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(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).

Query: bin_tree(g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: 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: !_1 | 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: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(left(T3, X7), ','(right(T3, X8), ','(bin_tree(X7), bin_tree(X8))))\n\nT3 is ground\nX7 is free\nX8 is free\nbin_tree(T3) !=? bin_tree(void)", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(left(T3, X7), ','(right(T3, X8), ','(bin_tree(X7), bin_tree(X8)))), SCOPE: 2, CLAUSE: 2 | ','(left(T3, X7), ','(right(T3, X8), ','(bin_tree(X7), bin_tree(X8)))), SCOPE: 2, CLAUSE: 3\n\nT3 is ground\nX7 is free\nX8 is free\nbin_tree(T3) !=? bin_tree(void)", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(left(T3, X7), ','(right(T3, X8), ','(bin_tree(X7), bin_tree(X8)))), SCOPE: 2, CLAUSE: 3\n\nT3 is ground\nX7 is free\nX8 is free\nbin_tree(T3) !=? bin_tree(void)", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(right(tree(T10, T11, T12), X8), ','(bin_tree(T11), bin_tree(X8)))\n\nT10 is ground\nT11 is ground\nT12 is ground\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(right(tree(T10, T11, T12), X8), ','(bin_tree(T11), bin_tree(X8))), SCOPE: 3, CLAUSE: 4 | ','(right(tree(T10, T11, T12), X8), ','(bin_tree(T11), bin_tree(X8))), SCOPE: 3, CLAUSE: 5\n\nT10 is ground\nT11 is ground\nT12 is ground\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(right(tree(T10, T11, T12), X8), ','(bin_tree(T11), bin_tree(X8))), SCOPE: 3, CLAUSE: 5\n\nT10 is ground\nT11 is ground\nT12 is ground\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(bin_tree(T20), bin_tree(T21))\n\nT20 is ground\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: bin_tree(T20)\n\nT20 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: bin_tree(T21)\n\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 17 [arrowhead = none ];
17 -> 2;

18 [label="EVAL with clause\nbin_tree(void) :- !_1.\nand substitutionT1 -> void", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 18 [arrowhead = none ];
18 -> 3;

19 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 19 [arrowhead = none ];
19 -> 4;

20 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 20 [arrowhead = none ];
20 -> 5;

21 [label="ONLY EVAL with clause\nbin_tree(X6) :- ','(left(X6, X7), ','(right(X6, X8), ','(bin_tree(X7), bin_tree(X8)))).\nand substitutionT1 -> T3,\nX6 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 21 [arrowhead = none ];
21 -> 7;

22 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 22 [arrowhead = none ];
22 -> 6;

23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 23 [arrowhead = none ];
23 -> 8;

24 [label="BACKTRACK\nfor clause: left(void, void)\nwith clash: (bin_tree(T3), bin_tree(void))", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 24 [arrowhead = none ];
24 -> 9;

25 [label="EVAL with clause\nleft(tree(X15, X16, X17), X16).\nand substitutionX15 -> T10,\nX16 -> T11,\nX17 -> T12,\nT3 -> tree(T10, T11, T12),\nX7 -> T11", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 25 [arrowhead = none ];
25 -> 10;

26 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 26 [arrowhead = none ];
26 -> 11;

27 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 27 [arrowhead = none ];
27 -> 12;

28 [label="BACKTRACK\nfor clause: right(void, void)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 28 [arrowhead = none ];
28 -> 13;

29 [label="ONLY EVAL with clause\nright(tree(X24, X25, X26), X26).\nand substitutionT10 -> T19,\nX24 -> T19,\nT11 -> T20,\nX25 -> T20,\nT12 -> T21,\nX26 -> T21,\nX8 -> T21", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 29 [arrowhead = none ];
29 -> 14;

30 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
14 -> 30 [arrowhead = none ];
30 -> 15;

31 [label="SPLIT 2\nnew knowledge:\nT20 is ground", fontsize=14, style = filled, fillcolor = violet];
14 -> 31 [arrowhead = none ];
31 -> 16;

