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

(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).

Queries:

bin_tree(g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

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: bin_tree(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: bin_tree(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", 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: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: ','(no(empty(T2)), ','(left(T2, X8), ','(right(T2, X9), ','(bin_tree(X8), bin_tree(X9)))))\n\nT2 is ground\nX8 is free; \nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(no(empty(T2)), ','(left(T2, X8), ','(right(T2, X9), ','(bin_tree(X8), bin_tree(X9))))), SCOPE: 2, CLAUSE: 7 | ','(no(empty(T2)), ','(left(T2, X8), ','(right(T2, X9), ','(bin_tree(X8), bin_tree(X9))))), SCOPE: 2, CLAUSE: 8 | ?_2\n\nT2 is ground\nX8 is free; \nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(empty(T3), ','(!_2, ','(failure(a), ','(left(T3, X8), ','(right(T3, X9), ','(bin_tree(X8), bin_tree(X9))))))) | ','(no(empty(T3)), ','(left(T3, X8), ','(right(T3, X9), ','(bin_tree(X8), bin_tree(X9))))), SCOPE: 2, CLAUSE: 8 | ?_2\n\nT3 is ground\nX8 is free; \nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(empty(T3), ','(!_2, ','(failure(a), ','(left(T3, X8), ','(right(T3, X9), ','(bin_tree(X8), bin_tree(X9))))))), SCOPE: 3, CLAUSE: 6 | ','(no(empty(T3)), ','(left(T3, X8), ','(right(T3, X9), ','(bin_tree(X8), bin_tree(X9))))), SCOPE: 2, CLAUSE: 8 | ?_2\n\nT3 is ground\nX8 is free; \nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(!_2, ','(failure(a), ','(left(void, X8), ','(right(void, X9), ','(bin_tree(X8), bin_tree(X9)))))) | ','(no(empty(void)), ','(left(void, X8), ','(right(void, X9), ','(bin_tree(X8), bin_tree(X9))))), SCOPE: 2, CLAUSE: 8 | ?_2\n\nX8 is free\nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(no(empty(T3)), ','(left(T3, X8), ','(right(T3, X9), ','(bin_tree(X8), bin_tree(X9))))), SCOPE: 2, CLAUSE: 8\n\nT3 is ground\nX8 is free; \nX9 is free; \nempty(T3) !=? empty(void)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(failure(a), ','(left(void, X8), ','(right(void, X9), ','(bin_tree(X8), bin_tree(X9)))))\n\nX8 is free\nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(failure(a), ','(left(void, X8), ','(right(void, X9), ','(bin_tree(X8), bin_tree(X9))))), SCOPE: 4, CLAUSE: 9\n\nX8 is free\nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(left(T4, X8), ','(right(T4, X9), ','(bin_tree(X8), bin_tree(X9))))\n\nT4 is ground\nX8 is free; \nX9 is free; \nempty(T4) !=? empty(void)", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(left(T4, X8), ','(right(T4, X9), ','(bin_tree(X8), bin_tree(X9)))), SCOPE: 5, CLAUSE: 2 | ','(left(T4, X8), ','(right(T4, X9), ','(bin_tree(X8), bin_tree(X9)))), SCOPE: 5, CLAUSE: 3\n\nT4 is ground\nX8 is free; \nX9 is free; \nempty(T4) !=? empty(void)", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(left(T4, X8), ','(right(T4, X9), ','(bin_tree(X8), bin_tree(X9)))), SCOPE: 5, CLAUSE: 3\n\nT4 is ground\nX8 is free; \nX9 is free; \nempty(T4) !=? empty(void)\nleft(T4, X8) !=? left(void, void)", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(right(tree(T5, T6, T7), X9), ','(bin_tree(T6), bin_tree(X9)))\n\nT5 is ground\nT6 is ground\nT7 is ground\nX8 is free; \nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\nX8 is free\nX17 is free; \nX18 is free; \nX19 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(right(tree(T5, T6, T7), X9), ','(bin_tree(T6), bin_tree(X9))), SCOPE: 6, CLAUSE: 4 | ','(right(tree(T5, T6, T7), X9), ','(bin_tree(T6), bin_tree(X9))), SCOPE: 6, CLAUSE: 5\n\nT5 is ground\nT6 is ground\nT7 is ground\nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: ','(right(tree(T5, T6, T7), X9), ','(bin_tree(T6), bin_tree(X9))), SCOPE: 6, CLAUSE: 5\n\nT5 is ground\nT6 is ground\nT7 is ground\nX9 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(bin_tree(T9), bin_tree(T10))\n\nT9 is ground\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: bin_tree(T9)\n\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: bin_tree(T10)\n\nT10 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 27 [arrowhead = none ];
27 -> 2;

28 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 28 [arrowhead = none ];
28 -> 3;

29 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
2 -> 29 [arrowhead = none ];
29 -> 4;

30 [label="EVAL with clause\nbin_tree(void).\nand substitution\nT1 -> void", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 30 [arrowhead = none ];
30 -> 5;

