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) 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];
8 [label="8: !_1 | bin_tree(void), SCOPE: 1, CLAUSE: 1\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: bin_tree(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground\nbin_tree(T1) !=? bin_tree(void)", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(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];
21 [label="21: ','(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];
22 [label="22: ','(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];
26 [label="26: ','(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];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: ','(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];
31 [label="31: ','(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];
36 [label="B: ','(bin_tree(T20), bin_tree(T21))\n\nT20 is ground\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="42: bin_tree(T20)\n\nT20 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
44 [label="44: bin_tree(T21)\n\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
45 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 45 [arrowhead = none ];
45 -> 5;

46 [label="EVAL with clause\nbin_tree(void) :- !_1.\nand substitutionT1 -> void", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 46 [arrowhead = none ];
46 -> 8;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 47 [arrowhead = none ];
47 -> 10;

48 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 48 [arrowhead = none ];
48 -> 14;

49 [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];
10 -> 49 [arrowhead = none ];
49 -> 19;

50 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 50 [arrowhead = none ];
50 -> 16;

51 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
19 -> 51 [arrowhead = none ];
51 -> 21;

52 [label="BACKTRACK\nfor clause: left(void, void)\nwith clash: (bin_tree(T3), bin_tree(void))", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 52 [arrowhead = none ];
52 -> 22;

53 [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];
22 -> 53 [arrowhead = none ];
53 -> 26;

54 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
22 -> 54 [arrowhead = none ];
54 -> 28;

55 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
26 -> 55 [arrowhead = none ];
55 -> 30;

56 [label="BACKTRACK\nfor clause: right(void, void)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 56 [arrowhead = none ];
56 -> 31;

57 [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];
31 -> 57 [arrowhead = none ];
57 -> 36;

58 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
36 -> 58 [arrowhead = none ];
58 -> 42;

59 [label="SPLIT 2\nnew knowledge:\nT20 is ground", fontsize=14, style = filled, fillcolor = violet];
36 -> 59 [arrowhead = none ];
59 -> 44;

60 [label="INSTANCE with matching:\nT1 -> T20", fontsize=14, style = filled, fillcolor = lightgrey];
42 -> 60 [arrowhead = none , style=dashed];
60 -> 2[style=dashed];

61 [label="INSTANCE with matching:\nT1 -> T21", fontsize=14, style = filled, fillcolor = lightgrey];
44 -> 61 [arrowhead = none , style=dashed];
61 -> 2[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(T19, T20, T21)) → U1_g(T19, T20, T21, pB_in_gg(T20, T21))
pB_in_gg(T20, T21) → U2_gg(T20, T21, bin_treeA_in_g(T20))
U2_gg(T20, T21, bin_treeA_out_g(T20)) → U3_gg(T20, T21, bin_treeA_in_g(T21))
U3_gg(T20, T21, bin_treeA_out_g(T21)) → pB_out_gg(T20, T21)
U1_g(T19, T20, T21, pB_out_gg(T20, T21)) → bin_treeA_out_g(tree(T19, T20, T21))

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(T19, T20, T21)) → U1_G(T19, T20, T21, pB_in_gg(T20, T21))
BIN_TREEA_IN_G(tree(T19, T20, T21)) → PB_IN_GG(T20, T21)
PB_IN_GG(T20, T21) → U2_GG(T20, T21, bin_treeA_in_g(T20))
PB_IN_GG(T20, T21) → BIN_TREEA_IN_G(T20)
U2_GG(T20, T21, bin_treeA_out_g(T20)) → U3_GG(T20, T21, bin_treeA_in_g(T21))
U2_GG(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, pB_in_gg(T20, T21))
pB_in_gg(T20, T21) → U2_gg(T20, T21, bin_treeA_in_g(T20))
U2_gg(T20, T21, bin_treeA_out_g(T20)) → U3_gg(T20, T21, bin_treeA_in_g(T21))
U3_gg(T20, T21, bin_treeA_out_g(T21)) → pB_out_gg(T20, T21)
U1_g(T19, T20, T21, pB_out_gg(T20, T21)) → bin_treeA_out_g(tree(T19, T20, T21))

