Left Termination of the query pattern len_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

len([], 0).
len(Xs, s(N)) :- ','(no(empty(Xs)), ','(tail(Xs, Ys), len(Ys, N))).
tail([], []).
tail(.(X, Xs), Xs).
empty([]).
no(X) :- ','(X, ','(!, failure(a))).
no(X1).
failure(b).

Queries:

len(g,a).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

len1(.(T21, T22), s(T10)) :- len1(T22, T10).

Clauses:

lenc1([], 0).
lenc1(.(T21, T22), s(T10)) :- lenc1(T22, T10).

Afs:

len1(x1, x2)  =  len1(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
len1_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

LEN1_IN_GA(.(T21, T22), s(T10)) → U1_GA(T21, T22, T10, len1_in_ga(T22, T10))
LEN1_IN_GA(.(T21, T22), s(T10)) → LEN1_IN_GA(T22, T10)

R is empty.
The argument filtering Pi contains the following mapping:
len1_in_ga(x1, x2)  =  len1_in_ga(x1)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x2, x4)

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:

LEN1_IN_GA(.(T21, T22), s(T10)) → U1_GA(T21, T22, T10, len1_in_ga(T22, T10))
LEN1_IN_GA(.(T21, T22), s(T10)) → LEN1_IN_GA(T22, T10)

R is empty.
The argument filtering Pi contains the following mapping:
len1_in_ga(x1, x2)  =  len1_in_ga(x1)
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)
U1_GA(x1, x2, x3, x4)  =  U1_GA(x1, x2, 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 1 less node.

(6) Obligation:

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

LEN1_IN_GA(.(T21, T22), s(T10)) → LEN1_IN_GA(T22, T10)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
s(x1)  =  s(x1)
LEN1_IN_GA(x1, x2)  =  LEN1_IN_GA(x1)

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

(7) PiDPToQDPProof (SOUND 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:

LEN1_IN_GA(.(T21, T22)) → LEN1_IN_GA(T22)

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:

• LEN1_IN_GA(.(T21, T22)) → LEN1_IN_GA(T22)
  The graph contains the following edges 1 > 1

(10) YES