NO

Prover = TRS(tech=GUIDED_UNF, nb_unfoldings=unlimited, unfold_variables=true, strategy=LEFTMOST_NE)

** BEGIN proof argument **
The following rule was generated while unfolding the analyzed TRS:
[iteration = 3] f(g(s(0)),g(s(0)),_0) -> f(g(s(0)),_0,g(s(0)))
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {_0->g(s(0))} and theta2 = {}.
We have r|p = f(g(s(0)),_0,g(s(0))) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = f(g(s(0)),g(s(0)),g(s(0))) loops w.r.t. the analyzed TRS.
** END proof argument **

** BEGIN proof description **
## Searching for a generalized rewrite rule
(a rule whose right-hand side contains a variable
that does not occur in the left-hand side)...
No generalized rewrite rule found!

## Searching for a loop by unfolding (unfolding of variable subterms: ON)...
# Iteration 0: no loop detected, 1 unfolded rule generated.
# Iteration 1: no loop detected, 5 unfolded rules generated.
# Iteration 2: no loop detected, 18 unfolded rules generated.
# Iteration 3: loop detected, 11 unfolded rules generated.
Here is the successful unfolding. Let IR be the TRS under analysis.
L0 = f^#(g(_0),s(0),_1) -> f^#(g(s(0)),_1,g(_0)) is in U_IR^0.
Let p0 = [0, 0].
The subterm at position p0 in the left-hand side of the rule of L0 unifies with
the subterm at position p0 in the right-hand side of the rule of L0.
==> L1 = f^#(g(s(0)),s(0),_0) -> f^#(g(s(0)),_0,g(s(0))) is in U_IR^1.
Let p1 = [1, 0].
We unfold the rule of L1 backwards at position p1
with the rule g(0) -> 0.
==> L2 = f^#(g(s(0)),s(g(0)),_0) -> f^#(g(s(0)),_0,g(s(0))) is in U_IR^2.
Let p2 = [1].
We unfold the rule of L2 backwards at position p2
with the rule g(s(_0)) -> s(g(_0)).
==> L3 = f^#(g(s(0)),g(s(0)),g(s(0))) -> f^#(g(s(0)),g(s(0)),g(s(0))) is in U_IR^3.

** END proof description **

Proof stopped at iteration 3
Number of unfolded rules generated by this proof = 35
Number of unfolded rules generated by all the parallel proofs = 36