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)) :- ','(p(X, P), f(c(P, s(Y)))).
g(c(X2, 0)) :- !.
g(c(X, Y)) :- ','(p(Y, P), g(c(s(X), P))).
p(0, 0).
p(s(X), X).

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 | ?_2\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: !_2 | f(c(0, T3)), SCOPE: 2, CLAUSE: 2 | ?_2\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: f(T2), SCOPE: 2, CLAUSE: 2\n\nT2 is ground\nX6 is free; \nf(T2) !=? f(c(0, X6))", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(p(T4, X11), f(c(X11, s(T5))))\n\nT4 is ground\nT5 is ground\nX6 is free; \nX11 is free; \nf(c(T4, T5)) !=? f(c(0, X6))", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: \n\nX6 is free\nX9 is free; \nX10 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: ','(p(T4, X11), f(c(X11, s(T5)))), SCOPE: 3, CLAUSE: 5 | ','(p(T4, X11), f(c(X11, s(T5)))), SCOPE: 3, CLAUSE: 6\n\nT4 is ground\nT5 is ground\nX6 is free; \nX11 is free; \nf(c(T4, T5)) !=? f(c(0, X6))", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: ','(p(T4, X11), f(c(X11, s(T5)))), SCOPE: 3, CLAUSE: 6\n\nT4 is ground\nT5 is ground\nX6 is free; \nX11 is free; \nf(c(T4, T5)) !=? f(c(0, X6))\np(T4, X11) !=? p(0, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: f(c(T6, s(T5)))\n\nT5 is ground\nT6 is ground\nX6 is free; \nX11 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: \n\nX6 is free\nX11 is free; \nX13 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: g(T2), SCOPE: 4, CLAUSE: 3 | g(T2), SCOPE: 4, CLAUSE: 4 | ?_4\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: !_4 | g(c(T7, 0)), SCOPE: 4, CLAUSE: 4 | ?_4\n\nT7 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: g(T2), SCOPE: 4, CLAUSE: 4\n\nT2 is ground\nX15 is free; \ng(T2) !=? g(c(X15, 0))", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: ','(p(T9, X20), g(c(s(T8), X20)))\n\nT8 is ground\nT9 is ground\nX15 is free; \nX20 is free; \ng(c(T8, T9)) !=? g(c(X15, 0))", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: \n\nX15 is free\nX18 is free; \nX19 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: ','(p(T9, X20), g(c(s(T8), X20))), SCOPE: 5, CLAUSE: 5 | ','(p(T9, X20), g(c(s(T8), X20))), SCOPE: 5, CLAUSE: 6\n\nT8 is ground\nT9 is ground\nX15 is free; \nX20 is free; \ng(c(T8, T9)) !=? g(c(X15, 0))", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: ','(p(T9, X20), g(c(s(T8), X20))), SCOPE: 5, CLAUSE: 6\n\nT8 is ground\nT9 is ground\nX15 is free; \nX20 is free; \ng(c(T8, T9)) !=? g(c(X15, 0))\np(T9, X20) !=? p(0, 0)", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: g(c(s(T8), T10))\n\nT8 is ground\nT10 is ground\nX15 is free; \nX20 is free; ", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: \n\nX15 is free\nX20 is free; \nX22 is free; ", 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\nh(X4) :- ','(f(X4), g(X4)).\nand substitution\nT1 -> T2\nX4 -> T2", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 29 [arrowhead = none ];
29 -> 3;

30 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
3 -> 30 [arrowhead = none ];
30 -> 4;

31 [label="SPLIT 2", fontsize=14, style = filled, fillcolor = violet];
3 -> 31 [arrowhead = none ];
31 -> 5;

32 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
4 -> 32 [arrowhead = none ];
32 -> 6;

33 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
5 -> 33 [arrowhead = none ];
33 -> 17;

34 [label="EVAL with clause\nf(c(0, X6)) :- !.\nand substitution\nX6 -> T3\nT2 -> c(0, T3)", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 34 [arrowhead = none ];
34 -> 7;

35 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 35 [arrowhead = none ];
35 -> 8;

36 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 36 [arrowhead = none ];
36 -> 9;

37 [label="EVAL with clause\nf(c(X9, X10)) :- ','(p(X9, X11), f(c(X11, s(X10)))).\nand substitution\nX9 -> T4\nX10 -> T5\nT2 -> c(T4, T5)", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 37 [arrowhead = none ];
37 -> 11;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
8 -> 38 [arrowhead = none ];
38 -> 12;

39 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 39 [arrowhead = none ];
39 -> 10;

40 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
11 -> 40 [arrowhead = none ];
40 -> 13;

