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

(0) Obligation:

Clauses:

hbal_tree(zero, nil).
hbal_tree(s(zero), t(x, nil, nil)).
hbal_tree(s(s(X)), t(x, L, R)) :- ','(distr(s(X), X, DL, DR), ','(hbal_tree(DL, L), hbal_tree(DR, R))).
distr(D1, X1, D1, D1).
distr(D1, D2, D1, D2).
distr(D1, D2, D2, D1).

Query: hbal_tree(g,a)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: hbal_tree(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: hbal_tree(T1, T2), SCOPE: 1, CLAUSE: 0 | hbal_tree(T1, T2), SCOPE: 1, CLAUSE: 1 | hbal_tree(T1, T2), SCOPE: 1, CLAUSE: 2\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | hbal_tree(zero, T2), SCOPE: 1, CLAUSE: 1 | hbal_tree(zero, T2), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: hbal_tree(T1, T2), SCOPE: 1, CLAUSE: 1 | hbal_tree(T1, T2), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\nhbal_tree(T1, T2) !=? hbal_tree(zero, nil)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: hbal_tree(zero, T2), SCOPE: 1, CLAUSE: 1 | hbal_tree(zero, T2), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: hbal_tree(zero, T2), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: true | hbal_tree(s(zero), T2), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: hbal_tree(T1, T2), SCOPE: 1, CLAUSE: 2\n\nT1 is ground\nhbal_tree(T1, T2) !=? hbal_tree(zero, nil)\nhbal_tree(T1, T2) !=? hbal_tree(s(zero), t(x, nil, nil))", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: hbal_tree(s(zero), T2), SCOPE: 1, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(distr(s(T6), T6, X14, X15), ','(hbal_tree(X14, T9), hbal_tree(X15, T10)))\n\nT6 is ground\nX14 is free\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(distr(s(T6), T6, X14, X15), ','(hbal_tree(X14, T9), hbal_tree(X15, T10))), SCOPE: 2, CLAUSE: 3 | ','(distr(s(T6), T6, X14, X15), ','(hbal_tree(X14, T9), hbal_tree(X15, T10))), SCOPE: 2, CLAUSE: 4 | ','(distr(s(T6), T6, X14, X15), ','(hbal_tree(X14, T9), hbal_tree(X15, T10))), SCOPE: 2, CLAUSE: 5\n\nT6 is ground\nX14 is free\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(distr(s(T6), T6, X14, X15), ','(hbal_tree(X14, T9), hbal_tree(X15, T10))), SCOPE: 2, CLAUSE: 3\n\nT6 is ground\nX14 is free\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(distr(s(T6), T6, X14, X15), ','(hbal_tree(X14, T9), hbal_tree(X15, T10))), SCOPE: 2, CLAUSE: 4 | ','(distr(s(T6), T6, X14, X15), ','(hbal_tree(X14, T9), hbal_tree(X15, T10))), SCOPE: 2, CLAUSE: 5\n\nT6 is ground\nX14 is free\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(hbal_tree(s(T15), T9), hbal_tree(s(T15), T10))\n\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: hbal_tree(s(T15), T9)\n\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: hbal_tree(s(T15), T16)\n\nT15 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(distr(s(T6), T6, X14, X15), ','(hbal_tree(X14, T9), hbal_tree(X15, T10))), SCOPE: 2, CLAUSE: 4\n\nT6 is ground\nX14 is free\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(distr(s(T6), T6, X14, X15), ','(hbal_tree(X14, T9), hbal_tree(X15, T10))), SCOPE: 2, CLAUSE: 5\n\nT6 is ground\nX14 is free\nX15 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(hbal_tree(s(T21), T9), hbal_tree(T21, T10))\n\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: hbal_tree(s(T21), T9)\n\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: hbal_tree(T21, T22)\n\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(hbal_tree(T25, T9), hbal_tree(s(T25), T10))\n\nT25 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: hbal_tree(T25, T9)\n\nT25 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: hbal_tree(s(T25), T26)\n\nT25 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 28 [arrowhead = none ];
28 -> 2;

