Left Termination of the query pattern even_in_1(g) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

even(0).
even(s(X)) :- odd(X).
odd(s(X)) :- even(X).

Queries:

even(g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

even1(0).
even1(s(s(T6))) :- even1(T6).

Queries:

even1(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:
even1_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

even1_in_g(0) → even1_out_g(0)
even1_in_g(s(s(T6))) → U1_g(T6, even1_in_g(T6))
U1_g(T6, even1_out_g(T6)) → even1_out_g(s(s(T6)))

The argument filtering Pi contains the following mapping:
even1_in_g(x1)  =  even1_in_g(x1)
0  =  0
even1_out_g(x1)  =  even1_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:

even1_in_g(0) → even1_out_g(0)
even1_in_g(s(s(T6))) → U1_g(T6, even1_in_g(T6))
U1_g(T6, even1_out_g(T6)) → even1_out_g(s(s(T6)))

The argument filtering Pi contains the following mapping:
even1_in_g(x1)  =  even1_in_g(x1)
0  =  0
even1_out_g(x1)  =  even1_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:

EVEN1_IN_G(s(s(T6))) → U1_G(T6, even1_in_g(T6))
EVEN1_IN_G(s(s(T6))) → EVEN1_IN_G(T6)

The TRS R consists of the following rules:

even1_in_g(0) → even1_out_g(0)
even1_in_g(s(s(T6))) → U1_g(T6, even1_in_g(T6))
U1_g(T6, even1_out_g(T6)) → even1_out_g(s(s(T6)))

The argument filtering Pi contains the following mapping:
even1_in_g(x1)  =  even1_in_g(x1)
0  =  0
even1_out_g(x1)  =  even1_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
EVEN1_IN_G(x1)  =  EVEN1_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:

EVEN1_IN_G(s(s(T6))) → U1_G(T6, even1_in_g(T6))
EVEN1_IN_G(s(s(T6))) → EVEN1_IN_G(T6)

The TRS R consists of the following rules:

even1_in_g(0) → even1_out_g(0)
even1_in_g(s(s(T6))) → U1_g(T6, even1_in_g(T6))
U1_g(T6, even1_out_g(T6)) → even1_out_g(s(s(T6)))

The argument filtering Pi contains the following mapping:
even1_in_g(x1)  =  even1_in_g(x1)
0  =  0
even1_out_g(x1)  =  even1_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
EVEN1_IN_G(x1)  =  EVEN1_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:

EVEN1_IN_G(s(s(T6))) → EVEN1_IN_G(T6)

The TRS R consists of the following rules:

even1_in_g(0) → even1_out_g(0)
even1_in_g(s(s(T6))) → U1_g(T6, even1_in_g(T6))
U1_g(T6, even1_out_g(T6)) → even1_out_g(s(s(T6)))

The argument filtering Pi contains the following mapping:
even1_in_g(x1)  =  even1_in_g(x1)
0  =  0
even1_out_g(x1)  =  even1_out_g
s(x1)  =  s(x1)
U1_g(x1, x2)  =  U1_g(x2)
EVEN1_IN_G(x1)  =  EVEN1_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:

EVEN1_IN_G(s(s(T6))) → EVEN1_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:

EVEN1_IN_G(s(s(T6))) → EVEN1_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:

• EVEN1_IN_G(s(s(T6))) → EVEN1_IN_G(T6)
  The graph contains the following edges 1 > 1

(14) YES