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

(0) Obligation:

Clauses:

f(X) :- g(s(s(s(X)))).
f(s(X)) :- f(X).
g(s(s(s(s(X))))) :- f(X).

Query: f(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="A: f(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: f(T1), SCOPE: 1, CLAUSE: 0 | f(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: g(s(s(s(T3)))) | f(T3), SCOPE: 1, CLAUSE: 1\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: g(s(s(s(T3)))), SCOPE: 2, CLAUSE: 2 | ?_2 | f(T3), SCOPE: 1, CLAUSE: 1\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: g(s(s(s(T3)))), SCOPE: 2, CLAUSE: 2\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: ?_2 | f(T3), SCOPE: 1, CLAUSE: 1\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: f(T8)\n\nT8 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: f(T3), SCOPE: 1, CLAUSE: 1\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: f(T12)\n\nT12 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 12 [arrowhead = none ];
12 -> 2;

13 [label="ONLY EVAL with clause\nf(X2) :- g(s(s(s(X2)))).\nand substitutionT1 -> T3,\nX2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 13 [arrowhead = none ];
13 -> 3;

14 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 14 [arrowhead = none ];
14 -> 4;

15 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 15 [arrowhead = none ];
15 -> 5;

16 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 16 [arrowhead = none ];
16 -> 6;

17 [label="EVAL with clause\ng(s(s(s(s(X7))))) :- f(X7).\nand substitutionX7 -> T8,\nT3 -> s(T8)", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 17 [arrowhead = none ];
17 -> 7;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 18 [arrowhead = none ];
18 -> 8;

19 [label="FAILURE", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 19 [arrowhead = none ];
19 -> 9;

20 [label="INSTANCE with matching:\nT1 -> T8", fontsize=14, style = filled, fillcolor = lightgrey];
7 -> 20 [arrowhead = none , style=dashed];
20 -> 1[style=dashed];

21 [label="EVAL with clause\nf(s(X11)) :- f(X11).\nand substitutionX11 -> T12,\nT3 -> s(T12)", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 21 [arrowhead = none ];
21 -> 10;

22 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 22 [arrowhead = none ];
22 -> 11;

23 [label="INSTANCE with matching:\nT1 -> T12", fontsize=14, style = filled, fillcolor = lightgrey];
10 -> 23 [arrowhead = none , style=dashed];
23 -> 1[style=dashed];

}

(2) Obligation:

Triples:

fA(s(X1)) :- fA(X1).
fA(s(X1)) :- fA(X1).

Clauses:

fcA(s(X1)) :- fcA(X1).
fcA(s(X1)) :- fcA(X1).

Afs:

fA(x1)  =  fA(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:
fA_in: (b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

FA_IN_G(s(X1)) → U1_G(X1, fA_in_g(X1))
FA_IN_G(s(X1)) → FA_IN_G(X1)

R is empty.
Pi is empty.
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:

FA_IN_G(s(X1)) → U1_G(X1, fA_in_g(X1))
FA_IN_G(s(X1)) → FA_IN_G(X1)

R is empty.
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 1 less node.

(6) Obligation:

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

FA_IN_G(s(X1)) → FA_IN_G(X1)

R is empty.
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:

FA_IN_G(s(X1)) → FA_IN_G(X1)

R is empty.
Q is empty.
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:

• FA_IN_G(s(X1)) → FA_IN_G(X1)
  The graph contains the following edges 1 > 1

(10) YES