29 [label="EVAL with clause\nhbal_tree(zero, nil).\nand substitutionT1 -> zero,\nT2 -> nil", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 29 [arrowhead = none ];
29 -> 3;

30 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 30 [arrowhead = none ];
30 -> 4;

31 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 31 [arrowhead = none ];
31 -> 5;

32 [label="EVAL with clause\nhbal_tree(s(zero), t(x, nil, nil)).\nand substitutionT1 -> s(zero),\nT2 -> t(x, nil, nil)", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 32 [arrowhead = none ];
32 -> 8;

33 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 33 [arrowhead = none ];
33 -> 9;

34 [label="BACKTRACK\nfor clause: hbal_tree(s(zero), t(x, nil, nil))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 34 [arrowhead = none ];
34 -> 6;

35 [label="BACKTRACK\nfor clause: hbal_tree(s(s(X)), t(x, L, R)) :- ','(distr(s(X), X, DL, DR), ','(hbal_tree(DL, L), hbal_tree(DR, R)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 35 [arrowhead = none ];
35 -> 7;

36 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
8 -> 36 [arrowhead = none ];
36 -> 10;

37 [label="EVAL with clause\nhbal_tree(s(s(X11)), t(x, X12, X13)) :- ','(distr(s(X11), X11, X14, X15), ','(hbal_tree(X14, X12), hbal_tree(X15, X13))).\nand substitutionX11 -> T6,\nT1 -> s(s(T6)),\nX12 -> T9,\nX13 -> T10,\nT2 -> t(x, T9, T10),\nT7 -> T9,\nT8 -> T10", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 37 [arrowhead = none ];
37 -> 12;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 38 [arrowhead = none ];
38 -> 13;

39 [label="BACKTRACK\nfor clause: hbal_tree(s(s(X)), t(x, L, R)) :- ','(distr(s(X), X, DL, DR), ','(hbal_tree(DL, L), hbal_tree(DR, R)))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 39 [arrowhead = none ];
39 -> 11;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 40 [arrowhead = none ];
40 -> 14;

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

42 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 42 [arrowhead = none ];
42 -> 16;

43 [label="ONLY EVAL with clause\ndistr(X24, X25, X24, X24).\nand substitutionT6 -> T15,\nX24 -> s(T15),\nX25 -> T15,\nX14 -> s(T15),\nX15 -> s(T15)", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 43 [arrowhead = none ];
43 -> 17;

44 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
16 -> 44 [arrowhead = none ];
44 -> 20;

45 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
16 -> 45 [arrowhead = none ];
45 -> 21;

46 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
17 -> 46 [arrowhead = none ];
46 -> 18;

47 [label="SPLIT 2\nnew knowledge:\nT15 is ground\nT9 is ground\nreplacements:T10 -> T16", fontsize=14, style = filled, fillcolor = violet];
17 -> 47 [arrowhead = none ];
47 -> 19;

48 [label="INSTANCE with matching:\nT1 -> s(T15)\nT2 -> T9", fontsize=14, style = filled, fillcolor = lightgrey];
18 -> 48 [arrowhead = none , style=dashed];
48 -> 1[style=dashed];

49 [label="INSTANCE with matching:\nT1 -> s(T15)\nT2 -> T16", fontsize=14, style = filled, fillcolor = lightgrey];
19 -> 49 [arrowhead = none , style=dashed];
49 -> 1[style=dashed];

50 [label="ONLY EVAL with clause\ndistr(X34, X35, X34, X35).\nand substitutionT6 -> T21,\nX34 -> s(T21),\nX35 -> T21,\nX14 -> s(T21),\nX15 -> T21", fontsize=14, style = filled, fillcolor = lightblue];
20 -> 50 [arrowhead = none ];
50 -> 22;

51 [label="ONLY EVAL with clause\ndistr(X40, X41, X41, X40).\nand substitutionT6 -> T25,\nX40 -> s(T25),\nX41 -> T25,\nX14 -> T25,\nX15 -> s(T25)", fontsize=14, style = filled, fillcolor = lightblue];
21 -> 51 [arrowhead = none ];
51 -> 25;

