Left Termination of the query pattern h_in_1(g) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

h(X) :- ','(f(X), g(X)).
f(c(0, X1)).
f(c(X, Y)) :- ','(no(zero(X)), ','(p(X, P), f(c(P, s(Y))))).
g(c(X2, 0)).
g(c(X, Y)) :- ','(no(zero(Y)), ','(p(Y, P), g(c(s(X), P)))).
p(0, 0).
p(s(X), X).
zero(0).
no(X) :- ','(X, ','(!, failure(a))).
no(X3).
failure(b).

Queries:

h(g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: h(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: h(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(f(T2), g(T2))\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: f(T2)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: g(T2)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: f(T2), SCOPE: 2, CLAUSE: 1 | f(T2), SCOPE: 2, CLAUSE: 2\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: f(T2), SCOPE: 2, CLAUSE: 1\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: f(T2), SCOPE: 2, CLAUSE: 2\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\nX8 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: ','(no(zero(T4)), ','(p(T4, X13), f(c(X13, s(T5)))))\n\nT4 is ground\nT5 is ground\nX13 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\nX11 is free\nX12 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(no(zero(T4)), ','(p(T4, X13), f(c(X13, s(T5))))), SCOPE: 3, CLAUSE: 8 | ','(no(zero(T4)), ','(p(T4, X13), f(c(X13, s(T5))))), SCOPE: 3, CLAUSE: 9 | ?_3\n\nT4 is ground\nT5 is ground\nX13 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: ','(zero(T6), ','(!_3, ','(failure(a), ','(p(T6, X13), f(c(X13, s(T5))))))) | ','(no(zero(T6)), ','(p(T6, X13), f(c(X13, s(T5))))), SCOPE: 3, CLAUSE: 9 | ?_3\n\nT5 is ground\nT6 is ground\nX13 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: ','(zero(T6), ','(!_3, ','(failure(a), ','(p(T6, X13), f(c(X13, s(T5))))))), SCOPE: 4, CLAUSE: 7 | ','(no(zero(T6)), ','(p(T6, X13), f(c(X13, s(T5))))), SCOPE: 3, CLAUSE: 9 | ?_3\n\nT5 is ground\nT6 is ground\nX13 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: ','(!_3, ','(failure(a), ','(p(0, X13), f(c(X13, s(T5)))))) | ','(no(zero(0)), ','(p(0, X13), f(c(X13, s(T5))))), SCOPE: 3, CLAUSE: 9 | ?_3\n\nT5 is ground\nX13 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: ','(no(zero(T6)), ','(p(T6, X13), f(c(X13, s(T5))))), SCOPE: 3, CLAUSE: 9\n\nT5 is ground\nT6 is ground\nX13 is free; \nzero(T6) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: ','(failure(a), ','(p(0, X13), f(c(X13, s(T5)))))\n\nT5 is ground\nX13 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: ','(failure(a), ','(p(0, X13), f(c(X13, s(T5))))), SCOPE: 5, CLAUSE: 10\n\nT5 is ground\nX13 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(p(T7, X13), f(c(X13, s(T5))))\n\nT5 is ground\nT7 is ground\nX13 is free; \nzero(T7) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: ','(p(T7, X13), f(c(X13, s(T5)))), SCOPE: 6, CLAUSE: 5 | ','(p(T7, X13), f(c(X13, s(T5)))), SCOPE: 6, CLAUSE: 6\n\nT5 is ground\nT7 is ground\nX13 is free; \nzero(T7) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(p(T7, X13), f(c(X13, s(T5)))), SCOPE: 6, CLAUSE: 6\n\nT5 is ground\nT7 is ground\nX13 is free; \nzero(T7) !=? zero(0)\np(T7, X13) !=? p(0, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: f(c(T8, s(T5)))\n\nT5 is ground\nT8 is ground\nX13 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: \n\nX13 is free\nX19 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: g(T2), SCOPE: 7, CLAUSE: 3 | g(T2), SCOPE: 7, CLAUSE: 4\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: g(T2), SCOPE: 7, CLAUSE: 3\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="29: g(T2), SCOPE: 7, CLAUSE: 4\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
30 [label="30: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: \n\nX22 is free", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="32: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
33 [label="33: ','(no(zero(T11)), ','(p(T11, X27), g(c(s(T10), X27))))\n\nT10 is ground\nT11 is ground\nX27 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: \n\nX25 is free\nX26 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
35 [label="35: ','(no(zero(T11)), ','(p(T11, X27), g(c(s(T10), X27)))), SCOPE: 8, CLAUSE: 8 | ','(no(zero(T11)), ','(p(T11, X27), g(c(s(T10), X27)))), SCOPE: 8, CLAUSE: 9 | ?_8\n\nT10 is ground\nT11 is ground\nX27 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: ','(zero(T12), ','(!_8, ','(failure(a), ','(p(T12, X27), g(c(s(T10), X27)))))) | ','(no(zero(T12)), ','(p(T12, X27), g(c(s(T10), X27)))), SCOPE: 8, CLAUSE: 9 | ?_8\n\nT10 is ground\nT12 is ground\nX27 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
37 [label="37: ','(zero(T12), ','(!_8, ','(failure(a), ','(p(T12, X27), g(c(s(T10), X27)))))), SCOPE: 9, CLAUSE: 7 | ','(no(zero(T12)), ','(p(T12, X27), g(c(s(T10), X27)))), SCOPE: 8, CLAUSE: 9 | ?_8\n\nT10 is ground\nT12 is ground\nX27 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
38 [label="38: ','(!_8, ','(failure(a), ','(p(0, X27), g(c(s(T10), X27))))) | ','(no(zero(0)), ','(p(0, X27), g(c(s(T10), X27)))), SCOPE: 8, CLAUSE: 9 | ?_8\n\nT10 is ground\nX27 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
39 [label="39: ','(no(zero(T12)), ','(p(T12, X27), g(c(s(T10), X27)))), SCOPE: 8, CLAUSE: 9\n\nT10 is ground\nT12 is ground\nX27 is free; \nzero(T12) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
40 [label="40: ','(failure(a), ','(p(0, X27), g(c(s(T10), X27))))\n\nT10 is ground\nX27 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
41 [label="41: ','(failure(a), ','(p(0, X27), g(c(s(T10), X27)))), SCOPE: 10, CLAUSE: 10\n\nT10 is ground\nX27 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
42 [label="42: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
43 [label="43: ','(p(T13, X27), g(c(s(T10), X27)))\n\nT10 is ground\nT13 is ground\nX27 is free; \nzero(T13) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
44 [label="44: ','(p(T13, X27), g(c(s(T10), X27))), SCOPE: 11, CLAUSE: 5 | ','(p(T13, X27), g(c(s(T10), X27))), SCOPE: 11, CLAUSE: 6\n\nT10 is ground\nT13 is ground\nX27 is free; \nzero(T13) !=? zero(0)", fontsize=16, style=filled, fillcolor=lightyellow];
45 [label="45: ','(p(T13, X27), g(c(s(T10), X27))), SCOPE: 11, CLAUSE: 6\n\nT10 is ground\nT13 is ground\nX27 is free; \nzero(T13) !=? zero(0)\np(T13, X27) !=? p(0, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
46 [label="46: g(c(s(T10), T14))\n\nT10 is ground\nT14 is ground\nX27 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
47 [label="47: \n\nX27 is free\nX33 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
48 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 48 [arrowhead = none ];
48 -> 2;

