Left Termination of the query pattern f_in_1(g) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 PiDP
    ↳7 UsableRulesProof (⇔)
    ↳8 PiDP
    ↳9 PiDPToQDPProof (⇔)
    ↳10 QDP
    ↳11 MRRProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 PrologToPiTRSProof (⇐)
    ↳16 PiTRS
    ↳17 DependencyPairsProof (⇔)
    ↳18 PiDP
    ↳19 DependencyGraphProof (⇔)
    ↳20 PiDP
    ↳21 UsableRulesProof (⇔)
    ↳22 PiDP
    ↳23 PiDPToQDPProof (⇔)
    ↳24 QDP

(0) Obligation:

Clauses:

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

Queries:

f(g).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
f_in: (b)
g_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

The argument filtering Pi contains the following mapping:
f_in_g(x1)  =  f_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
g_in_g(x1)  =  g_in_g(x1)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x2)
U2_g(x1, x2)  =  U2_g(x2)
f_out_g(x1)  =  f_out_g
g_out_g(x1)  =  g_out_g

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

The argument filtering Pi contains the following mapping:
f_in_g(x1)  =  f_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
g_in_g(x1)  =  g_in_g(x1)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x2)
U2_g(x1, x2)  =  U2_g(x2)
f_out_g(x1)  =  f_out_g
g_out_g(x1)  =  g_out_g

(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:

F_IN_G(X) → U1_G(X, g_in_g(s(s(s(X)))))
F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → U3_G(X, f_in_g(X))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → U2_G(X, f_in_g(X))
F_IN_G(s(X)) → F_IN_G(X)

The TRS R consists of the following rules:

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

The argument filtering Pi contains the following mapping:
f_in_g(x1)  =  f_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
g_in_g(x1)  =  g_in_g(x1)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x2)
U2_g(x1, x2)  =  U2_g(x2)
f_out_g(x1)  =  f_out_g
g_out_g(x1)  =  g_out_g
F_IN_G(x1)  =  F_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
G_IN_G(x1)  =  G_IN_G(x1)
U3_G(x1, x2)  =  U3_G(x2)
U2_G(x1, x2)  =  U2_G(x2)

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

(4) Obligation:

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

F_IN_G(X) → U1_G(X, g_in_g(s(s(s(X)))))
F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → U3_G(X, f_in_g(X))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → U2_G(X, f_in_g(X))
F_IN_G(s(X)) → F_IN_G(X)

The TRS R consists of the following rules:

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

The argument filtering Pi contains the following mapping:
f_in_g(x1)  =  f_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
g_in_g(x1)  =  g_in_g(x1)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x2)
U2_g(x1, x2)  =  U2_g(x2)
f_out_g(x1)  =  f_out_g
g_out_g(x1)  =  g_out_g
F_IN_G(x1)  =  F_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
G_IN_G(x1)  =  G_IN_G(x1)
U3_G(x1, x2)  =  U3_G(x2)
U2_G(x1, x2)  =  U2_G(x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → F_IN_G(X)

The TRS R consists of the following rules:

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

The argument filtering Pi contains the following mapping:
f_in_g(x1)  =  f_in_g(x1)
U1_g(x1, x2)  =  U1_g(x2)
g_in_g(x1)  =  g_in_g(x1)
s(x1)  =  s(x1)
U3_g(x1, x2)  =  U3_g(x2)
U2_g(x1, x2)  =  U2_g(x2)
f_out_g(x1)  =  f_out_g
g_out_g(x1)  =  g_out_g
F_IN_G(x1)  =  F_IN_G(x1)
G_IN_G(x1)  =  G_IN_G(x1)

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

(7) UsableRulesProof (EQUIVALENT transformation)

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

(8) Obligation:

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

F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → F_IN_G(X)

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

(9) PiDPToQDPProof (EQUIVALENT transformation)

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

(10) Obligation:

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

F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → F_IN_G(X)

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

(11) 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:

F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → F_IN_G(X)


Used ordering: Polynomial interpretation [POLO]:

POL(F_IN_G(x1)) = 7 + x1   
POL(G_IN_G(x1)) = x1   
POL(s(x1)) = 2 + x1   

(12) Obligation:

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
f_in: (b)
g_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(16) Obligation:

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

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

Pi is empty.

(17) 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:

F_IN_G(X) → U1_G(X, g_in_g(s(s(s(X)))))
F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → U3_G(X, f_in_g(X))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → U2_G(X, f_in_g(X))
F_IN_G(s(X)) → F_IN_G(X)

The TRS R consists of the following rules:

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

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

(18) Obligation:

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

F_IN_G(X) → U1_G(X, g_in_g(s(s(s(X)))))
F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → U3_G(X, f_in_g(X))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → U2_G(X, f_in_g(X))
F_IN_G(s(X)) → F_IN_G(X)

The TRS R consists of the following rules:

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(20) Obligation:

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

F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → F_IN_G(X)

The TRS R consists of the following rules:

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

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

(21) UsableRulesProof (EQUIVALENT transformation)

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

(22) Obligation:

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

F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → F_IN_G(X)

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

(23) PiDPToQDPProof (EQUIVALENT transformation)

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

(24) Obligation:

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

F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → F_IN_G(X)

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