52 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
22 -> 52 [arrowhead = none ];
52 -> 23;

53 [label="SPLIT 2\nnew knowledge:\nT21 is ground\nT9 is ground\nreplacements:T10 -> T22", fontsize=14, style = filled, fillcolor = violet];
22 -> 53 [arrowhead = none ];
53 -> 24;

54 [label="INSTANCE with matching:\nT1 -> s(T21)\nT2 -> T9", fontsize=14, style = filled, fillcolor = lightgrey];
23 -> 54 [arrowhead = none , style=dashed];
54 -> 1[style=dashed];

55 [label="INSTANCE with matching:\nT1 -> T21\nT2 -> T22", fontsize=14, style = filled, fillcolor = lightgrey];
24 -> 55 [arrowhead = none , style=dashed];
55 -> 1[style=dashed];

56 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
25 -> 56 [arrowhead = none ];
56 -> 26;

57 [label="SPLIT 2\nnew knowledge:\nT25 is ground\nT9 is ground\nreplacements:T10 -> T26", fontsize=14, style = filled, fillcolor = violet];
25 -> 57 [arrowhead = none ];
57 -> 27;

58 [label="INSTANCE with matching:\nT1 -> T25\nT2 -> T9", fontsize=14, style = filled, fillcolor = lightgrey];
26 -> 58 [arrowhead = none , style=dashed];
58 -> 1[style=dashed];

59 [label="INSTANCE with matching:\nT1 -> s(T25)\nT2 -> T26", fontsize=14, style = filled, fillcolor = lightgrey];
27 -> 59 [arrowhead = none , style=dashed];
59 -> 1[style=dashed];

}

(2) Obligation:

Triples:

hbal_treeA(s(s(X1)), t(x, X2, X3)) :- hbal_treeA(s(X1), X2).
hbal_treeA(s(s(X1)), t(x, X2, X3)) :- ','(hbal_treecA(s(X1), X2), hbal_treeA(s(X1), X3)).
hbal_treeA(s(s(X1)), t(x, X2, X3)) :- hbal_treeA(s(X1), X2).
hbal_treeA(s(s(X1)), t(x, X2, X3)) :- ','(hbal_treecA(s(X1), X2), hbal_treeA(X1, X3)).
hbal_treeA(s(s(X1)), t(x, X2, X3)) :- hbal_treeA(X1, X2).
hbal_treeA(s(s(X1)), t(x, X2, X3)) :- ','(hbal_treecA(X1, X2), hbal_treeA(s(X1), X3)).

Clauses:

hbal_treecA(zero, nil).
hbal_treecA(s(zero), t(x, nil, nil)).
hbal_treecA(s(s(X1)), t(x, X2, X3)) :- ','(hbal_treecA(s(X1), X2), hbal_treecA(s(X1), X3)).
hbal_treecA(s(s(X1)), t(x, X2, X3)) :- ','(hbal_treecA(s(X1), X2), hbal_treecA(X1, X3)).
hbal_treecA(s(s(X1)), t(x, X2, X3)) :- ','(hbal_treecA(X1, X2), hbal_treecA(s(X1), X3)).

Afs:

hbal_treeA(x1, x2)  =  hbal_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:
hbal_treeA_in: (b,f)
hbal_treecA_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → U1_GA(X1, X2, X3, hbal_treeA_in_ga(s(X1), X2))
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → HBAL_TREEA_IN_GA(s(X1), X2)
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → U2_GA(X1, X2, X3, hbal_treecA_in_ga(s(X1), X2))
U2_GA(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → U3_GA(X1, X2, X3, hbal_treeA_in_ga(s(X1), X3))
U2_GA(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → HBAL_TREEA_IN_GA(s(X1), X3)
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → U5_GA(X1, X2, X3, hbal_treeA_in_ga(X1, X2))
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → HBAL_TREEA_IN_GA(X1, X2)
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → U6_GA(X1, X2, X3, hbal_treecA_in_ga(X1, X2))
U6_GA(X1, X2, X3, hbal_treecA_out_ga(X1, X2)) → U7_GA(X1, X2, X3, hbal_treeA_in_ga(s(X1), X3))
U6_GA(X1, X2, X3, hbal_treecA_out_ga(X1, X2)) → HBAL_TREEA_IN_GA(s(X1), X3)
U2_GA(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → U4_GA(X1, X2, X3, hbal_treeA_in_ga(X1, X3))
U2_GA(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → HBAL_TREEA_IN_GA(X1, X3)

