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) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: bin_tree(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: bin_tree(T1), SCOPE: 1, CLAUSE: 0 | bin_tree(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true | bin_tree(void), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: bin_tree(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nbin_tree(T1) !=? bin_tree(void)", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: bin_tree(void), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="B: ','(bin_tree(T6), bin_tree(T7))\n\nT6 is ground\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: ','(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];
25 [label="25: ','(bin_tree(T6), bin_tree(T7)), SCOPE: 2, CLAUSE: 0\n\nT6 is ground\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: ','(bin_tree(T6), bin_tree(T7)), SCOPE: 2, CLAUSE: 1\n\nT6 is ground\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: bin_tree(T7)\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="C: ','(','(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];
39 [label="39: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="41: bin_tree(T15)\n\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="42: ','(bin_tree(T16), bin_tree(T7))\n\nT7 is ground\nT16 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 43 [arrowhead = none ];
43 -> 5;

44 [label="EVAL with clause\nbin_tree(void).\nand substitutionT1 -> void", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 44 [arrowhead = none ];
44 -> 9;

45 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 45 [arrowhead = none ];
45 -> 11;

46 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 46 [arrowhead = none ];
46 -> 16;

47 [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];
11 -> 47 [arrowhead = none ];
47 -> 20;

48 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 48 [arrowhead = none ];
48 -> 21;

49 [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];
16 -> 49 [arrowhead = none ];
49 -> 18;

50 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
20 -> 50 [arrowhead = none ];
50 -> 23;

51 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 51 [arrowhead = none ];
51 -> 25;

52 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
23 -> 52 [arrowhead = none ];
52 -> 26;

53 [label="EVAL with clause\nbin_tree(void).\nand substitutionT6 -> void", fontsize=14, style = filled, fillcolor = lightblue];
25 -> 53 [arrowhead = none ];
53 -> 27;

54 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 54 [arrowhead = none ];
54 -> 28;

55 [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];
26 -> 55 [arrowhead = none ];
55 -> 38;

56 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 56 [arrowhead = none ];
56 -> 39;

57 [label="INSTANCE with matching:\nT1 -> T7", fontsize=14, style = filled, fillcolor = lightgrey];
27 -> 57 [arrowhead = none , style=dashed];
57 -> 2[style=dashed];

58 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
38 -> 58 [arrowhead = none ];
58 -> 41;

59 [label="SPLIT 2\nnew knowledge:\nT15 is ground", fontsize=14, style = filled, fillcolor = violet];
38 -> 59 [arrowhead = none ];
59 -> 42;

60 [label="INSTANCE with matching:\nT1 -> T15", fontsize=14, style = filled, fillcolor = lightgrey];
41 -> 60 [arrowhead = none , style=dashed];
60 -> 2[style=dashed];

61 [label="INSTANCE with matching:\nT6 -> T16", fontsize=14, style = filled, fillcolor = lightgrey];
42 -> 61 [arrowhead = none , style=dashed];
61 -> 20[style=dashed];

}

(2) 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(T5, T6, T7)) → U1_g(T5, T6, T7, pB_in_gg(T6, T7))
pB_in_gg(void, T7) → U2_gg(T7, bin_treeA_in_g(T7))
U2_gg(T7, bin_treeA_out_g(T7)) → pB_out_gg(void, T7)
pB_in_gg(tree(T14, T15, T16), T7) → U3_gg(T14, T15, T16, T7, pC_in_ggg(T15, T16, T7))
pC_in_ggg(T15, T16, T7) → U4_ggg(T15, T16, T7, bin_treeA_in_g(T15))
U4_ggg(T15, T16, T7, bin_treeA_out_g(T15)) → U5_ggg(T15, T16, T7, pB_in_gg(T16, T7))
U5_ggg(T15, T16, T7, pB_out_gg(T16, T7)) → pC_out_ggg(T15, T16, T7)
U3_gg(T14, T15, T16, T7, pC_out_ggg(T15, T16, T7)) → pB_out_gg(tree(T14, T15, T16), T7)
U1_g(T5, T6, T7, pB_out_gg(T6, T7)) → bin_treeA_out_g(tree(T5, T6, T7))

Pi is empty.