49 [label="EVAL with clause\nh(X5) :- ','(f(X5), g(X5)).\nand substitution\nT1 -> T2\nX5 -> T2", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 49 [arrowhead = none ];
49 -> 3;

50 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
3 -> 50 [arrowhead = none ];
50 -> 4;

51 [label="SPLIT 2", fontsize=14, style = filled, fillcolor = violet];
3 -> 51 [arrowhead = none ];
51 -> 5;

52 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
4 -> 52 [arrowhead = none ];
52 -> 6;

53 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 53 [arrowhead = none ];
53 -> 27;

54 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
6 -> 54 [arrowhead = none ];
54 -> 7;

55 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
6 -> 55 [arrowhead = none ];
55 -> 8;

56 [label="EVAL with clause\nf(c(0, X8)).\nand substitution\nX8 -> T3\nT2 -> c(0, T3)", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 56 [arrowhead = none ];
56 -> 9;

57 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 57 [arrowhead = none ];
57 -> 10;

58 [label="EVAL with clause\nf(c(X11, X12)) :- ','(no(zero(X11)), ','(p(X11, X13), f(c(X13, s(X12))))).\nand substitution\nX11 -> T4\nX12 -> T5\nT2 -> c(T4, T5)", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 58 [arrowhead = none ];
58 -> 12;

59 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 59 [arrowhead = none ];
59 -> 13;

60 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 60 [arrowhead = none ];
60 -> 11;