41 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (f(c(T4, T5)), f(c(0, X6)))", fontsize=14, style = filled, fillcolor = white];
13 -> 41 [arrowhead = none ];
41 -> 14;

42 [label="EVAL with clause\np(s(X13), X13).\nand substitution\nX13 -> T6\nT4 -> s(T6)\nX11 -> T6", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 42 [arrowhead = none ];
42 -> 15;

43 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 43 [arrowhead = none ];
43 -> 16;

44 [label="INSTANCE with matching:\nT2 -> c(T6, s(T5))", fontsize=14, style = filled, fillcolor = lightgrey];
15 -> 44 [arrowhead = none , style=dashed];
44 -> 4[style=dashed];

45 [label="EVAL with clause\ng(c(X15, 0)) :- !.\nand substitution\nX15 -> T7\nT2 -> c(T7, 0)", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 45 [arrowhead = none ];
45 -> 18;

46 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 46 [arrowhead = none ];
46 -> 19;

47 [label="CUT", fontsize=14, style = filled, fillcolor = lightgreen];
18 -> 47 [arrowhead = none ];
47 -> 20;

48 [label="EVAL with clause\ng(c(X18, X19)) :- ','(p(X19, X20), g(c(s(X18), X20))).\nand substitution\nX18 -> T8\nX19 -> T9\nT2 -> c(T8, T9)", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 48 [arrowhead = none ];
48 -> 22;

49 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
19 -> 49 [arrowhead = none ];
49 -> 23;

50 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
20 -> 50 [arrowhead = none ];
50 -> 21;

51 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
22 -> 51 [arrowhead = none ];
51 -> 24;

52 [label="BACKTRACK\nfor clause: p(0, 0)\nwith clash: (g(c(T8, T9)), g(c(X15, 0)))", fontsize=14, style = filled, fillcolor = white];
24 -> 52 [arrowhead = none ];
52 -> 25;

53 [label="EVAL with clause\np(s(X22), X22).\nand substitution\nX22 -> T10\nT9 -> s(T10)\nX20 -> T10", fontsize=14, style = filled, fillcolor = lightblue];
25 -> 53 [arrowhead = none ];
53 -> 26;

54 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightblue];
25 -> 54 [arrowhead = none ];
54 -> 27;

55 [label="INSTANCE with matching:\nT2 -> c(s(T8), T10)", fontsize=14, style = filled, fillcolor = lightgrey];
26 -> 55 [arrowhead = none , style=dashed];
55 -> 5[style=dashed];

}

(2) Obligation:

Clauses:

f4(c(0, T3)).
f4(c(s(T6), T5)) :- f4(c(T6, s(T5))).
g5(c(T7, 0)).
g5(c(T8, s(T10))) :- g5(c(s(T8), T10)).
h1(T2) :- f4(T2).
h1(c(T7, 0)) :- f4(c(T7, 0)).
h1(c(T8, s(T10))) :- ','(f4(c(T8, s(T10))), g5(c(s(T8), T10))).

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(T6), T5)) → U1_g(T6, T5, f4_in_g(c(T6, s(T5))))
U1_g(T6, T5, f4_out_g(c(T6, s(T5)))) → f4_out_g(c(s(T6), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T7, 0)) → U4_g(T7, f4_in_g(c(T7, 0)))
U4_g(T7, f4_out_g(c(T7, 0))) → h1_out_g(c(T7, 0))
h1_in_g(c(T8, s(T10))) → U5_g(T8, T10, f4_in_g(c(T8, s(T10))))
U5_g(T8, T10, f4_out_g(c(T8, s(T10)))) → U6_g(T8, T10, g5_in_g(c(s(T8), T10)))
g5_in_g(c(T7, 0)) → g5_out_g(c(T7, 0))
g5_in_g(c(T8, s(T10))) → U2_g(T8, T10, g5_in_g(c(s(T8), T10)))
U2_g(T8, T10, g5_out_g(c(s(T8), T10))) → g5_out_g(c(T8, s(T10)))
U6_g(T8, T10, g5_out_g(c(s(T8), T10))) → h1_out_g(c(T8, s(T10)))

The argument filtering Pi contains the following mapping:
h1_in_g(x1)  =  h1_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
f4_in_g(x1)  =  f4_in_g(x1)
c(x1, x2)  =  c(x1, x2)
0  =  0
f4_out_g(x1)  =  f4_out_g
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
h1_out_g(x1)  =  h1_out_g
U4_g(x1, x2)  =  U4_g(x2)
U5_g(x1, x2, x3)  =  U5_g(x1, x2, x3)
U6_g(x1, x2, x3)  =  U6_g(x3)
g5_in_g(x1)  =  g5_in_g(x1)
g5_out_g(x1)  =  g5_out_g
U2_g(x1, x2, x3)  =  U2_g(x3)