(3) 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(T5, T6, T7)) → U1_G(T5, T6, T7, pB_in_gg(T6, T7))
BIN_TREEA_IN_G(tree(T5, T6, T7)) → PB_IN_GG(T6, T7)
PB_IN_GG(void, T7) → U2_GG(T7, bin_treeA_in_g(T7))
PB_IN_GG(void, T7) → BIN_TREEA_IN_G(T7)
PB_IN_GG(tree(T14, T15, T16), T7) → U3_GG(T14, T15, T16, T7, pC_in_ggg(T15, T16, T7))
PB_IN_GG(tree(T14, T15, T16), T7) → PC_IN_GGG(T15, T16, T7)
PC_IN_GGG(T15, T16, T7) → U4_GGG(T15, T16, T7, bin_treeA_in_g(T15))
PC_IN_GGG(T15, T16, T7) → BIN_TREEA_IN_G(T15)
U4_GGG(T15, T16, T7, bin_treeA_out_g(T15)) → U5_GGG(T15, T16, T7, pB_in_gg(T16, T7))
U4_GGG(T15, T16, T7, bin_treeA_out_g(T15)) → PB_IN_GG(T16, T7)

The TRS R consists of the following rules:

bin_treeA_in_g(void) → bin_treeA_out_g(void)
bin_treeA_in_g(tree(T5, T6, T7)) → U1_g(T5, T6, T7, pB_in_gg(T6, T7))
pB_in_gg(void, T7) → U2_gg(T7, bin_treeA_in_g(T7))
U2_gg(T7, bin_treeA_out_g(T7)) → pB_out_gg(void, T7)
pB_in_gg(tree(T14, T15, T16), T7) → U3_gg(T14, T15, T16, T7, pC_in_ggg(T15, T16, T7))
pC_in_ggg(T15, T16, T7) → U4_ggg(T15, T16, T7, bin_treeA_in_g(T15))
U4_ggg(T15, T16, T7, bin_treeA_out_g(T15)) → U5_ggg(T15, T16, T7, pB_in_gg(T16, T7))
U5_ggg(T15, T16, T7, pB_out_gg(T16, T7)) → pC_out_ggg(T15, T16, T7)
U3_gg(T14, T15, T16, T7, pC_out_ggg(T15, T16, T7)) → pB_out_gg(tree(T14, T15, T16), T7)
U1_g(T5, T6, T7, pB_out_gg(T6, T7)) → bin_treeA_out_g(tree(T5, T6, T7))

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

(4) Obligation:

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

BIN_TREEA_IN_G(tree(T5, T6, T7)) → U1_G(T5, T6, T7, pB_in_gg(T6, T7))
BIN_TREEA_IN_G(tree(T5, T6, T7)) → PB_IN_GG(T6, T7)
PB_IN_GG(void, T7) → U2_GG(T7, bin_treeA_in_g(T7))
PB_IN_GG(void, T7) → BIN_TREEA_IN_G(T7)
PB_IN_GG(tree(T14, T15, T16), T7) → U3_GG(T14, T15, T16, T7, pC_in_ggg(T15, T16, T7))
PB_IN_GG(tree(T14, T15, T16), T7) → PC_IN_GGG(T15, T16, T7)
PC_IN_GGG(T15, T16, T7) → U4_GGG(T15, T16, T7, bin_treeA_in_g(T15))
PC_IN_GGG(T15, T16, T7) → BIN_TREEA_IN_G(T15)
U4_GGG(T15, T16, T7, bin_treeA_out_g(T15)) → U5_GGG(T15, T16, T7, pB_in_gg(T16, T7))
U4_GGG(T15, T16, T7, bin_treeA_out_g(T15)) → PB_IN_GG(T16, T7)

The TRS R consists of the following rules:

bin_treeA_in_g(void) → bin_treeA_out_g(void)
bin_treeA_in_g(tree(T5, T6, T7)) → U1_g(T5, T6, T7, pB_in_gg(T6, T7))
pB_in_gg(void, T7) → U2_gg(T7, bin_treeA_in_g(T7))
U2_gg(T7, bin_treeA_out_g(T7)) → pB_out_gg(void, T7)
pB_in_gg(tree(T14, T15, T16), T7) → U3_gg(T14, T15, T16, T7, pC_in_ggg(T15, T16, T7))
pC_in_ggg(T15, T16, T7) → U4_ggg(T15, T16, T7, bin_treeA_in_g(T15))
U4_ggg(T15, T16, T7, bin_treeA_out_g(T15)) → U5_ggg(T15, T16, T7, pB_in_gg(T16, T7))
U5_ggg(T15, T16, T7, pB_out_gg(T16, T7)) → pC_out_ggg(T15, T16, T7)
U3_gg(T14, T15, T16, T7, pC_out_ggg(T15, T16, T7)) → pB_out_gg(tree(T14, T15, T16), T7)
U1_g(T5, T6, T7, pB_out_gg(T6, T7)) → bin_treeA_out_g(tree(T5, T6, T7))

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(T5, T6, T7)) → PB_IN_GG(T6, T7)
PB_IN_GG(void, T7) → BIN_TREEA_IN_G(T7)
PB_IN_GG(tree(T14, T15, T16), T7) → PC_IN_GGG(T15, T16, T7)
PC_IN_GGG(T15, T16, T7) → U4_GGG(T15, T16, T7, bin_treeA_in_g(T15))
U4_GGG(T15, T16, T7, bin_treeA_out_g(T15)) → PB_IN_GG(T16, T7)
PC_IN_GGG(T15, T16, T7) → BIN_TREEA_IN_G(T15)