61 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 61 [arrowhead = none ];
61 -> 14;

62 [label="EVAL with clause\nno(X15) :- ','(X15, ','(!, failure(a))).\nand substitution\nT4 -> T6\nX15 -> zero(T6)", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 62 [arrowhead = none ];
62 -> 15;

63 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
15 -> 63 [arrowhead = none ];
63 -> 16;

64 [label="EVAL with clause\nzero(0).\nand substitution\nT6 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 64 [arrowhead = none ];
64 -> 17;

65 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 65 [arrowhead = none ];
65 -> 18;

66 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 66 [arrowhead = none ];
66 -> 19;

67 [label="EVAL with clause\nno(X17).\nand substitution\nT6 -> T7\nX17 -> zero(T7)", fontsize=14, style = filled, fillcolor = lightblue];
18 -> 67 [arrowhead = none ];
67 -> 22;

68 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
19 -> 68 [arrowhead = none ];
68 -> 20;

69 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = white];
20 -> 69 [arrowhead = none ];
69 -> 21;

70 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
22 -> 70 [arrowhead = none ];
70 -> 23;

71 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (zero(T7), zero(0))", fontsize=14, style = filled, fillcolor = white];
23 -> 71 [arrowhead = none ];
71 -> 24;

72 [label="EVAL with clause\np(s(X19), X19).\nand substitution\nX19 -> T8\nT7 -> s(T8)\nX13 -> T8", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 72 [arrowhead = none ];
72 -> 25;

73 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
24 -> 73 [arrowhead = none ];
73 -> 26;

74 [label="INSTANCE with matching:\nT2 -> c(T8, s(T5))", fontsize=14, style = filled, fillcolor = lightgrey];
25 -> 74 [arrowhead = none , style=dashed];
74 -> 4[style=dashed];

75 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
27 -> 75 [arrowhead = none ];
75 -> 28;

76 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
27 -> 76 [arrowhead = none ];
76 -> 29;

77 [label="EVAL with clause\ng(c(X22, 0)).\nand substitution\nX22 -> T9\nT2 -> c(T9, 0)", fontsize=14, style = filled, fillcolor = lightblue];
28 -> 77 [arrowhead = none ];
77 -> 30;

78 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
28 -> 78 [arrowhead = none ];
78 -> 31;

79 [label="EVAL with clause\ng(c(X25, X26)) :- ','(no(zero(X26)), ','(p(X26, X27), g(c(s(X25), X27)))).\nand substitution\nX25 -> T10\nX26 -> T11\nT2 -> c(T10, T11)", fontsize=14, style = filled, fillcolor = lightblue];
29 -> 79 [arrowhead = none ];
79 -> 33;

80 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
29 -> 80 [arrowhead = none ];
80 -> 34;

81 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
30 -> 81 [arrowhead = none ];
81 -> 32;

82 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
33 -> 82 [arrowhead = none ];
82 -> 35;

83 [label="EVAL with clause\nno(X29) :- ','(X29, ','(!, failure(a))).\nand substitution\nT11 -> T12\nX29 -> zero(T12)", fontsize=14, style = filled, fillcolor = lightblue];
35 -> 83 [arrowhead = none ];
83 -> 36;

84 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
36 -> 84 [arrowhead = none ];
84 -> 37;

85 [label="EVAL with clause\nzero(0).\nand substitution\nT12 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
37 -> 85 [arrowhead = none ];
85 -> 38;

86 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
37 -> 86 [arrowhead = none ];
86 -> 39;

87 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
38 -> 87 [arrowhead = none ];
87 -> 40;

88 [label="EVAL with clause\nno(X31).\nand substitution\nT12 -> T13\nX31 -> zero(T13)", fontsize=14, style = filled, fillcolor = lightblue];
39 -> 88 [arrowhead = none ];
88 -> 43;

89 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
40 -> 89 [arrowhead = none ];
89 -> 41;

90 [label="BACKTRACK\nfor clause: failure(b)because of non-unification", fontsize=14, style = filled, fillcolor = white];
41 -> 90 [arrowhead = none ];
90 -> 42;

91 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
43 -> 91 [arrowhead = none ];
91 -> 44;

92 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (zero(T13), zero(0))", fontsize=14, style = filled, fillcolor = white];
44 -> 92 [arrowhead = none ];
92 -> 45;