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(T19, T20, T21)) → U1_G(T19, T20, T21, pB_in_gg(T20, T21))
BIN_TREEA_IN_G(tree(T19, T20, T21)) → PB_IN_GG(T20, T21)
PB_IN_GG(T20, T21) → U2_GG(T20, T21, bin_treeA_in_g(T20))
PB_IN_GG(T20, T21) → BIN_TREEA_IN_G(T20)
U2_GG(T20, T21, bin_treeA_out_g(T20)) → U3_GG(T20, T21, bin_treeA_in_g(T21))
U2_GG(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, pB_in_gg(T20, T21))
pB_in_gg(T20, T21) → U2_gg(T20, T21, bin_treeA_in_g(T20))
U2_gg(T20, T21, bin_treeA_out_g(T20)) → U3_gg(T20, T21, bin_treeA_in_g(T21))
U3_gg(T20, T21, bin_treeA_out_g(T21)) → pB_out_gg(T20, T21)
U1_g(T19, T20, T21, pB_out_gg(T20, T21)) → bin_treeA_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_TREEA_IN_G(tree(T19, T20, T21)) → PB_IN_GG(T20, T21)
PB_IN_GG(T20, T21) → U2_GG(T20, T21, bin_treeA_in_g(T20))
U2_GG(T20, T21, bin_treeA_out_g(T20)) → BIN_TREEA_IN_G(T21)
PB_IN_GG(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, pB_in_gg(T20, T21))
pB_in_gg(T20, T21) → U2_gg(T20, T21, bin_treeA_in_g(T20))
U2_gg(T20, T21, bin_treeA_out_g(T20)) → U3_gg(T20, T21, bin_treeA_in_g(T21))
U3_gg(T20, T21, bin_treeA_out_g(T21)) → pB_out_gg(T20, T21)
U1_g(T19, T20, T21, pB_out_gg(T20, T21)) → bin_treeA_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_TREEA_IN_G(tree(T19, T20, T21)) → PB_IN_GG(T20, T21)
PB_IN_GG(T20, T21) → U2_GG(T20, T21, bin_treeA_in_g(T20))
U2_GG(T20, T21, bin_treeA_out_g(T20)) → BIN_TREEA_IN_G(T21)
PB_IN_GG(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, pB_in_gg(T20, T21))
pB_in_gg(T20, T21) → U2_gg(T20, T21, bin_treeA_in_g(T20))
U2_gg(T20, T21, bin_treeA_out_g(T20)) → U3_gg(T20, T21, bin_treeA_in_g(T21))
U3_gg(T20, T21, bin_treeA_out_g(T21)) → pB_out_gg(T20, T21)
U1_g(T19, T20, T21, pB_out_gg(T20, T21)) → bin_treeA_out_g(tree(T19, T20, T21))

The set Q consists of the following terms:

bin_treeA_in_g(x0)
pB_in_gg(x0, x1)
U2_gg(x0, x1, x2)
U3_gg(x0, x1, x2)
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(T20, T21) → BIN_TREEA_IN_G(T20)
  The graph contains the following edges 1 >= 1

• PB_IN_GG(T20, T21) → U2_GG(T20, T21, bin_treeA_in_g(T20))
  The graph contains the following edges 1 >= 1, 2 >= 2

• U2_GG(T20, T21, bin_treeA_out_g(T20)) → BIN_TREEA_IN_G(T21)
  The graph contains the following edges 2 >= 1

• BIN_TREEA_IN_G(tree(T19, T20, T21)) → PB_IN_GG(T20, T21)
  The graph contains the following edges 1 > 1, 1 > 2

(10) YES