Left Termination of the query pattern less_in_2(a, g) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

less(0, s(X1)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

less(a,g).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

less1(s(s(T21)), s(s(T20))) :- less1(T21, T20).
less1(s(s(T40)), s(s(T39))) :- less1(T40, T39).

Clauses:

lessc1(0, s(T4)).
lessc1(s(0), s(s(T14))).
lessc1(s(s(T21)), s(s(T20))) :- lessc1(T21, T20).
lessc1(s(0), s(s(T33))).
lessc1(s(s(T40)), s(s(T39))) :- lessc1(T40, T39).

Afs:

less1(x1, x2)  =  less1(x2)

(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:
less1_in: (f,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

LESS1_IN_AG(s(s(T21)), s(s(T20))) → U1_AG(T21, T20, less1_in_ag(T21, T20))
LESS1_IN_AG(s(s(T21)), s(s(T20))) → LESS1_IN_AG(T21, T20)
LESS1_IN_AG(s(s(T40)), s(s(T39))) → U2_AG(T40, T39, less1_in_ag(T40, T39))

R is empty.
The argument filtering Pi contains the following mapping:
less1_in_ag(x1, x2)  =  less1_in_ag(x2)
s(x1)  =  s(x1)
LESS1_IN_AG(x1, x2)  =  LESS1_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
U2_AG(x1, x2, x3)  =  U2_AG(x2, x3)

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:

LESS1_IN_AG(s(s(T21)), s(s(T20))) → U1_AG(T21, T20, less1_in_ag(T21, T20))
LESS1_IN_AG(s(s(T21)), s(s(T20))) → LESS1_IN_AG(T21, T20)
LESS1_IN_AG(s(s(T40)), s(s(T39))) → U2_AG(T40, T39, less1_in_ag(T40, T39))

R is empty.
The argument filtering Pi contains the following mapping:
less1_in_ag(x1, x2)  =  less1_in_ag(x2)
s(x1)  =  s(x1)
LESS1_IN_AG(x1, x2)  =  LESS1_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
U2_AG(x1, x2, x3)  =  U2_AG(x2, x3)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.

(6) Obligation:

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

LESS1_IN_AG(s(s(T21)), s(s(T20))) → LESS1_IN_AG(T21, T20)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
LESS1_IN_AG(x1, x2)  =  LESS1_IN_AG(x2)

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:

LESS1_IN_AG(s(s(T20))) → LESS1_IN_AG(T20)

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:

• LESS1_IN_AG(s(s(T20))) → LESS1_IN_AG(T20)
  The graph contains the following edges 1 > 1

(10) YES