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

(0) Obligation:

Clauses:

even(X) :- ','(eq(Y, f(e, f(o, Y))), c(Y, X)).
c(f(e, X1), 0).
c(f(X2, X), s(Y)) :- c(X, Y).
eq(X, X).

Query: even(g)

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
even_in: (b)
c_in: (f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

even_in_g(X) → U1_g(X, eq_in_aa(Y, f(e, f(o, Y))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_g(X, eq_out_aa(Y, f(e, f(o, Y)))) → U2_g(X, c_in_ag(Y, X))
c_in_ag(f(e, X1), 0) → c_out_ag(f(e, X1), 0)
c_in_ag(f(X2, X), s(Y)) → U3_ag(X2, X, Y, c_in_ag(X, Y))
U3_ag(X2, X, Y, c_out_ag(X, Y)) → c_out_ag(f(X2, X), s(Y))
U2_g(X, c_out_ag(Y, X)) → even_out_g(X)

The argument filtering Pi contains the following mapping:
even_in_g(x1)  =  even_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_g(x1, x2)  =  U2_g(x2)
c_in_ag(x1, x2)  =  c_in_ag(x2)
0  =  0
c_out_ag(x1, x2)  =  c_out_ag
s(x1)  =  s(x1)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
even_out_g(x1)  =  even_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:

even_in_g(X) → U1_g(X, eq_in_aa(Y, f(e, f(o, Y))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_g(X, eq_out_aa(Y, f(e, f(o, Y)))) → U2_g(X, c_in_ag(Y, X))
c_in_ag(f(e, X1), 0) → c_out_ag(f(e, X1), 0)
c_in_ag(f(X2, X), s(Y)) → U3_ag(X2, X, Y, c_in_ag(X, Y))
U3_ag(X2, X, Y, c_out_ag(X, Y)) → c_out_ag(f(X2, X), s(Y))
U2_g(X, c_out_ag(Y, X)) → even_out_g(X)

The argument filtering Pi contains the following mapping:
even_in_g(x1)  =  even_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_g(x1, x2)  =  U2_g(x2)
c_in_ag(x1, x2)  =  c_in_ag(x2)
0  =  0
c_out_ag(x1, x2)  =  c_out_ag
s(x1)  =  s(x1)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
even_out_g(x1)  =  even_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:

EVEN_IN_G(X) → U1_G(X, eq_in_aa(Y, f(e, f(o, Y))))
EVEN_IN_G(X) → EQ_IN_AA(Y, f(e, f(o, Y)))
U1_G(X, eq_out_aa(Y, f(e, f(o, Y)))) → U2_G(X, c_in_ag(Y, X))
U1_G(X, eq_out_aa(Y, f(e, f(o, Y)))) → C_IN_AG(Y, X)
C_IN_AG(f(X2, X), s(Y)) → U3_AG(X2, X, Y, c_in_ag(X, Y))
C_IN_AG(f(X2, X), s(Y)) → C_IN_AG(X, Y)

The TRS R consists of the following rules:

even_in_g(X) → U1_g(X, eq_in_aa(Y, f(e, f(o, Y))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_g(X, eq_out_aa(Y, f(e, f(o, Y)))) → U2_g(X, c_in_ag(Y, X))
c_in_ag(f(e, X1), 0) → c_out_ag(f(e, X1), 0)
c_in_ag(f(X2, X), s(Y)) → U3_ag(X2, X, Y, c_in_ag(X, Y))
U3_ag(X2, X, Y, c_out_ag(X, Y)) → c_out_ag(f(X2, X), s(Y))
U2_g(X, c_out_ag(Y, X)) → even_out_g(X)

The argument filtering Pi contains the following mapping:
even_in_g(x1)  =  even_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_g(x1, x2)  =  U2_g(x2)
c_in_ag(x1, x2)  =  c_in_ag(x2)
0  =  0
c_out_ag(x1, x2)  =  c_out_ag
s(x1)  =  s(x1)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
even_out_g(x1)  =  even_out_g
EVEN_IN_G(x1)  =  EVEN_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
EQ_IN_AA(x1, x2)  =  EQ_IN_AA
U2_G(x1, x2)  =  U2_G(x2)
C_IN_AG(x1, x2)  =  C_IN_AG(x2)
U3_AG(x1, x2, x3, x4)  =  U3_AG(x4)