31 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
3 -> 31 [arrowhead = none ];
31 -> 6;

32 [label="EVAL with clause\nbin_tree(X7) :- ','(no(empty(X7)), ','(left(X7, X8), ','(right(X7, X9), ','(bin_tree(X8), bin_tree(X9))))).\nand substitution\nT1 -> T2\nX7 -> T2", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 32 [arrowhead = none ];
32 -> 8;

33 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 33 [arrowhead = none ];
33 -> 7;

34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 34 [arrowhead = none ];
34 -> 9;

35 [label="EVAL with clause\nno(X11) :- ','(X11, ','(!, failure(a))).\nand substitution\nT2 -> T3\nX11 -> empty(T3)", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 35 [arrowhead = none ];
35 -> 10;

36 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
10 -> 36 [arrowhead = none ];
36 -> 11;

37 [label="EVAL with clause\nempty(void).\nand substitution\nT3 -> void", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 37 [arrowhead = none ];
37 -> 12;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 38 [arrowhead = none ];
38 -> 13;

39 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 39 [arrowhead = none ];
39 -> 14;

40 [label="EVAL with clause\nno(X13).\nand substitution\nT3 -> T4\nX13 -> empty(T4)", fontsize=14, style = filled, fillcolor = lightblue];
13 -> 40 [arrowhead = none ];
40 -> 17;

41 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
14 -> 41 [arrowhead = none ];
41 -> 15;

42 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = white];
15 -> 42 [arrowhead = none ];
42 -> 16;

43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
17 -> 43 [arrowhead = none ];
43 -> 18;

44 [label="BACKTRACK\nfor clause: left(void, void)\nwith clash: (empty(T4), empty(void))", fontsize=14, style = filled, fillcolor = white];
18 -> 44 [arrowhead = none ];
44 -> 19;

45 [label="EVAL with clause\nleft(tree(X17, X18, X19), X18).\nand substitution\nX17 -> T5\nX18 -> T6\nX19 -> T7\nT4 -> tree(T5, T6, T7)\nX8 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 45 [arrowhead = none ];
45 -> 20;

46 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 46 [arrowhead = none ];
46 -> 21;

47 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 47 [arrowhead = none ];
47 -> 22;

48 [label="BACKTRACK\nfor clause: right(void, void)because of non-unification", fontsize=14, style = filled, fillcolor = white];
22 -> 48 [arrowhead = none ];
48 -> 23;

49 [label="EVAL with clause\nright(tree(X23, X24, X25), X25).\nand substitution\nT5 -> T8\nX23 -> T8\nT6 -> T9\nX24 -> T9\nT7 -> T10\nX25 -> T10\nX9 -> T10", fontsize=14, style = filled, fillcolor = lightblue];
23 -> 49 [arrowhead = none ];
49 -> 24;

50 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
24 -> 50 [arrowhead = none ];
50 -> 25;

51 [label="SPLIT 2", fontsize=14, style = filled, fillcolor = violet];
24 -> 51 [arrowhead = none ];
51 -> 26;

52 [label="INSTANCE with matching:\nT1 -> T9", fontsize=14, style = filled, fillcolor = lightgrey];
25 -> 52 [arrowhead = none , style=dashed];
52 -> 1[style=dashed];

53 [label="INSTANCE with matching:\nT1 -> T10", fontsize=14, style = filled, fillcolor = lightgrey];
26 -> 53 [arrowhead = none , style=dashed];
53 -> 1[style=dashed];

}

(2) Obligation:

Clauses:

bin_tree1(void).
bin_tree1(tree(T8, T9, T10)) :- bin_tree1(T9).
bin_tree1(tree(T8, T9, T10)) :- ','(bin_tree1(T9), bin_tree1(T10)).

Queries:

bin_tree1(g).

(3) PrologToPiTRSProof (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)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

bin_tree1_in_g(void) → bin_tree1_out_g(void)
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T8, T9, T10, bin_tree1_in_g(T9))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → bin_tree1_out_g(tree(T8, T9, T10))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → U2_g(T8, T9, T10, bin_tree1_in_g(T10))
U2_g(T8, T9, T10, bin_tree1_out_g(T10)) → bin_tree1_out_g(tree(T8, T9, T10))

The argument filtering Pi contains the following mapping:
bin_tree1_in_g(x1)  =  bin_tree1_in_g(x1)
void  =  void
bin_tree1_out_g(x1)  =  bin_tree1_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_tree1_in_g(void) → bin_tree1_out_g(void)
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T8, T9, T10, bin_tree1_in_g(T9))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → bin_tree1_out_g(tree(T8, T9, T10))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → U2_g(T8, T9, T10, bin_tree1_in_g(T10))
U2_g(T8, T9, T10, bin_tree1_out_g(T10)) → bin_tree1_out_g(tree(T8, T9, T10))

