Left Termination of the query pattern num_in_1(g) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

num(0).
num(s(X)) :- num(X).

Queries:

num(g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

num1(0).
num1(s(0)).
num1(s(s(T6))) :- num1(T6).

Queries:

num1(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:
num1_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

num1_in_g(0) → num1_out_g(0)
num1_in_g(s(0)) → num1_out_g(s(0))
num1_in_g(s(s(T6))) → U1_g(T6, num1_in_g(T6))
U1_g(T6, num1_out_g(T6)) → num1_out_g(s(s(T6)))

The argument filtering Pi contains the following mapping:
num1_in_g(x1)  =  num1_in_g(x1)
0  =  0
num1_out_g(x1)  =  num1_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

num1_in_g(0) → num1_out_g(0)
num1_in_g(s(0)) → num1_out_g(s(0))
num1_in_g(s(s(T6))) → U1_g(T6, num1_in_g(T6))
U1_g(T6, num1_out_g(T6)) → num1_out_g(s(s(T6)))

The argument filtering Pi contains the following mapping:
num1_in_g(x1)  =  num1_in_g(x1)
0  =  0
num1_out_g(x1)  =  num1_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)

(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:

NUM1_IN_G(s(s(T6))) → U1_G(T6, num1_in_g(T6))
NUM1_IN_G(s(s(T6))) → NUM1_IN_G(T6)

The TRS R consists of the following rules:

num1_in_g(0) → num1_out_g(0)
num1_in_g(s(0)) → num1_out_g(s(0))
num1_in_g(s(s(T6))) → U1_g(T6, num1_in_g(T6))
U1_g(T6, num1_out_g(T6)) → num1_out_g(s(s(T6)))

The argument filtering Pi contains the following mapping:
num1_in_g(x1)  =  num1_in_g(x1)
0  =  0
num1_out_g(x1)  =  num1_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
NUM1_IN_G(x1)  =  NUM1_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(6) Obligation:

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

NUM1_IN_G(s(s(T6))) → U1_G(T6, num1_in_g(T6))
NUM1_IN_G(s(s(T6))) → NUM1_IN_G(T6)

The TRS R consists of the following rules:

num1_in_g(0) → num1_out_g(0)
num1_in_g(s(0)) → num1_out_g(s(0))
num1_in_g(s(s(T6))) → U1_g(T6, num1_in_g(T6))
U1_g(T6, num1_out_g(T6)) → num1_out_g(s(s(T6)))

The argument filtering Pi contains the following mapping:
num1_in_g(x1)  =  num1_in_g(x1)
0  =  0
num1_out_g(x1)  =  num1_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
NUM1_IN_G(x1)  =  NUM1_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

NUM1_IN_G(s(s(T6))) → NUM1_IN_G(T6)

The TRS R consists of the following rules:

num1_in_g(0) → num1_out_g(0)
num1_in_g(s(0)) → num1_out_g(s(0))
num1_in_g(s(s(T6))) → U1_g(T6, num1_in_g(T6))
U1_g(T6, num1_out_g(T6)) → num1_out_g(s(s(T6)))

The argument filtering Pi contains the following mapping:
num1_in_g(x1)  =  num1_in_g(x1)
0  =  0
num1_out_g(x1)  =  num1_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
NUM1_IN_G(x1)  =  NUM1_IN_G(x1)

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

(9) UsableRulesProof (EQUIVALENT transformation)

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

(10) Obligation:

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

NUM1_IN_G(s(s(T6))) → NUM1_IN_G(T6)

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

(11) PiDPToQDPProof (EQUIVALENT transformation)

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

(12) Obligation:

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

NUM1_IN_G(s(s(T6))) → NUM1_IN_G(T6)

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

(13) 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:

• NUM1_IN_G(s(s(T6))) → NUM1_IN_G(T6)
  The graph contains the following edges 1 > 1

(14) YES