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

(4) Obligation:

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

EVEN_IN_G(X) → U1_G(X, eq_in_aa(Y, f(e, f(o, Y))))
EVEN_IN_G(X) → EQ_IN_AA(Y, f(e, f(o, Y)))
U1_G(X, eq_out_aa(Y, f(e, f(o, Y)))) → U2_G(X, c_in_ag(Y, X))
U1_G(X, eq_out_aa(Y, f(e, f(o, Y)))) → C_IN_AG(Y, X)
C_IN_AG(f(X2, X), s(Y)) → U3_AG(X2, X, Y, c_in_ag(X, Y))
C_IN_AG(f(X2, X), s(Y)) → C_IN_AG(X, Y)

The TRS R consists of the following rules:

even_in_g(X) → U1_g(X, eq_in_aa(Y, f(e, f(o, Y))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_g(X, eq_out_aa(Y, f(e, f(o, Y)))) → U2_g(X, c_in_ag(Y, X))
c_in_ag(f(e, X1), 0) → c_out_ag(f(e, X1), 0)
c_in_ag(f(X2, X), s(Y)) → U3_ag(X2, X, Y, c_in_ag(X, Y))
U3_ag(X2, X, Y, c_out_ag(X, Y)) → c_out_ag(f(X2, X), s(Y))
U2_g(X, c_out_ag(Y, X)) → even_out_g(X)

The argument filtering Pi contains the following mapping:
even_in_g(x1)  =  even_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_g(x1, x2)  =  U2_g(x2)
c_in_ag(x1, x2)  =  c_in_ag(x2)
0  =  0
c_out_ag(x1, x2)  =  c_out_ag
s(x1)  =  s(x1)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
even_out_g(x1)  =  even_out_g
EVEN_IN_G(x1)  =  EVEN_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
EQ_IN_AA(x1, x2)  =  EQ_IN_AA
U2_G(x1, x2)  =  U2_G(x2)
C_IN_AG(x1, x2)  =  C_IN_AG(x2)
U3_AG(x1, x2, x3, x4)  =  U3_AG(x4)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.

(6) Obligation:

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

C_IN_AG(f(X2, X), s(Y)) → C_IN_AG(X, Y)

The TRS R consists of the following rules:

even_in_g(X) → U1_g(X, eq_in_aa(Y, f(e, f(o, Y))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_g(X, eq_out_aa(Y, f(e, f(o, Y)))) → U2_g(X, c_in_ag(Y, X))
c_in_ag(f(e, X1), 0) → c_out_ag(f(e, X1), 0)
c_in_ag(f(X2, X), s(Y)) → U3_ag(X2, X, Y, c_in_ag(X, Y))
U3_ag(X2, X, Y, c_out_ag(X, Y)) → c_out_ag(f(X2, X), s(Y))
U2_g(X, c_out_ag(Y, X)) → even_out_g(X)

The argument filtering Pi contains the following mapping:
even_in_g(x1)  =  even_in_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_g(x1, x2)  =  U2_g(x2)
c_in_ag(x1, x2)  =  c_in_ag(x2)
0  =  0
c_out_ag(x1, x2)  =  c_out_ag
s(x1)  =  s(x1)
U3_ag(x1, x2, x3, x4)  =  U3_ag(x4)
even_out_g(x1)  =  even_out_g
C_IN_AG(x1, x2)  =  C_IN_AG(x2)

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:

C_IN_AG(f(X2, X), s(Y)) → C_IN_AG(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
C_IN_AG(x1, x2)  =  C_IN_AG(x2)

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

(9) PiDPToQDPProof (SOUND 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:

C_IN_AG(s(Y)) → C_IN_AG(Y)

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

(11) 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:

• C_IN_AG(s(Y)) → C_IN_AG(Y)
  The graph contains the following edges 1 > 1

(12) YES