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 = 9] f(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f(s(s(s(s(s(s(s(s(id(_0))))))))),_1,_1)
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {} and theta2 = {_0->id(_0)}.
We have r|p = f(s(s(s(s(s(s(s(s(id(_0))))))))),_1,_1) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = f(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) 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^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1)]
  TRS = {f(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1), id(s(_0)) -> s(id(_0)), id(0) -> 0}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... Too many coefficients (14)! 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, 1 unfolded rule generated.
        # Iteration 3: nontermination not detected, 1 unfolded rule generated.
        # Iteration 4: nontermination not detected, 1 unfolded rule generated.
        # Iteration 5: nontermination not detected, 1 unfolded rule generated.
        # Iteration 6: nontermination not detected, 1 unfolded rule generated.
        # Iteration 7: nontermination not detected, 1 unfolded rule generated.
        # Iteration 8: nontermination not detected, 1 unfolded rule generated.
        # Iteration 9: nontermination detected, 1 unfolded rule generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1) [trans] is in U_IR^0.
        We build a unit triple from L0.
        ==> L1 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^1.
        Let p1 = [0].
        We unfold the rule of L1 forwards at position p1
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L2 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(id(s(s(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^2.
        Let p2 = [0, 0].
        We unfold the rule of L2 forwards at position p2
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L3 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(id(s(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^3.
        Let p3 = [0, 0, 0].
        We unfold the rule of L3 forwards at position p3
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L4 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(id(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^4.
        Let p4 = [0, 0, 0, 0].
        We unfold the rule of L4 forwards at position p4
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L5 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(id(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^5.
        Let p5 = [0, 0, 0, 0, 0].
        We unfold the rule of L5 forwards at position p5
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L6 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(id(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^6.
        Let p6 = [0, 0, 0, 0, 0, 0].
        We unfold the rule of L6 forwards at position p6
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L7 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(s(id(s(s(_0))))))))),_1,_1) [unit] is in U_IR^7.
        Let p7 = [0, 0, 0, 0, 0, 0, 0].
        We unfold the rule of L7 forwards at position p7
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L8 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(s(s(id(s(_0))))))))),_1,_1) [unit] is in U_IR^8.
        Let p8 = [0, 0, 0, 0, 0, 0, 0, 0].
        We unfold the rule of L8 forwards at position p8
        with the rule id(s(_0)) -> s(id(_0)).
        ==> L9 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(s(s(s(id(_0))))))))),_1,_1) [unit] is in U_IR^9.
  This DP problem is infinite.

** END proof description **

Proof stopped at iteration 9
Number of unfolded rules generated by this proof = 10
Number of unfolded rules generated by all the parallel proofs = 2378