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] a__f(c,g(c),mark(g(b))) -> a__f(c,g(c),mark(g(b)))
Let l be the left-hand side and r be the right-hand side of this rule.
Let p = epsilon, theta1 = {} and theta2 = {}.
We have r|p = a__f(c,g(c),mark(g(b))) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = a__f(c,g(c),mark(g(b))) 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 = [mark^#(g(_0)) -> mark^#(_0)]
  TRS = {a__f(_0,g(_0),_1) -> a__f(_1,_1,_1), a__g(b) -> c, a__b -> c, mark(f(_0,_1,_2)) -> a__f(_0,_1,_2), mark(g(_0)) -> a__g(mark(_0)), mark(b) -> a__b, mark(c) -> c, a__f(_0,_1,_2) -> f(_0,_1,_2), a__g(_0) -> g(_0), a__b -> b}
    ## Trying with homeomorphic embeddings... Success!
  This DP problem is finite.
  ## DP problem: 
  Dependency pairs = [a__f^#(_0,g(_0),_1) -> a__f^#(_1,_1,_1)]
  TRS = {a__f(_0,g(_0),_1) -> a__f(_1,_1,_1), a__g(b) -> c, a__b -> c, mark(f(_0,_1,_2)) -> a__f(_0,_1,_2), mark(g(_0)) -> a__g(mark(_0)), mark(b) -> a__b, mark(c) -> c, a__f(_0,_1,_2) -> f(_0,_1,_2), a__g(_0) -> g(_0), a__b -> b}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... Too many coefficients (26)! Aborting!
    ## Trying with lexicographic path orders... Failed!
    ## Trying to prove nontermination by unfolding the dependency pairs with the rules of the TRS
      # max_depth=2, 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, 0 unfolded rule generated.
        Nontermination not detected!
      # max_depth=2, unfold_variables=true:
        # Iteration 0: nontermination not detected, 1 unfolded rule generated.
        # Iteration 1: nontermination not detected, 1 unfolded rule generated.
        # Iteration 2: nontermination not detected, 4 unfolded rules generated.
        # Iteration 3: nontermination not detected, 9 unfolded rules generated.
        # Iteration 4: nontermination not detected, 4 unfolded rules generated.
        # Iteration 5: nontermination not detected, 4 unfolded rules generated.
        # Iteration 6: nontermination not detected, 0 unfolded rule generated.
        Nontermination not detected!
      # max_depth=3, 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, 0 unfolded rule generated.
        Nontermination not detected!
      # max_depth=3, unfold_variables=true:
        # Iteration 0: nontermination not detected, 1 unfolded rule generated.
        # Iteration 1: nontermination not detected, 1 unfolded rule generated.
        # Iteration 2: nontermination not detected, 6 unfolded rules generated.
        # Iteration 3: nontermination not detected, 14 unfolded rules generated.
        # Iteration 4: nontermination not detected, 13 unfolded rules generated.
        # Iteration 5: nontermination not detected, 20 unfolded rules generated.
        # Iteration 6: nontermination not detected, 21 unfolded rules generated.
        # Iteration 7: nontermination not detected, 25 unfolded rules generated.
        # Iteration 8: nontermination not detected, 28 unfolded rules generated.
        # Iteration 9: nontermination detected, 15 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = a__f^#(_0,g(_0),_1) -> a__f^#(_1,_1,_1) [trans] is in U_IR^0.
        We build a unit triple from L0.
        ==> L1 = a__f^#(_0,g(_0),_1) -> a__f^#(_1,_1,_1) [unit] is in U_IR^1.
        Let p1 = [0].
        We unfold the rule of L1 forwards at position p1
        with the rule mark(g(_0)) -> a__g(mark(_0)).
        ==> L2 = a__f^#(_0,g(_0),mark(g(_1))) -> a__f^#(a__g(mark(_1)),mark(g(_1)),mark(g(_1))) [unit] is in U_IR^2.
        Let p2 = [0, 0].
        We unfold the rule of L2 forwards at position p2
        with the rule mark(b) -> a__b.
        ==> L3 = a__f^#(_0,g(_0),mark(g(b))) -> a__f^#(a__g(a__b),mark(g(b)),mark(g(b))) [unit] is in U_IR^3.
        Let p3 = [0, 0].
        We unfold the rule of L3 forwards at position p3
        with the rule a__b -> b.
        ==> L4 = a__f^#(_0,g(_0),mark(g(b))) -> a__f^#(a__g(b),mark(g(b)),mark(g(b))) [unit] is in U_IR^4.
        Let p4 = [0].
        We unfold the rule of L4 forwards at position p4
        with the rule a__g(b) -> c.
        ==> L5 = a__f^#(c,g(c),mark(g(b))) -> a__f^#(c,mark(g(b)),mark(g(b))) [unit] is in U_IR^5.
        Let p5 = [1].
        We unfold the rule of L5 forwards at position p5
        with the rule mark(g(_0)) -> a__g(mark(_0)).
        ==> L6 = a__f^#(c,g(c),mark(g(b))) -> a__f^#(c,a__g(mark(b)),mark(g(b))) [unit] is in U_IR^6.
        Let p6 = [1].
        We unfold the rule of L6 forwards at position p6
        with the rule a__g(_0) -> g(_0).
        ==> L7 = a__f^#(c,g(c),mark(g(b))) -> a__f^#(c,g(mark(b)),mark(g(b))) [unit] is in U_IR^7.
        Let p7 = [1, 0].
        We unfold the rule of L7 forwards at position p7
        with the rule mark(b) -> a__b.
        ==> L8 = a__f^#(c,g(c),mark(g(b))) -> a__f^#(c,g(a__b),mark(g(b))) [unit] is in U_IR^8.
        Let p8 = [1, 0].
        We unfold the rule of L8 forwards at position p8
        with the rule a__b -> c.
        ==> L9 = a__f^#(c,g(c),mark(g(b))) -> a__f^#(c,g(c),mark(g(b))) [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 = 171
Number of unfolded rules generated by all the parallel proofs = 481