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(T6), T5)) → U1_g(T6, T5, f4_in_g(c(T6, s(T5))))
U1_g(T6, T5, f4_out_g(c(T6, s(T5)))) → f4_out_g(c(s(T6), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T7, 0)) → U4_g(T7, f4_in_g(c(T7, 0)))
U4_g(T7, f4_out_g(c(T7, 0))) → h1_out_g(c(T7, 0))
h1_in_g(c(T8, s(T10))) → U5_g(T8, T10, f4_in_g(c(T8, s(T10))))
U5_g(T8, T10, f4_out_g(c(T8, s(T10)))) → U6_g(T8, T10, g5_in_g(c(s(T8), T10)))
g5_in_g(c(T7, 0)) → g5_out_g(c(T7, 0))
g5_in_g(c(T8, s(T10))) → U2_g(T8, T10, g5_in_g(c(s(T8), T10)))
U2_g(T8, T10, g5_out_g(c(s(T8), T10))) → g5_out_g(c(T8, s(T10)))
U6_g(T8, T10, g5_out_g(c(s(T8), T10))) → h1_out_g(c(T8, s(T10)))

The argument filtering Pi contains the following mapping:
h1_in_g(x1)  =  h1_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
f4_in_g(x1)  =  f4_in_g(x1)
c(x1, x2)  =  c(x1, x2)
0  =  0
f4_out_g(x1)  =  f4_out_g
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
h1_out_g(x1)  =  h1_out_g
U4_g(x1, x2)  =  U4_g(x2)
U5_g(x1, x2, x3)  =  U5_g(x1, x2, x3)
U6_g(x1, x2, x3)  =  U6_g(x3)
g5_in_g(x1)  =  g5_in_g(x1)
g5_out_g(x1)  =  g5_out_g
U2_g(x1, x2, x3)  =  U2_g(x3)

(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(T6), T5)) → U1_G(T6, T5, f4_in_g(c(T6, s(T5))))
F4_IN_G(c(s(T6), T5)) → F4_IN_G(c(T6, s(T5)))
H1_IN_G(c(T7, 0)) → U4_G(T7, f4_in_g(c(T7, 0)))
H1_IN_G(c(T7, 0)) → F4_IN_G(c(T7, 0))
H1_IN_G(c(T8, s(T10))) → U5_G(T8, T10, f4_in_g(c(T8, s(T10))))
H1_IN_G(c(T8, s(T10))) → F4_IN_G(c(T8, s(T10)))
U5_G(T8, T10, f4_out_g(c(T8, s(T10)))) → U6_G(T8, T10, g5_in_g(c(s(T8), T10)))
U5_G(T8, T10, f4_out_g(c(T8, s(T10)))) → G5_IN_G(c(s(T8), T10))
G5_IN_G(c(T8, s(T10))) → U2_G(T8, T10, g5_in_g(c(s(T8), T10)))
G5_IN_G(c(T8, s(T10))) → G5_IN_G(c(s(T8), T10))

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(T6), T5)) → U1_g(T6, T5, f4_in_g(c(T6, s(T5))))
U1_g(T6, T5, f4_out_g(c(T6, s(T5)))) → f4_out_g(c(s(T6), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T7, 0)) → U4_g(T7, f4_in_g(c(T7, 0)))
U4_g(T7, f4_out_g(c(T7, 0))) → h1_out_g(c(T7, 0))
h1_in_g(c(T8, s(T10))) → U5_g(T8, T10, f4_in_g(c(T8, s(T10))))
U5_g(T8, T10, f4_out_g(c(T8, s(T10)))) → U6_g(T8, T10, g5_in_g(c(s(T8), T10)))
g5_in_g(c(T7, 0)) → g5_out_g(c(T7, 0))
g5_in_g(c(T8, s(T10))) → U2_g(T8, T10, g5_in_g(c(s(T8), T10)))
U2_g(T8, T10, g5_out_g(c(s(T8), T10))) → g5_out_g(c(T8, s(T10)))
U6_g(T8, T10, g5_out_g(c(s(T8), T10))) → h1_out_g(c(T8, s(T10)))

