24 -> 6;

25 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 25 [arrowhead = none ];
25 -> 9;

26 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 26 [arrowhead = none ];
26 -> 10;

27 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 27 [arrowhead = none ];
27 -> 11;

28 [label="EVAL with clause\nbin_tree(void).\nand substitutionT6 -> void", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 28 [arrowhead = none ];
28 -> 12;

29 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 29 [arrowhead = none ];
29 -> 13;

30 [label="EVAL with clause\nbin_tree(tree(X17, X18, X19)) :- ','(bin_tree(X18), bin_tree(X19)).\nand substitutionX17 -> T14,\nX18 -> T15,\nX19 -> T16,\nT6 -> tree(T14, T15, T16)", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 30 [arrowhead = none ];
30 -> 14;

31 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 31 [arrowhead = none ];
31 -> 15;

32 [label="INSTANCE with matching:\nT1 -> T7", fontsize=14, style = filled, fillcolor = lightgrey];
12 -> 32 [arrowhead = none , style=dashed];
32 -> 1[style=dashed];

33 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
14 -> 33 [arrowhead = none ];
33 -> 16;

34 [label="SPLIT 2\nnew knowledge:\nT15 is ground", fontsize=14, style = filled, fillcolor = violet];
14 -> 34 [arrowhead = none ];
34 -> 17;

35 [label="INSTANCE with matching:\nT1 -> T15", fontsize=14, style = filled, fillcolor = lightgrey];
16 -> 35 [arrowhead = none , style=dashed];
35 -> 1[style=dashed];

36 [label="INSTANCE with matching:\nT6 -> T16", fontsize=14, style = filled, fillcolor = lightgrey];
17 -> 36 [arrowhead = none , style=dashed];
36 -> 7[style=dashed];

}

(2) Obligation:

Triples:

pB(void, X1) :- bin_treeA(X1).
pB(tree(X1, X2, X3), X4) :- bin_treeA(X2).
pB(tree(X1, X2, X3), X4) :- ','(bin_treecA(X2), pB(X3, X4)).
bin_treeA(tree(X1, X2, X3)) :- pB(X2, X3).

Clauses:

bin_treecA(void).
bin_treecA(tree(X1, X2, X3)) :- qcB(X2, X3).
qcB(void, X1) :- bin_treecA(X1).
qcB(tree(X1, X2, X3), X4) :- ','(bin_treecA(X2), qcB(X3, X4)).

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)
pB_in: (b,b)
bin_treecA_in: (b)
qcB_in: (b,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)) → U5_G(X1, X2, X3, pB_in_gg(X2, X3))
BIN_TREEA_IN_G(tree(X1, X2, X3)) → PB_IN_GG(X2, X3)
PB_IN_GG(void, X1) → U1_GG(X1, bin_treeA_in_g(X1))
PB_IN_GG(void, X1) → BIN_TREEA_IN_G(X1)
PB_IN_GG(tree(X1, X2, X3), X4) → U2_GG(X1, X2, X3, X4, bin_treeA_in_g(X2))
PB_IN_GG(tree(X1, X2, X3), X4) → BIN_TREEA_IN_G(X2)
PB_IN_GG(tree(X1, X2, X3), X4) → U3_GG(X1, X2, X3, X4, bin_treecA_in_g(X2))
U3_GG(X1, X2, X3, X4, bin_treecA_out_g(X2)) → U4_GG(X1, X2, X3, X4, pB_in_gg(X3, X4))
U3_GG(X1, X2, X3, X4, bin_treecA_out_g(X2)) → PB_IN_GG(X3, X4)

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)) → U7_g(X1, X2, X3, qcB_in_gg(X2, X3))
qcB_in_gg(void, X1) → U8_gg(X1, bin_treecA_in_g(X1))
U8_gg(X1, bin_treecA_out_g(X1)) → qcB_out_gg(void, X1)
qcB_in_gg(tree(X1, X2, X3), X4) → U9_gg(X1, X2, X3, X4, bin_treecA_in_g(X2))
U9_gg(X1, X2, X3, X4, bin_treecA_out_g(X2)) → U10_gg(X1, X2, X3, X4, qcB_in_gg(X3, X4))
U10_gg(X1, X2, X3, X4, qcB_out_gg(X3, X4)) → qcB_out_gg(tree(X1, X2, X3), X4)
U7_g(X1, X2, X3, qcB_out_gg(X2, 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)) → U5_G(X1, X2, X3, pB_in_gg(X2, X3))
BIN_TREEA_IN_G(tree(X1, X2, X3)) → PB_IN_GG(X2, X3)
PB_IN_GG(void, X1) → U1_GG(X1, bin_treeA_in_g(X1))
PB_IN_GG(void, X1) → BIN_TREEA_IN_G(X1)
PB_IN_GG(tree(X1, X2, X3), X4) → U2_GG(X1, X2, X3, X4, bin_treeA_in_g(X2))
PB_IN_GG(tree(X1, X2, X3), X4) → BIN_TREEA_IN_G(X2)
PB_IN_GG(tree(X1, X2, X3), X4) → U3_GG(X1, X2, X3, X4, bin_treecA_in_g(X2))
U3_GG(X1, X2, X3, X4, bin_treecA_out_g(X2)) → U4_GG(X1, X2, X3, X4, pB_in_gg(X3, X4))
U3_GG(X1, X2, X3, X4, bin_treecA_out_g(X2)) → PB_IN_GG(X3, X4)