32 [label="INSTANCE with matching:\nT1 -> T20", fontsize=14, style = filled, fillcolor = lightgrey];
15 -> 32 [arrowhead = none , style=dashed];
32 -> 1[style=dashed];

33 [label="INSTANCE with matching:\nT1 -> T21", fontsize=14, style = filled, fillcolor = lightgrey];
16 -> 33 [arrowhead = none , style=dashed];
33 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

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

Query: bin_treeA(g)

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
bin_treeA_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

bin_treeA_in_g(void) → bin_treeA_out_g(void)
bin_treeA_in_g(tree(T19, T20, T21)) → U1_g(T19, T20, T21, bin_treeA_in_g(T20))
U1_g(T19, T20, T21, bin_treeA_out_g(T20)) → bin_treeA_out_g(tree(T19, T20, T21))
U1_g(T19, T20, T21, bin_treeA_out_g(T20)) → U2_g(T19, T20, T21, bin_treeA_in_g(T21))
U2_g(T19, T20, T21, bin_treeA_out_g(T21)) → bin_treeA_out_g(tree(T19, T20, T21))

The argument filtering Pi contains the following mapping:
bin_treeA_in_g(x1)  =  bin_treeA_in_g(x1)
void  =  void
bin_treeA_out_g(x1)  =  bin_treeA_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)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

bin_treeA_in_g(void) → bin_treeA_out_g(void)
bin_treeA_in_g(tree(T19, T20, T21)) → U1_g(T19, T20, T21, bin_treeA_in_g(T20))
U1_g(T19, T20, T21, bin_treeA_out_g(T20)) → bin_treeA_out_g(tree(T19, T20, T21))
U1_g(T19, T20, T21, bin_treeA_out_g(T20)) → U2_g(T19, T20, T21, bin_treeA_in_g(T21))
U2_g(T19, T20, T21, bin_treeA_out_g(T21)) → bin_treeA_out_g(tree(T19, T20, T21))

The argument filtering Pi contains the following mapping:
bin_treeA_in_g(x1)  =  bin_treeA_in_g(x1)
void  =  void
bin_treeA_out_g(x1)  =  bin_treeA_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)

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

BIN_TREEA_IN_G(tree(T19, T20, T21)) → U1_G(T19, T20, T21, bin_treeA_in_g(T20))
BIN_TREEA_IN_G(tree(T19, T20, T21)) → BIN_TREEA_IN_G(T20)
U1_G(T19, T20, T21, bin_treeA_out_g(T20)) → U2_G(T19, T20, T21, bin_treeA_in_g(T21))
U1_G(T19, T20, T21, bin_treeA_out_g(T20)) → BIN_TREEA_IN_G(T21)

The TRS R consists of the following rules:

bin_treeA_in_g(void) → bin_treeA_out_g(void)
bin_treeA_in_g(tree(T19, T20, T21)) → U1_g(T19, T20, T21, bin_treeA_in_g(T20))
U1_g(T19, T20, T21, bin_treeA_out_g(T20)) → bin_treeA_out_g(tree(T19, T20, T21))
U1_g(T19, T20, T21, bin_treeA_out_g(T20)) → U2_g(T19, T20, T21, bin_treeA_in_g(T21))
U2_g(T19, T20, T21, bin_treeA_out_g(T21)) → bin_treeA_out_g(tree(T19, T20, T21))

The argument filtering Pi contains the following mapping:
bin_treeA_in_g(x1)  =  bin_treeA_in_g(x1)
void  =  void
bin_treeA_out_g(x1)  =  bin_treeA_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)
BIN_TREEA_IN_G(x1)  =  BIN_TREEA_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x3, x4)
U2_G(x1, x2, x3, x4)  =  U2_G(x4)

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

(6) Obligation:

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

BIN_TREEA_IN_G(tree(T19, T20, T21)) → U1_G(T19, T20, T21, bin_treeA_in_g(T20))
BIN_TREEA_IN_G(tree(T19, T20, T21)) → BIN_TREEA_IN_G(T20)
U1_G(T19, T20, T21, bin_treeA_out_g(T20)) → U2_G(T19, T20, T21, bin_treeA_in_g(T21))
U1_G(T19, T20, T21, bin_treeA_out_g(T20)) → BIN_TREEA_IN_G(T21)