The TRS R consists of the following rules:

bin_treeA_in_g(void) → bin_treeA_out_g(void)
bin_treeA_in_g(tree(T5, T6, T7)) → U1_g(T5, T6, T7, pB_in_gg(T6, T7))
pB_in_gg(void, T7) → U2_gg(T7, bin_treeA_in_g(T7))
U2_gg(T7, bin_treeA_out_g(T7)) → pB_out_gg(void, T7)
pB_in_gg(tree(T14, T15, T16), T7) → U3_gg(T14, T15, T16, T7, pC_in_ggg(T15, T16, T7))
pC_in_ggg(T15, T16, T7) → U4_ggg(T15, T16, T7, bin_treeA_in_g(T15))
U4_ggg(T15, T16, T7, bin_treeA_out_g(T15)) → U5_ggg(T15, T16, T7, pB_in_gg(T16, T7))
U5_ggg(T15, T16, T7, pB_out_gg(T16, T7)) → pC_out_ggg(T15, T16, T7)
U3_gg(T14, T15, T16, T7, pC_out_ggg(T15, T16, T7)) → pB_out_gg(tree(T14, T15, T16), T7)
U1_g(T5, T6, T7, pB_out_gg(T6, T7)) → bin_treeA_out_g(tree(T5, T6, T7))

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(T5, T6, T7)) → PB_IN_GG(T6, T7)
PB_IN_GG(void, T7) → BIN_TREEA_IN_G(T7)
PB_IN_GG(tree(T14, T15, T16), T7) → PC_IN_GGG(T15, T16, T7)
PC_IN_GGG(T15, T16, T7) → U4_GGG(T15, T16, T7, bin_treeA_in_g(T15))
U4_GGG(T15, T16, T7, bin_treeA_out_g(T15)) → PB_IN_GG(T16, T7)
PC_IN_GGG(T15, T16, T7) → BIN_TREEA_IN_G(T15)

The TRS R consists of the following rules:

bin_treeA_in_g(void) → bin_treeA_out_g(void)
bin_treeA_in_g(tree(T5, T6, T7)) → U1_g(T5, T6, T7, pB_in_gg(T6, T7))
pB_in_gg(void, T7) → U2_gg(T7, bin_treeA_in_g(T7))
U2_gg(T7, bin_treeA_out_g(T7)) → pB_out_gg(void, T7)
pB_in_gg(tree(T14, T15, T16), T7) → U3_gg(T14, T15, T16, T7, pC_in_ggg(T15, T16, T7))
pC_in_ggg(T15, T16, T7) → U4_ggg(T15, T16, T7, bin_treeA_in_g(T15))
U4_ggg(T15, T16, T7, bin_treeA_out_g(T15)) → U5_ggg(T15, T16, T7, pB_in_gg(T16, T7))
U5_ggg(T15, T16, T7, pB_out_gg(T16, T7)) → pC_out_ggg(T15, T16, T7)
U3_gg(T14, T15, T16, T7, pC_out_ggg(T15, T16, T7)) → pB_out_gg(tree(T14, T15, T16), T7)
U1_g(T5, T6, T7, pB_out_gg(T6, T7)) → bin_treeA_out_g(tree(T5, T6, T7))

The set Q consists of the following terms:

bin_treeA_in_g(x0)
pB_in_gg(x0, x1)
U2_gg(x0, x1)
pC_in_ggg(x0, x1, x2)
U4_ggg(x0, x1, x2, x3)
U5_ggg(x0, x1, x2, x3)
U3_gg(x0, x1, x2, x3, x4)
U1_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(void, T7) → BIN_TREEA_IN_G(T7)
  The graph contains the following edges 2 >= 1

• PB_IN_GG(tree(T14, T15, T16), T7) → PC_IN_GGG(T15, T16, T7)
  The graph contains the following edges 1 > 1, 1 > 2, 2 >= 3

• PC_IN_GGG(T15, T16, T7) → BIN_TREEA_IN_G(T15)
  The graph contains the following edges 1 >= 1

• BIN_TREEA_IN_G(tree(T5, T6, T7)) → PB_IN_GG(T6, T7)
  The graph contains the following edges 1 > 1, 1 > 2

• U4_GGG(T15, T16, T7, bin_treeA_out_g(T15)) → PB_IN_GG(T16, T7)
  The graph contains the following edges 2 >= 1, 3 >= 2

• PC_IN_GGG(T15, T16, T7) → U4_GGG(T15, T16, T7, bin_treeA_in_g(T15))
  The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3

(10) YES