The TRS R consists of the following rules:

hbal_treecA_in_ga(zero, nil) → hbal_treecA_out_ga(zero, nil)
hbal_treecA_in_ga(s(zero), t(x, nil, nil)) → hbal_treecA_out_ga(s(zero), t(x, nil, nil))
hbal_treecA_in_ga(s(s(X1)), t(x, X2, X3)) → U9_ga(X1, X2, X3, hbal_treecA_in_ga(s(X1), X2))
hbal_treecA_in_ga(s(s(X1)), t(x, X2, X3)) → U12_ga(X1, X2, X3, hbal_treecA_in_ga(X1, X2))
U12_ga(X1, X2, X3, hbal_treecA_out_ga(X1, X2)) → U13_ga(X1, X2, X3, hbal_treecA_in_ga(s(X1), X3))
U13_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))
U9_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → U10_ga(X1, X2, X3, hbal_treecA_in_ga(s(X1), X3))
U10_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))
U9_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → U11_ga(X1, X2, X3, hbal_treecA_in_ga(X1, X3))
U11_ga(X1, X2, X3, hbal_treecA_out_ga(X1, X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))

The argument filtering Pi contains the following mapping:
hbal_treeA_in_ga(x1, x2)  =  hbal_treeA_in_ga(x1)
s(x1)  =  s(x1)
hbal_treecA_in_ga(x1, x2)  =  hbal_treecA_in_ga(x1)
zero  =  zero
hbal_treecA_out_ga(x1, x2)  =  hbal_treecA_out_ga(x1, x2)
U9_ga(x1, x2, x3, x4)  =  U9_ga(x1, x4)
U12_ga(x1, x2, x3, x4)  =  U12_ga(x1, x4)
U13_ga(x1, x2, x3, x4)  =  U13_ga(x1, x2, x4)
U10_ga(x1, x2, x3, x4)  =  U10_ga(x1, x2, x4)
U11_ga(x1, x2, x3, x4)  =  U11_ga(x1, x2, x4)
t(x1, x2, x3)  =  t(x1, x2, x3)
x  =  x
HBAL_TREEA_IN_GA(x1, x2)  =  HBAL_TREEA_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x4)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x4)
U3_GA(x1, x2, x3, x4)  =  U3_GA(x1, x4)
U5_GA(x1, x2, x3, x4)  =  U5_GA(x1, x4)
U6_GA(x1, x2, x3, x4)  =  U6_GA(x1, x4)
U7_GA(x1, x2, x3, x4)  =  U7_GA(x1, x4)
U4_GA(x1, x2, x3, x4)  =  U4_GA(x1, x4)

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:

HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → U1_GA(X1, X2, X3, hbal_treeA_in_ga(s(X1), X2))
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → HBAL_TREEA_IN_GA(s(X1), X2)
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → U2_GA(X1, X2, X3, hbal_treecA_in_ga(s(X1), X2))
U2_GA(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → U3_GA(X1, X2, X3, hbal_treeA_in_ga(s(X1), X3))
U2_GA(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → HBAL_TREEA_IN_GA(s(X1), X3)
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → U5_GA(X1, X2, X3, hbal_treeA_in_ga(X1, X2))
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → HBAL_TREEA_IN_GA(X1, X2)
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → U6_GA(X1, X2, X3, hbal_treecA_in_ga(X1, X2))
U6_GA(X1, X2, X3, hbal_treecA_out_ga(X1, X2)) → U7_GA(X1, X2, X3, hbal_treeA_in_ga(s(X1), X3))
U6_GA(X1, X2, X3, hbal_treecA_out_ga(X1, X2)) → HBAL_TREEA_IN_GA(s(X1), X3)
U2_GA(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → U4_GA(X1, X2, X3, hbal_treeA_in_ga(X1, X3))
U2_GA(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → HBAL_TREEA_IN_GA(X1, X3)

