NO

Prover = TRS(tech=GUIDED_UNF_TRIPLES, nb_unfoldings=unlimited, unfold_variables=false, max_nb_coefficients=12, max_nb_unfolded_rules=-1, strategy=LEFTMOST_NE)

** BEGIN proof argument **
The following rule was generated while unfolding the analyzed TRS:
[iteration = 5] f(g(0,_0),s(0),g(0,_0)) -> f(g(0,_0),_0,g(0,_0))
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {_0->s(0)} and theta2 = {}.
We have r|p = f(g(0,_0),_0,g(0,_0)) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = f(g(0,s(0)),s(0),g(0,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!

## Applying the DP framework...
## Round 1:
  ## DP problem: 
  Dependency pairs = [f^#(0,s(0),_0) -> f^#(_0,+(_0,_0),_0)]
  TRS = {+(_0,0) -> _0, +(_0,s(_1)) -> s(+(_0,_1)), f(0,s(0),_0) -> f(_0,+(_0,_0),_0), g(_0,_1) -> _0, g(_0,_1) -> _1}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... Too many coefficients (20)! Aborting!
    ## Trying with lexicographic path orders... Failed!
    ## Trying to prove nontermination by unfolding the dependency pairs with the rules of the TRS
      # max_depth=20, unfold_variables=false:
        # Iteration 0: nontermination not detected, 1 unfolded rule generated.
        # Iteration 1: nontermination not detected, 1 unfolded rule generated.
        # Iteration 2: nontermination not detected, 7 unfolded rules generated.
        # Iteration 3: nontermination not detected, 65 unfolded rules generated.
        # Iteration 4: nontermination not detected, 805 unfolded rules generated.
        # Iteration 5: nontermination detected, 6983 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = f^#(0,s(0),_0) -> f^#(_0,+(_0,_0),_0) [trans] is in U_IR^0.
        We build a unit triple from L0.
        ==> L1 = f^#(0,s(0),_0) -> f^#(_0,+(_0,_0),_0) [unit] is in U_IR^1.
        Let p1 = [0].
        We unfold the rule of L1 backwards at position p1
        with the rule g(_0,_1) -> _0.
        ==> L2 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),+(g(0,_0),g(0,_0)),g(0,_0)) [unit] is in U_IR^2.
        Let p2 = [1, 0].
        We unfold the rule of L2 forwards at position p2
        with the rule g(_0,_1) -> _1.
        ==> L3 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),+(_0,g(0,_0)),g(0,_0)) [unit] is in U_IR^3.
        Let p3 = [1, 1].
        We unfold the rule of L3 forwards at position p3
        with the rule g(_0,_1) -> _0.
        ==> L4 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),+(_0,0),g(0,_0)) [unit] is in U_IR^4.
        Let p4 = [1].
        We unfold the rule of L4 forwards at position p4
        with the rule +(_0,0) -> _0.
        ==> L5 = f^#(g(0,_0),s(0),g(0,_0)) -> f^#(g(0,_0),_0,g(0,_0)) [unit] is in U_IR^5.
  This DP problem is infinite.

** END proof description **

Proof stopped at iteration 5
Number of unfolded rules generated by this proof = 7862
Number of unfolded rules generated by all the parallel proofs = 13610