The argument filtering Pi contains the following mapping:
bin_tree1_in_g(x1)  =  bin_tree1_in_g(x1)
void  =  void
bin_tree1_out_g(x1)  =  bin_tree1_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_TREE1_IN_G(tree(T8, T9, T10)) → U1_G(T8, T9, T10, bin_tree1_in_g(T9))
BIN_TREE1_IN_G(tree(T8, T9, T10)) → BIN_TREE1_IN_G(T9)
U1_G(T8, T9, T10, bin_tree1_out_g(T9)) → U2_G(T8, T9, T10, bin_tree1_in_g(T10))
U1_G(T8, T9, T10, bin_tree1_out_g(T9)) → BIN_TREE1_IN_G(T10)

The TRS R consists of the following rules:

bin_tree1_in_g(void) → bin_tree1_out_g(void)
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T8, T9, T10, bin_tree1_in_g(T9))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → bin_tree1_out_g(tree(T8, T9, T10))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → U2_g(T8, T9, T10, bin_tree1_in_g(T10))
U2_g(T8, T9, T10, bin_tree1_out_g(T10)) → bin_tree1_out_g(tree(T8, T9, T10))

The argument filtering Pi contains the following mapping:
bin_tree1_in_g(x1)  =  bin_tree1_in_g(x1)
void  =  void
bin_tree1_out_g(x1)  =  bin_tree1_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_TREE1_IN_G(x1)  =  BIN_TREE1_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_TREE1_IN_G(tree(T8, T9, T10)) → U1_G(T8, T9, T10, bin_tree1_in_g(T9))
BIN_TREE1_IN_G(tree(T8, T9, T10)) → BIN_TREE1_IN_G(T9)
U1_G(T8, T9, T10, bin_tree1_out_g(T9)) → U2_G(T8, T9, T10, bin_tree1_in_g(T10))
U1_G(T8, T9, T10, bin_tree1_out_g(T9)) → BIN_TREE1_IN_G(T10)

The TRS R consists of the following rules:

bin_tree1_in_g(void) → bin_tree1_out_g(void)
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T8, T9, T10, bin_tree1_in_g(T9))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → bin_tree1_out_g(tree(T8, T9, T10))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → U2_g(T8, T9, T10, bin_tree1_in_g(T10))
U2_g(T8, T9, T10, bin_tree1_out_g(T10)) → bin_tree1_out_g(tree(T8, T9, T10))

The argument filtering Pi contains the following mapping:
bin_tree1_in_g(x1)  =  bin_tree1_in_g(x1)
void  =  void
bin_tree1_out_g(x1)  =  bin_tree1_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_TREE1_IN_G(x1)  =  BIN_TREE1_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(T8, T9, T10, bin_tree1_out_g(T9)) → BIN_TREE1_IN_G(T10)
BIN_TREE1_IN_G(tree(T8, T9, T10)) → U1_G(T8, T9, T10, bin_tree1_in_g(T9))
BIN_TREE1_IN_G(tree(T8, T9, T10)) → BIN_TREE1_IN_G(T9)

The TRS R consists of the following rules:

bin_tree1_in_g(void) → bin_tree1_out_g(void)
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T8, T9, T10, bin_tree1_in_g(T9))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → bin_tree1_out_g(tree(T8, T9, T10))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → U2_g(T8, T9, T10, bin_tree1_in_g(T10))
U2_g(T8, T9, T10, bin_tree1_out_g(T10)) → bin_tree1_out_g(tree(T8, T9, T10))

The argument filtering Pi contains the following mapping:
bin_tree1_in_g(x1)  =  bin_tree1_in_g(x1)
void  =  void
bin_tree1_out_g(x1)  =  bin_tree1_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_TREE1_IN_G(x1)  =  BIN_TREE1_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(T10, bin_tree1_out_g) → BIN_TREE1_IN_G(T10)
BIN_TREE1_IN_G(tree(T8, T9, T10)) → U1_G(T10, bin_tree1_in_g(T9))
BIN_TREE1_IN_G(tree(T8, T9, T10)) → BIN_TREE1_IN_G(T9)

The TRS R consists of the following rules:

bin_tree1_in_g(void) → bin_tree1_out_g
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T10, bin_tree1_in_g(T9))
U1_g(T10, bin_tree1_out_g) → bin_tree1_out_g
U1_g(T10, bin_tree1_out_g) → U2_g(bin_tree1_in_g(T10))
U2_g(bin_tree1_out_g) → bin_tree1_out_g

The set Q consists of the following terms:

bin_tree1_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_TREE1_IN_G(tree(T8, T9, T10)) → U1_G(T10, bin_tree1_in_g(T9))
  The graph contains the following edges 1 > 1

• BIN_TREE1_IN_G(tree(T8, T9, T10)) → BIN_TREE1_IN_G(T9)
  The graph contains the following edges 1 > 1

• U1_G(T10, bin_tree1_out_g) → BIN_TREE1_IN_G(T10)
  The graph contains the following edges 1 >= 1

(12) TRUE