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)) → U7_g(X1, X2, X3, qcB_in_gg(X2, X3))
qcB_in_gg(void, X1) → U8_gg(X1, bin_treecA_in_g(X1))
U8_gg(X1, bin_treecA_out_g(X1)) → qcB_out_gg(void, X1)
qcB_in_gg(tree(X1, X2, X3), X4) → U9_gg(X1, X2, X3, X4, bin_treecA_in_g(X2))
U9_gg(X1, X2, X3, X4, bin_treecA_out_g(X2)) → U10_gg(X1, X2, X3, X4, qcB_in_gg(X3, X4))
U10_gg(X1, X2, X3, X4, qcB_out_gg(X3, X4)) → qcB_out_gg(tree(X1, X2, X3), X4)
U7_g(X1, X2, X3, qcB_out_gg(X2, 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 4 less nodes.

(6) Obligation:

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

BIN_TREEA_IN_G(tree(X1, X2, X3)) → PB_IN_GG(X2, X3)
PB_IN_GG(void, X1) → BIN_TREEA_IN_G(X1)
PB_IN_GG(tree(X1, X2, X3), X4) → BIN_TREEA_IN_G(X2)
PB_IN_GG(tree(X1, X2, X3), X4) → U3_GG(X1, X2, X3, X4, bin_treecA_in_g(X2))
U3_GG(X1, X2, X3, X4, bin_treecA_out_g(X2)) → PB_IN_GG(X3, X4)

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)) → U7_g(X1, X2, X3, qcB_in_gg(X2, X3))
qcB_in_gg(void, X1) → U8_gg(X1, bin_treecA_in_g(X1))
U8_gg(X1, bin_treecA_out_g(X1)) → qcB_out_gg(void, X1)
qcB_in_gg(tree(X1, X2, X3), X4) → U9_gg(X1, X2, X3, X4, bin_treecA_in_g(X2))
U9_gg(X1, X2, X3, X4, bin_treecA_out_g(X2)) → U10_gg(X1, X2, X3, X4, qcB_in_gg(X3, X4))
U10_gg(X1, X2, X3, X4, qcB_out_gg(X3, X4)) → qcB_out_gg(tree(X1, X2, X3), X4)
U7_g(X1, X2, X3, qcB_out_gg(X2, 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)) → PB_IN_GG(X2, X3)
PB_IN_GG(void, X1) → BIN_TREEA_IN_G(X1)
PB_IN_GG(tree(X1, X2, X3), X4) → BIN_TREEA_IN_G(X2)
PB_IN_GG(tree(X1, X2, X3), X4) → U3_GG(X1, X2, X3, X4, bin_treecA_in_g(X2))
U3_GG(X1, X2, X3, X4, bin_treecA_out_g(X2)) → PB_IN_GG(X3, X4)

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)) → U7_g(X1, X2, X3, qcB_in_gg(X2, X3))
qcB_in_gg(void, X1) → U8_gg(X1, bin_treecA_in_g(X1))
U8_gg(X1, bin_treecA_out_g(X1)) → qcB_out_gg(void, X1)
qcB_in_gg(tree(X1, X2, X3), X4) → U9_gg(X1, X2, X3, X4, bin_treecA_in_g(X2))
U9_gg(X1, X2, X3, X4, bin_treecA_out_g(X2)) → U10_gg(X1, X2, X3, X4, qcB_in_gg(X3, X4))
U10_gg(X1, X2, X3, X4, qcB_out_gg(X3, X4)) → qcB_out_gg(tree(X1, X2, X3), X4)
U7_g(X1, X2, X3, qcB_out_gg(X2, X3)) → bin_treecA_out_g(tree(X1, X2, X3))

The set Q consists of the following terms:

bin_treecA_in_g(x0)
qcB_in_gg(x0, x1)
U8_gg(x0, x1)
U9_gg(x0, x1, x2, x3, x4)
U10_gg(x0, x1, x2, x3, x4)
U7_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:

• PB_IN_GG(tree(X1, X2, X3), X4) → U3_GG(X1, X2, X3, X4, bin_treecA_in_g(X2))
  The graph contains the following edges 1 > 1, 1 > 2, 1 > 3, 2 >= 4

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

• U3_GG(X1, X2, X3, X4, bin_treecA_out_g(X2)) → PB_IN_GG(X3, X4)
  The graph contains the following edges 3 >= 1, 4 >= 2

• PB_IN_GG(void, X1) → BIN_TREEA_IN_G(X1)
  The graph contains the following edges 2 >= 1

• PB_IN_GG(tree(X1, X2, X3), X4) → BIN_TREEA_IN_G(X2)
  The graph contains the following edges 1 > 1

(10) YES