93 [label="EVAL with clause\np(s(X33), X33).\nand substitution\nX33 -> T14\nT13 -> s(T14)\nX27 -> T14", fontsize=14, style = filled, fillcolor = lightblue];
45 -> 93 [arrowhead = none ];
93 -> 46;

94 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
45 -> 94 [arrowhead = none ];
94 -> 47;

95 [label="INSTANCE with matching:\nT2 -> c(s(T10), T14)", fontsize=14, style = filled, fillcolor = lightgrey];
46 -> 95 [arrowhead = none , style=dashed];
95 -> 5[style=dashed];

}

(2) Obligation:

Clauses:

f4(c(0, T3)).
f4(c(s(T8), T5)) :- f4(c(T8, s(T5))).
g5(c(T9, 0)).
g5(c(T10, s(T14))) :- g5(c(s(T10), T14)).
h1(T2) :- f4(T2).
h1(c(T9, 0)) :- f4(c(T9, 0)).
h1(c(T10, s(T14))) :- ','(f4(c(T10, s(T14))), g5(c(s(T10), T14))).

Queries:

h1(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:
h1_in: (b)
f4_in: (b)
g5_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

h1_in_g(T2) → U3_g(T2, f4_in_g(T2))
f4_in_g(c(0, T3)) → f4_out_g(c(0, T3))
f4_in_g(c(s(T8), T5)) → U1_g(T8, T5, f4_in_g(c(T8, s(T5))))
U1_g(T8, T5, f4_out_g(c(T8, s(T5)))) → f4_out_g(c(s(T8), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T9, 0)) → U4_g(T9, f4_in_g(c(T9, 0)))
U4_g(T9, f4_out_g(c(T9, 0))) → h1_out_g(c(T9, 0))
h1_in_g(c(T10, s(T14))) → U5_g(T10, T14, f4_in_g(c(T10, s(T14))))
U5_g(T10, T14, f4_out_g(c(T10, s(T14)))) → U6_g(T10, T14, g5_in_g(c(s(T10), T14)))
g5_in_g(c(T9, 0)) → g5_out_g(c(T9, 0))
g5_in_g(c(T10, s(T14))) → U2_g(T10, T14, g5_in_g(c(s(T10), T14)))
U2_g(T10, T14, g5_out_g(c(s(T10), T14))) → g5_out_g(c(T10, s(T14)))
U6_g(T10, T14, g5_out_g(c(s(T10), T14))) → h1_out_g(c(T10, s(T14)))

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

h1_in_g(T2) → U3_g(T2, f4_in_g(T2))
f4_in_g(c(0, T3)) → f4_out_g(c(0, T3))
f4_in_g(c(s(T8), T5)) → U1_g(T8, T5, f4_in_g(c(T8, s(T5))))
U1_g(T8, T5, f4_out_g(c(T8, s(T5)))) → f4_out_g(c(s(T8), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T9, 0)) → U4_g(T9, f4_in_g(c(T9, 0)))
U4_g(T9, f4_out_g(c(T9, 0))) → h1_out_g(c(T9, 0))
h1_in_g(c(T10, s(T14))) → U5_g(T10, T14, f4_in_g(c(T10, s(T14))))
U5_g(T10, T14, f4_out_g(c(T10, s(T14)))) → U6_g(T10, T14, g5_in_g(c(s(T10), T14)))
g5_in_g(c(T9, 0)) → g5_out_g(c(T9, 0))
g5_in_g(c(T10, s(T14))) → U2_g(T10, T14, g5_in_g(c(s(T10), T14)))
U2_g(T10, T14, g5_out_g(c(s(T10), T14))) → g5_out_g(c(T10, s(T14)))
U6_g(T10, T14, g5_out_g(c(s(T10), T14))) → h1_out_g(c(T10, s(T14)))

Pi is empty.

(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:

H1_IN_G(T2) → U3_G(T2, f4_in_g(T2))
H1_IN_G(T2) → F4_IN_G(T2)
F4_IN_G(c(s(T8), T5)) → U1_G(T8, T5, f4_in_g(c(T8, s(T5))))
F4_IN_G(c(s(T8), T5)) → F4_IN_G(c(T8, s(T5)))
H1_IN_G(c(T9, 0)) → U4_G(T9, f4_in_g(c(T9, 0)))
H1_IN_G(c(T9, 0)) → F4_IN_G(c(T9, 0))
H1_IN_G(c(T10, s(T14))) → U5_G(T10, T14, f4_in_g(c(T10, s(T14))))
H1_IN_G(c(T10, s(T14))) → F4_IN_G(c(T10, s(T14)))
U5_G(T10, T14, f4_out_g(c(T10, s(T14)))) → U6_G(T10, T14, g5_in_g(c(s(T10), T14)))
U5_G(T10, T14, f4_out_g(c(T10, s(T14)))) → G5_IN_G(c(s(T10), T14))
G5_IN_G(c(T10, s(T14))) → U2_G(T10, T14, g5_in_g(c(s(T10), T14)))
G5_IN_G(c(T10, s(T14))) → G5_IN_G(c(s(T10), T14))

