NO

Prover = TRS(tech=PATTERN_RULES, nb_unfoldings=unlimited, max_nb_unfolded_rules=200)

** BEGIN proof argument **
The following pattern rule was generated by the strategy
presented in Sect. 3 of [Emmes, Enger, Giesl, IJCAR'12]:
[iteration = 4] f(tt,s(s(_0))){_0->s(_0)}^n{_0->0} -> f(tt,s(s(s(_0)))){_0->s(_0)}^n{_0->0}
We apply Theorem 8 of [Emmes, Enger, Giesl, IJCAR'12]
on this rule with m = 1, b = 1, pi = epsilon,
sigma' = {} and mu' = {}.
Hence the term f(tt,s(s(0))), obtained from
instantiating n with 0 in the left-hand side of
the rule, starts an infinite derivation 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^#(tt,_0) -> f^#(eq(toOne(_0),s(0)),s(_0))]
  TRS = {f(tt,_0) -> f(eq(toOne(_0),s(0)),s(_0)), eq(s(_0),s(_1)) -> eq(_0,_1), eq(0,0) -> tt, toOne(s(s(_0))) -> toOne(s(_0)), toOne(s(0)) -> s(0)}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... Too many coefficients (15)! Aborting!
    ## Trying with lexicographic path orders... Failed!
    ## Trying to prove nontermination by unfolding the dependency pairs with the rules of the TRS
        # Iteration 0: nontermination not detected, 1 unfolded rule generated.
        # Iteration 1: nontermination not detected, 5 unfolded rules generated.
        # Iteration 2: nontermination not detected, 16 unfolded rules generated.
        # Iteration 3: nontermination not detected, 43 unfolded rules generated.
        # Iteration 4: nontermination detected, 98 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        IR contains the dependency pair f^#(tt,_0) -> f^#(eq(toOne(_0),s(0)),s(_0)).
        We apply (I) of [Emmes, Enger, Giesl, IJCAR'12] to this dependency pair.
        ==> P0 = f^#(tt,_0){}^n{} -> f^#(eq(toOne(_0),s(0)),s(_0)){}^n{} is in U_IR^0.
        We apply (VI) of [Emmes, Enger, Giesl, IJCAR'12] to this pattern rule
        at position [0, 0] using the pattern rule
        toOne(s(s(_0))){_0->s(_0)}^n{_0->_1} -> toOne(s(_1)){_0->s(_0)}^n{_0->_1}
        obtained from IR.
        ==> P1 = f^#(tt,s(s(_0))){_0->s(_0)}^n{_0->_1} -> f^#(eq(toOne(s(_1)),s(0)),s(s(s(_0)))){_0->s(_0)}^n{_0->_1} is in U_IR^1.
        We apply (V) + (IX) of [Emmes, Enger, Giesl, IJCAR'12] to this pattern rule
        at position [0, 0] using the rule
        toOne(s(0)) -> s(0)
        of IR.
        ==> P2 = f^#(tt,s(s(_0))){_0->s(_0)}^n{_0->0} -> f^#(eq(s(0),s(0)),s(s(s(_0)))){_0->s(_0)}^n{_0->0} is in U_IR^2.
        We apply (IX) of [Emmes, Enger, Giesl, IJCAR'12] to this pattern rule
        at position [0] using the rule
        eq(s(_0),s(_1)) -> eq(_0,_1)
        of IR.
        ==> P3 = f^#(tt,s(s(_0))){_0->s(_0)}^n{_0->0} -> f^#(eq(0,0),s(s(s(_0)))){_0->s(_0)}^n{_0->0} is in U_IR^3.
        We apply (IX) of [Emmes, Enger, Giesl, IJCAR'12] to this pattern rule
        at position [0] using the rule
        eq(0,0) -> tt
        of IR.
        ==> P4 = f^#(tt,s(s(_0))){_0->s(_0)}^n{_0->0} -> f^#(tt,s(s(s(_0)))){_0->s(_0)}^n{_0->0} is in U_IR^4.
  This DP problem is infinite.

** END proof description **

Proof stopped at iteration 4
Number of unfolded rules generated by this proof = 163
Number of unfolded rules generated by all the parallel proofs = 689