The argument filtering Pi contains the following mapping:
h1_in_g(x1)  =  h1_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
f4_in_g(x1)  =  f4_in_g(x1)
c(x1, x2)  =  c(x1, x2)
0  =  0
f4_out_g(x1)  =  f4_out_g
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
h1_out_g(x1)  =  h1_out_g
U4_g(x1, x2)  =  U4_g(x2)
U5_g(x1, x2, x3)  =  U5_g(x1, x2, x3)
U6_g(x1, x2, x3)  =  U6_g(x3)
g5_in_g(x1)  =  g5_in_g(x1)
g5_out_g(x1)  =  g5_out_g
U2_g(x1, x2, x3)  =  U2_g(x3)
H1_IN_G(x1)  =  H1_IN_G(x1)
U3_G(x1, x2)  =  U3_G(x2)
F4_IN_G(x1)  =  F4_IN_G(x1)
U1_G(x1, x2, x3)  =  U1_G(x3)
U4_G(x1, x2)  =  U4_G(x2)
U5_G(x1, x2, x3)  =  U5_G(x1, x2, x3)
U6_G(x1, x2, x3)  =  U6_G(x3)
G5_IN_G(x1)  =  G5_IN_G(x1)
U2_G(x1, x2, x3)  =  U2_G(x3)

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(T6), T5)) → U1_G(T6, T5, f4_in_g(c(T6, s(T5))))
F4_IN_G(c(s(T6), T5)) → F4_IN_G(c(T6, s(T5)))
H1_IN_G(c(T7, 0)) → U4_G(T7, f4_in_g(c(T7, 0)))
H1_IN_G(c(T7, 0)) → F4_IN_G(c(T7, 0))
H1_IN_G(c(T8, s(T10))) → U5_G(T8, T10, f4_in_g(c(T8, s(T10))))
H1_IN_G(c(T8, s(T10))) → F4_IN_G(c(T8, s(T10)))
U5_G(T8, T10, f4_out_g(c(T8, s(T10)))) → U6_G(T8, T10, g5_in_g(c(s(T8), T10)))
U5_G(T8, T10, f4_out_g(c(T8, s(T10)))) → G5_IN_G(c(s(T8), T10))
G5_IN_G(c(T8, s(T10))) → U2_G(T8, T10, g5_in_g(c(s(T8), T10)))
G5_IN_G(c(T8, s(T10))) → G5_IN_G(c(s(T8), T10))

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(T6), T5)) → U1_g(T6, T5, f4_in_g(c(T6, s(T5))))
U1_g(T6, T5, f4_out_g(c(T6, s(T5)))) → f4_out_g(c(s(T6), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T7, 0)) → U4_g(T7, f4_in_g(c(T7, 0)))
U4_g(T7, f4_out_g(c(T7, 0))) → h1_out_g(c(T7, 0))
h1_in_g(c(T8, s(T10))) → U5_g(T8, T10, f4_in_g(c(T8, s(T10))))
U5_g(T8, T10, f4_out_g(c(T8, s(T10)))) → U6_g(T8, T10, g5_in_g(c(s(T8), T10)))
g5_in_g(c(T7, 0)) → g5_out_g(c(T7, 0))
g5_in_g(c(T8, s(T10))) → U2_g(T8, T10, g5_in_g(c(s(T8), T10)))
U2_g(T8, T10, g5_out_g(c(s(T8), T10))) → g5_out_g(c(T8, s(T10)))
U6_g(T8, T10, g5_out_g(c(s(T8), T10))) → h1_out_g(c(T8, s(T10)))

The argument filtering Pi contains the following mapping:
h1_in_g(x1)  =  h1_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
f4_in_g(x1)  =  f4_in_g(x1)
c(x1, x2)  =  c(x1, x2)
0  =  0
f4_out_g(x1)  =  f4_out_g
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
h1_out_g(x1)  =  h1_out_g
U4_g(x1, x2)  =  U4_g(x2)
U5_g(x1, x2, x3)  =  U5_g(x1, x2, x3)
U6_g(x1, x2, x3)  =  U6_g(x3)
g5_in_g(x1)  =  g5_in_g(x1)
g5_out_g(x1)  =  g5_out_g
U2_g(x1, x2, x3)  =  U2_g(x3)
H1_IN_G(x1)  =  H1_IN_G(x1)
U3_G(x1, x2)  =  U3_G(x2)
F4_IN_G(x1)  =  F4_IN_G(x1)
U1_G(x1, x2, x3)  =  U1_G(x3)
U4_G(x1, x2)  =  U4_G(x2)
U5_G(x1, x2, x3)  =  U5_G(x1, x2, x3)
U6_G(x1, x2, x3)  =  U6_G(x3)
G5_IN_G(x1)  =  G5_IN_G(x1)
U2_G(x1, x2, x3)  =  U2_G(x3)