The TRS R consists of the following rules:

hbal_treecA_in_ga(zero, nil) → hbal_treecA_out_ga(zero, nil)
hbal_treecA_in_ga(s(zero), t(x, nil, nil)) → hbal_treecA_out_ga(s(zero), t(x, nil, nil))
hbal_treecA_in_ga(s(s(X1)), t(x, X2, X3)) → U9_ga(X1, X2, X3, hbal_treecA_in_ga(s(X1), X2))
hbal_treecA_in_ga(s(s(X1)), t(x, X2, X3)) → U12_ga(X1, X2, X3, hbal_treecA_in_ga(X1, X2))
U12_ga(X1, X2, X3, hbal_treecA_out_ga(X1, X2)) → U13_ga(X1, X2, X3, hbal_treecA_in_ga(s(X1), X3))
U13_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))
U9_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → U10_ga(X1, X2, X3, hbal_treecA_in_ga(s(X1), X3))
U10_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))
U9_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → U11_ga(X1, X2, X3, hbal_treecA_in_ga(X1, X3))
U11_ga(X1, X2, X3, hbal_treecA_out_ga(X1, X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))

The argument filtering Pi contains the following mapping:
hbal_treeA_in_ga(x1, x2)  =  hbal_treeA_in_ga(x1)
s(x1)  =  s(x1)
hbal_treecA_in_ga(x1, x2)  =  hbal_treecA_in_ga(x1)
zero  =  zero
hbal_treecA_out_ga(x1, x2)  =  hbal_treecA_out_ga(x1, x2)
U9_ga(x1, x2, x3, x4)  =  U9_ga(x1, x4)
U12_ga(x1, x2, x3, x4)  =  U12_ga(x1, x4)
U13_ga(x1, x2, x3, x4)  =  U13_ga(x1, x2, x4)
U10_ga(x1, x2, x3, x4)  =  U10_ga(x1, x2, x4)
U11_ga(x1, x2, x3, x4)  =  U11_ga(x1, x2, x4)
t(x1, x2, x3)  =  t(x1, x2, x3)
x  =  x
HBAL_TREEA_IN_GA(x1, x2)  =  HBAL_TREEA_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x4)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x4)
U3_GA(x1, x2, x3, x4)  =  U3_GA(x1, x4)
U5_GA(x1, x2, x3, x4)  =  U5_GA(x1, x4)
U6_GA(x1, x2, x3, x4)  =  U6_GA(x1, x4)
U7_GA(x1, x2, x3, x4)  =  U7_GA(x1, x4)
U4_GA(x1, x2, x3, x4)  =  U4_GA(x1, x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.

(6) Obligation:

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

HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → U2_GA(X1, X2, X3, hbal_treecA_in_ga(s(X1), X2))
U2_GA(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → HBAL_TREEA_IN_GA(s(X1), X3)
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → HBAL_TREEA_IN_GA(s(X1), X2)
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → HBAL_TREEA_IN_GA(X1, X2)
HBAL_TREEA_IN_GA(s(s(X1)), t(x, X2, X3)) → U6_GA(X1, X2, X3, hbal_treecA_in_ga(X1, X2))
U6_GA(X1, X2, X3, hbal_treecA_out_ga(X1, X2)) → HBAL_TREEA_IN_GA(s(X1), X3)
U2_GA(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → HBAL_TREEA_IN_GA(X1, X3)

The TRS R consists of the following rules:

hbal_treecA_in_ga(zero, nil) → hbal_treecA_out_ga(zero, nil)
hbal_treecA_in_ga(s(zero), t(x, nil, nil)) → hbal_treecA_out_ga(s(zero), t(x, nil, nil))
hbal_treecA_in_ga(s(s(X1)), t(x, X2, X3)) → U9_ga(X1, X2, X3, hbal_treecA_in_ga(s(X1), X2))
hbal_treecA_in_ga(s(s(X1)), t(x, X2, X3)) → U12_ga(X1, X2, X3, hbal_treecA_in_ga(X1, X2))
U12_ga(X1, X2, X3, hbal_treecA_out_ga(X1, X2)) → U13_ga(X1, X2, X3, hbal_treecA_in_ga(s(X1), X3))
U13_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))
U9_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → U10_ga(X1, X2, X3, hbal_treecA_in_ga(s(X1), X3))
U10_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))
U9_ga(X1, X2, X3, hbal_treecA_out_ga(s(X1), X2)) → U11_ga(X1, X2, X3, hbal_treecA_in_ga(X1, X3))
U11_ga(X1, X2, X3, hbal_treecA_out_ga(X1, X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
hbal_treecA_in_ga(x1, x2)  =  hbal_treecA_in_ga(x1)
zero  =  zero
hbal_treecA_out_ga(x1, x2)  =  hbal_treecA_out_ga(x1, x2)
U9_ga(x1, x2, x3, x4)  =  U9_ga(x1, x4)
U12_ga(x1, x2, x3, x4)  =  U12_ga(x1, x4)
U13_ga(x1, x2, x3, x4)  =  U13_ga(x1, x2, x4)
U10_ga(x1, x2, x3, x4)  =  U10_ga(x1, x2, x4)
U11_ga(x1, x2, x3, x4)  =  U11_ga(x1, x2, x4)
t(x1, x2, x3)  =  t(x1, x2, x3)
x  =  x
HBAL_TREEA_IN_GA(x1, x2)  =  HBAL_TREEA_IN_GA(x1)
U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x4)
U6_GA(x1, x2, x3, x4)  =  U6_GA(x1, x4)

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

(7) PiDPToQDPProof (SOUND 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:

HBAL_TREEA_IN_GA(s(s(X1))) → U2_GA(X1, hbal_treecA_in_ga(s(X1)))
U2_GA(X1, hbal_treecA_out_ga(s(X1), X2)) → HBAL_TREEA_IN_GA(s(X1))
HBAL_TREEA_IN_GA(s(s(X1))) → HBAL_TREEA_IN_GA(s(X1))
HBAL_TREEA_IN_GA(s(s(X1))) → HBAL_TREEA_IN_GA(X1)
HBAL_TREEA_IN_GA(s(s(X1))) → U6_GA(X1, hbal_treecA_in_ga(X1))
U6_GA(X1, hbal_treecA_out_ga(X1, X2)) → HBAL_TREEA_IN_GA(s(X1))
U2_GA(X1, hbal_treecA_out_ga(s(X1), X2)) → HBAL_TREEA_IN_GA(X1)

The TRS R consists of the following rules:

hbal_treecA_in_ga(zero) → hbal_treecA_out_ga(zero, nil)
hbal_treecA_in_ga(s(zero)) → hbal_treecA_out_ga(s(zero), t(x, nil, nil))
hbal_treecA_in_ga(s(s(X1))) → U9_ga(X1, hbal_treecA_in_ga(s(X1)))
hbal_treecA_in_ga(s(s(X1))) → U12_ga(X1, hbal_treecA_in_ga(X1))
U12_ga(X1, hbal_treecA_out_ga(X1, X2)) → U13_ga(X1, X2, hbal_treecA_in_ga(s(X1)))
U13_ga(X1, X2, hbal_treecA_out_ga(s(X1), X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))
U9_ga(X1, hbal_treecA_out_ga(s(X1), X2)) → U10_ga(X1, X2, hbal_treecA_in_ga(s(X1)))
U10_ga(X1, X2, hbal_treecA_out_ga(s(X1), X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))
U9_ga(X1, hbal_treecA_out_ga(s(X1), X2)) → U11_ga(X1, X2, hbal_treecA_in_ga(X1))
U11_ga(X1, X2, hbal_treecA_out_ga(X1, X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))