The TRS R consists of the following rules:

bin_treeA_in_g(void) → bin_treeA_out_g(void)
bin_treeA_in_g(tree(T19, T20, T21)) → U1_g(T19, T20, T21, bin_treeA_in_g(T20))
U1_g(T19, T20, T21, bin_treeA_out_g(T20)) → bin_treeA_out_g(tree(T19, T20, T21))
U1_g(T19, T20, T21, bin_treeA_out_g(T20)) → U2_g(T19, T20, T21, bin_treeA_in_g(T21))
U2_g(T19, T20, T21, bin_treeA_out_g(T21)) → bin_treeA_out_g(tree(T19, T20, T21))

The argument filtering Pi contains the following mapping:
bin_treeA_in_g(x1)  =  bin_treeA_in_g(x1)
void  =  void
bin_treeA_out_g(x1)  =  bin_treeA_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)
BIN_TREEA_IN_G(x1)  =  BIN_TREEA_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x3, x4)
U2_G(x1, x2, x3, x4)  =  U2_G(x4)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(8) Obligation:

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

U1_G(T19, T20, T21, bin_treeA_out_g(T20)) → BIN_TREEA_IN_G(T21)
BIN_TREEA_IN_G(tree(T19, T20, T21)) → U1_G(T19, T20, T21, bin_treeA_in_g(T20))
BIN_TREEA_IN_G(tree(T19, T20, T21)) → BIN_TREEA_IN_G(T20)

The TRS R consists of the following rules:

bin_treeA_in_g(void) → bin_treeA_out_g(void)
bin_treeA_in_g(tree(T19, T20, T21)) → U1_g(T19, T20, T21, bin_treeA_in_g(T20))
U1_g(T19, T20, T21, bin_treeA_out_g(T20)) → bin_treeA_out_g(tree(T19, T20, T21))
U1_g(T19, T20, T21, bin_treeA_out_g(T20)) → U2_g(T19, T20, T21, bin_treeA_in_g(T21))
U2_g(T19, T20, T21, bin_treeA_out_g(T21)) → bin_treeA_out_g(tree(T19, T20, T21))

The argument filtering Pi contains the following mapping:
bin_treeA_in_g(x1)  =  bin_treeA_in_g(x1)
void  =  void
bin_treeA_out_g(x1)  =  bin_treeA_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)
BIN_TREEA_IN_G(x1)  =  BIN_TREEA_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x3, x4)

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

(9) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(10) Obligation:

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

U1_G(T21, bin_treeA_out_g) → BIN_TREEA_IN_G(T21)
BIN_TREEA_IN_G(tree(T19, T20, T21)) → U1_G(T21, bin_treeA_in_g(T20))
BIN_TREEA_IN_G(tree(T19, T20, T21)) → BIN_TREEA_IN_G(T20)

The TRS R consists of the following rules:

bin_treeA_in_g(void) → bin_treeA_out_g
bin_treeA_in_g(tree(T19, T20, T21)) → U1_g(T21, bin_treeA_in_g(T20))
U1_g(T21, bin_treeA_out_g) → bin_treeA_out_g
U1_g(T21, bin_treeA_out_g) → U2_g(bin_treeA_in_g(T21))
U2_g(bin_treeA_out_g) → bin_treeA_out_g

The set Q consists of the following terms:

bin_treeA_in_g(x0)
U1_g(x0, x1)
U2_g(x0)

We have to consider all (P,Q,R)-chains.

(11) 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:

• BIN_TREEA_IN_G(tree(T19, T20, T21)) → U1_G(T21, bin_treeA_in_g(T20))
  The graph contains the following edges 1 > 1

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

• U1_G(T21, bin_treeA_out_g) → BIN_TREEA_IN_G(T21)
  The graph contains the following edges 1 >= 1

(12) YES