The TRS R consists of the following rules:

h1_in_g(T2) → U3_g(T2, f4_in_g(T2))
f4_in_g(c(0, T3)) → f4_out_g(c(0, T3))
f4_in_g(c(s(T8), T5)) → U1_g(T8, T5, f4_in_g(c(T8, s(T5))))
U1_g(T8, T5, f4_out_g(c(T8, s(T5)))) → f4_out_g(c(s(T8), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T9, 0)) → U4_g(T9, f4_in_g(c(T9, 0)))
U4_g(T9, f4_out_g(c(T9, 0))) → h1_out_g(c(T9, 0))
h1_in_g(c(T10, s(T14))) → U5_g(T10, T14, f4_in_g(c(T10, s(T14))))
U5_g(T10, T14, f4_out_g(c(T10, s(T14)))) → U6_g(T10, T14, g5_in_g(c(s(T10), T14)))
g5_in_g(c(T9, 0)) → g5_out_g(c(T9, 0))
g5_in_g(c(T10, s(T14))) → U2_g(T10, T14, g5_in_g(c(s(T10), T14)))
U2_g(T10, T14, g5_out_g(c(s(T10), T14))) → g5_out_g(c(T10, s(T14)))
U6_g(T10, T14, g5_out_g(c(s(T10), T14))) → h1_out_g(c(T10, s(T14)))

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

(6) Obligation:

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

H1_IN_G(T2) → U3_G(T2, f4_in_g(T2))
H1_IN_G(T2) → F4_IN_G(T2)
F4_IN_G(c(s(T8), T5)) → U1_G(T8, T5, f4_in_g(c(T8, s(T5))))
F4_IN_G(c(s(T8), T5)) → F4_IN_G(c(T8, s(T5)))
H1_IN_G(c(T9, 0)) → U4_G(T9, f4_in_g(c(T9, 0)))
H1_IN_G(c(T9, 0)) → F4_IN_G(c(T9, 0))
H1_IN_G(c(T10, s(T14))) → U5_G(T10, T14, f4_in_g(c(T10, s(T14))))
H1_IN_G(c(T10, s(T14))) → F4_IN_G(c(T10, s(T14)))
U5_G(T10, T14, f4_out_g(c(T10, s(T14)))) → U6_G(T10, T14, g5_in_g(c(s(T10), T14)))
U5_G(T10, T14, f4_out_g(c(T10, s(T14)))) → G5_IN_G(c(s(T10), T14))
G5_IN_G(c(T10, s(T14))) → U2_G(T10, T14, g5_in_g(c(s(T10), T14)))
G5_IN_G(c(T10, s(T14))) → G5_IN_G(c(s(T10), T14))

The TRS R consists of the following rules:

h1_in_g(T2) → U3_g(T2, f4_in_g(T2))
f4_in_g(c(0, T3)) → f4_out_g(c(0, T3))
f4_in_g(c(s(T8), T5)) → U1_g(T8, T5, f4_in_g(c(T8, s(T5))))
U1_g(T8, T5, f4_out_g(c(T8, s(T5)))) → f4_out_g(c(s(T8), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T9, 0)) → U4_g(T9, f4_in_g(c(T9, 0)))
U4_g(T9, f4_out_g(c(T9, 0))) → h1_out_g(c(T9, 0))
h1_in_g(c(T10, s(T14))) → U5_g(T10, T14, f4_in_g(c(T10, s(T14))))
U5_g(T10, T14, f4_out_g(c(T10, s(T14)))) → U6_g(T10, T14, g5_in_g(c(s(T10), T14)))
g5_in_g(c(T9, 0)) → g5_out_g(c(T9, 0))
g5_in_g(c(T10, s(T14))) → U2_g(T10, T14, g5_in_g(c(s(T10), T14)))
U2_g(T10, T14, g5_out_g(c(s(T10), T14))) → g5_out_g(c(T10, s(T14)))
U6_g(T10, T14, g5_out_g(c(s(T10), T14))) → h1_out_g(c(T10, s(T14)))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 10 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