The set Q consists of the following terms:

hbal_treecA_in_ga(x0)
U12_ga(x0, x1)
U13_ga(x0, x1, x2)
U9_ga(x0, x1)
U10_ga(x0, x1, x2)
U11_ga(x0, x1, x2)

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

(9) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

HBAL_TREEA_IN_GA(s(s(X1))) → U2_GA(X1, hbal_treecA_in_ga(s(X1)))
U2_GA(X1, hbal_treecA_out_ga(s(X1), X2)) → HBAL_TREEA_IN_GA(s(X1))
HBAL_TREEA_IN_GA(s(s(X1))) → HBAL_TREEA_IN_GA(s(X1))
HBAL_TREEA_IN_GA(s(s(X1))) → HBAL_TREEA_IN_GA(X1)
HBAL_TREEA_IN_GA(s(s(X1))) → U6_GA(X1, hbal_treecA_in_ga(X1))
U6_GA(X1, hbal_treecA_out_ga(X1, X2)) → HBAL_TREEA_IN_GA(s(X1))
U2_GA(X1, hbal_treecA_out_ga(s(X1), X2)) → HBAL_TREEA_IN_GA(X1)

Used ordering: Polynomial interpretation [POLO]:

POL(HBAL_TREEA_IN_GA(x₁)) = 1 + 2·x₁   
POL(U10_ga(x₁, x₂, x₃)) = 2 + 2·x₁ + 2·x₂ + x₃   
POL(U11_ga(x₁, x₂, x₃)) = 1 + 2·x₁ + 2·x₂ + 2·x₃   
POL(U12_ga(x₁, x₂)) = 1 + 2·x₁ + 2·x₂   
POL(U13_ga(x₁, x₂, x₃)) = 2 + 2·x₁ + 2·x₂ + x₃   
POL(U2_GA(x₁, x₂)) = x₁ + 2·x₂   
POL(U6_GA(x₁, x₂)) = 2·x₁ + 2·x₂   
POL(U9_ga(x₁, x₂)) = 2 + 2·x₁ + x₂   
POL(hbal_treecA_in_ga(x₁)) = 2 + x₁   
POL(hbal_treecA_out_ga(x₁, x₂)) = 2 + x₁ + 2·x₂   
POL(nil) = 0   
POL(s(x₁)) = 1 + 2·x₁   
POL(t(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(x) = 0   
POL(zero) = 2   

(10) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

hbal_treecA_in_ga(zero) → hbal_treecA_out_ga(zero, nil)
hbal_treecA_in_ga(s(zero)) → hbal_treecA_out_ga(s(zero), t(x, nil, nil))
hbal_treecA_in_ga(s(s(X1))) → U9_ga(X1, hbal_treecA_in_ga(s(X1)))
hbal_treecA_in_ga(s(s(X1))) → U12_ga(X1, hbal_treecA_in_ga(X1))
U12_ga(X1, hbal_treecA_out_ga(X1, X2)) → U13_ga(X1, X2, hbal_treecA_in_ga(s(X1)))
U13_ga(X1, X2, hbal_treecA_out_ga(s(X1), X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))
U9_ga(X1, hbal_treecA_out_ga(s(X1), X2)) → U10_ga(X1, X2, hbal_treecA_in_ga(s(X1)))
U10_ga(X1, X2, hbal_treecA_out_ga(s(X1), X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))
U9_ga(X1, hbal_treecA_out_ga(s(X1), X2)) → U11_ga(X1, X2, hbal_treecA_in_ga(X1))
U11_ga(X1, X2, hbal_treecA_out_ga(X1, X3)) → hbal_treecA_out_ga(s(s(X1)), t(x, X2, X3))

The set Q consists of the following terms:

hbal_treecA_in_ga(x0)
U12_ga(x0, x1)
U13_ga(x0, x1, x2)
U9_ga(x0, x1)
U10_ga(x0, x1, x2)
U11_ga(x0, x1, x2)

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

(11) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(12) YES