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(T8, s(T10))) → G5_IN_G(c(s(T8), T10))

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(T6), T5)) → U1_g(T6, T5, f4_in_g(c(T6, s(T5))))
U1_g(T6, T5, f4_out_g(c(T6, s(T5)))) → f4_out_g(c(s(T6), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T7, 0)) → U4_g(T7, f4_in_g(c(T7, 0)))
U4_g(T7, f4_out_g(c(T7, 0))) → h1_out_g(c(T7, 0))
h1_in_g(c(T8, s(T10))) → U5_g(T8, T10, f4_in_g(c(T8, s(T10))))
U5_g(T8, T10, f4_out_g(c(T8, s(T10)))) → U6_g(T8, T10, g5_in_g(c(s(T8), T10)))
g5_in_g(c(T7, 0)) → g5_out_g(c(T7, 0))
g5_in_g(c(T8, s(T10))) → U2_g(T8, T10, g5_in_g(c(s(T8), T10)))
U2_g(T8, T10, g5_out_g(c(s(T8), T10))) → g5_out_g(c(T8, s(T10)))
U6_g(T8, T10, g5_out_g(c(s(T8), T10))) → h1_out_g(c(T8, s(T10)))

The argument filtering Pi contains the following mapping:
h1_in_g(x1)  =  h1_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
f4_in_g(x1)  =  f4_in_g(x1)
c(x1, x2)  =  c(x1, x2)
0  =  0
f4_out_g(x1)  =  f4_out_g
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
h1_out_g(x1)  =  h1_out_g
U4_g(x1, x2)  =  U4_g(x2)
U5_g(x1, x2, x3)  =  U5_g(x1, x2, x3)
U6_g(x1, x2, x3)  =  U6_g(x3)
g5_in_g(x1)  =  g5_in_g(x1)
g5_out_g(x1)  =  g5_out_g
U2_g(x1, x2, x3)  =  U2_g(x3)
G5_IN_G(x1)  =  G5_IN_G(x1)

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(T8, s(T10))) → G5_IN_G(c(s(T8), T10))

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(T8, s(T10))) → G5_IN_G(c(s(T8), T10))

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(T8, s(T10))) → G5_IN_G(c(s(T8), T10))

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(T6), T5)) → F4_IN_G(c(T6, 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(T6), T5)) → U1_g(T6, T5, f4_in_g(c(T6, s(T5))))
U1_g(T6, T5, f4_out_g(c(T6, s(T5)))) → f4_out_g(c(s(T6), T5))
U3_g(T2, f4_out_g(T2)) → h1_out_g(T2)
h1_in_g(c(T7, 0)) → U4_g(T7, f4_in_g(c(T7, 0)))
U4_g(T7, f4_out_g(c(T7, 0))) → h1_out_g(c(T7, 0))
h1_in_g(c(T8, s(T10))) → U5_g(T8, T10, f4_in_g(c(T8, s(T10))))
U5_g(T8, T10, f4_out_g(c(T8, s(T10)))) → U6_g(T8, T10, g5_in_g(c(s(T8), T10)))
g5_in_g(c(T7, 0)) → g5_out_g(c(T7, 0))
g5_in_g(c(T8, s(T10))) → U2_g(T8, T10, g5_in_g(c(s(T8), T10)))
U2_g(T8, T10, g5_out_g(c(s(T8), T10))) → g5_out_g(c(T8, s(T10)))
U6_g(T8, T10, g5_out_g(c(s(T8), T10))) → h1_out_g(c(T8, s(T10)))

The argument filtering Pi contains the following mapping:
h1_in_g(x1)  =  h1_in_g(x1)
U3_g(x1, x2)  =  U3_g(x2)
f4_in_g(x1)  =  f4_in_g(x1)
c(x1, x2)  =  c(x1, x2)
0  =  0
f4_out_g(x1)  =  f4_out_g
s(x1)  =  s(x1)
U1_g(x1, x2, x3)  =  U1_g(x3)
h1_out_g(x1)  =  h1_out_g
U4_g(x1, x2)  =  U4_g(x2)
U5_g(x1, x2, x3)  =  U5_g(x1, x2, x3)
U6_g(x1, x2, x3)  =  U6_g(x3)
g5_in_g(x1)  =  g5_in_g(x1)
g5_out_g(x1)  =  g5_out_g
U2_g(x1, x2, x3)  =  U2_g(x3)
F4_IN_G(x1)  =  F4_IN_G(x1)

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(T6), T5)) → F4_IN_G(c(T6, 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(T6), T5)) → F4_IN_G(c(T6, 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(T6), T5)) → F4_IN_G(c(T6, 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