G5_IN_G(c(T10, s(T14))) → G5_IN_G(c(s(T10), T14))

The TRS R consists of the following rules:

h1_in_g(T2) → U3_g(T2, f4_in_g(T2))
f4_in_g(c(0, T3)) → f4_out_g(c(0, T3))
f4_in_g(c(s(T8), T5)) → U1_g(T8, T5, f4_in_g(c(T8, s(T5))))
U1_g(T8, T5, f4_out_g(c(T8, s(T5)))) → f4_out_g(c(s(T8), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T9, 0)) → U4_g(T9, f4_in_g(c(T9, 0)))
U4_g(T9, f4_out_g(c(T9, 0))) → h1_out_g(c(T9, 0))
h1_in_g(c(T10, s(T14))) → U5_g(T10, T14, f4_in_g(c(T10, s(T14))))
U5_g(T10, T14, f4_out_g(c(T10, s(T14)))) → U6_g(T10, T14, g5_in_g(c(s(T10), T14)))
g5_in_g(c(T9, 0)) → g5_out_g(c(T9, 0))
g5_in_g(c(T10, s(T14))) → U2_g(T10, T14, g5_in_g(c(s(T10), T14)))
U2_g(T10, T14, g5_out_g(c(s(T10), T14))) → g5_out_g(c(T10, s(T14)))
U6_g(T10, T14, g5_out_g(c(s(T10), T14))) → h1_out_g(c(T10, s(T14)))

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

(10) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(11) Obligation:

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

G5_IN_G(c(T10, s(T14))) → G5_IN_G(c(s(T10), T14))

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

(12) PiDPToQDPProof (EQUIVALENT transformation)

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

(13) Obligation:

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

G5_IN_G(c(T10, s(T14))) → G5_IN_G(c(s(T10), T14))

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

(14) 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:

G5_IN_G(c(T10, s(T14))) → G5_IN_G(c(s(T10), T14))

Used ordering: Polynomial interpretation [POLO]:

POL(G5_IN_G(x₁)) = 2·x₁   
POL(c(x₁, x₂)) = x₁ + 2·x₂   
POL(s(x₁)) = 1 + x₁   

(15) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE

(18) Obligation:

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

F4_IN_G(c(s(T8), T5)) → F4_IN_G(c(T8, s(T5)))

The TRS R consists of the following rules:

h1_in_g(T2) → U3_g(T2, f4_in_g(T2))
f4_in_g(c(0, T3)) → f4_out_g(c(0, T3))
f4_in_g(c(s(T8), T5)) → U1_g(T8, T5, f4_in_g(c(T8, s(T5))))
U1_g(T8, T5, f4_out_g(c(T8, s(T5)))) → f4_out_g(c(s(T8), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T9, 0)) → U4_g(T9, f4_in_g(c(T9, 0)))
U4_g(T9, f4_out_g(c(T9, 0))) → h1_out_g(c(T9, 0))
h1_in_g(c(T10, s(T14))) → U5_g(T10, T14, f4_in_g(c(T10, s(T14))))
U5_g(T10, T14, f4_out_g(c(T10, s(T14)))) → U6_g(T10, T14, g5_in_g(c(s(T10), T14)))
g5_in_g(c(T9, 0)) → g5_out_g(c(T9, 0))
g5_in_g(c(T10, s(T14))) → U2_g(T10, T14, g5_in_g(c(s(T10), T14)))
U2_g(T10, T14, g5_out_g(c(s(T10), T14))) → g5_out_g(c(T10, s(T14)))
U6_g(T10, T14, g5_out_g(c(s(T10), T14))) → h1_out_g(c(T10, s(T14)))

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

(19) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(20) Obligation:

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

F4_IN_G(c(s(T8), T5)) → F4_IN_G(c(T8, s(T5)))

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

(21) PiDPToQDPProof (EQUIVALENT transformation)

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

(22) Obligation:

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

F4_IN_G(c(s(T8), T5)) → F4_IN_G(c(T8, s(T5)))

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

(23) 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:

F4_IN_G(c(s(T8), T5)) → F4_IN_G(c(T8, s(T5)))

Used ordering: Polynomial interpretation [POLO]:

POL(F4_IN_G(x₁)) = 2·x₁   
POL(c(x₁, x₂)) = 2·x₁ + x₂   
POL(s(x₁)) = 1 + x₁   

(24) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE