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] active(f(b,mark(c),c)) -> active(f(b,mark(c),c))
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 = active(f(b,mark(c),c)) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = active(f(b,mark(c),c)) 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 = [active^#(f(b,_0,c)) -> mark^#(f(_0,c,_0)), mark^#(f(_0,_1,_2)) -> active^#(f(_0,mark(_1),_2)), mark^#(f(_0,_1,_2)) -> mark^#(_1)]
  TRS = {active(f(b,_0,c)) -> mark(f(_0,c,_0)), active(c) -> mark(b), mark(f(_0,_1,_2)) -> active(f(_0,mark(_1),_2)), mark(b) -> active(b), mark(c) -> active(c), f(mark(_0),_1,_2) -> f(_0,_1,_2), f(_0,mark(_1),_2) -> f(_0,_1,_2), f(_0,_1,mark(_2)) -> f(_0,_1,_2), f(active(_0),_1,_2) -> f(_0,_1,_2), f(_0,active(_1),_2) -> f(_0,_1,_2), f(_0,_1,active(_2)) -> f(_0,_1,_2)}
    ## 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
      # max_depth=3, unfold_variables=false:
        # Iteration 0: nontermination not detected, 3 unfolded rules generated.
        # Iteration 1: nontermination not detected, 8 unfolded rules generated.
        # Iteration 2: nontermination not detected, 48 unfolded rules generated.
        # Iteration 3: nontermination not detected, 134 unfolded rules generated.
        # Iteration 4: nontermination not detected, 279 unfolded rules generated.
        # Iteration 5: nontermination detected, 26 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = active^#(f(b,_0,c)) -> mark^#(f(_0,c,_0)) [trans] is in U_IR^0.
        D = mark^#(f(_0,_1,_2)) -> active^#(f(_0,mark(_1),_2)) is a dependency pair of IR.
        We build a composed triple from L0 and D.
        ==> L1 = [active^#(f(b,_0,c)) -> mark^#(f(_0,c,_0)), mark^#(f(_1,_2,_3)) -> active^#(f(_1,mark(_2),_3))] [comp] is in U_IR^1.
        Let p1 = [0].
        We unfold the first rule of L1 forwards at position p1
        with the rule f(_0,_1,mark(_2)) -> f(_0,_1,_2).
        ==> L2 = [active^#(f(b,mark(_0),c)) -> mark^#(f(mark(_0),c,_0)), mark^#(f(_1,_2,_3)) -> active^#(f(_1,mark(_2),_3))] [comp] is in U_IR^2.
        Let p2 = [0, 0].
        We unfold the first rule of L2 forwards at position p2
        with the rule mark(c) -> active(c).
        ==> L3 = [active^#(f(b,mark(c),c)) -> mark^#(f(active(c),c,c)), mark^#(f(_0,_1,_2)) -> active^#(f(_0,mark(_1),_2))] [comp] is in U_IR^3.
        Let p3 = [0, 0].
        We unfold the first rule of L3 forwards at position p3
        with the rule active(c) -> mark(b).
        ==> L4 = [active^#(f(b,mark(c),c)) -> mark^#(f(mark(b),c,c)), mark^#(f(_0,_1,_2)) -> active^#(f(_0,mark(_1),_2))] [comp] is in U_IR^4.
        Let p4 = [0].
        We unfold the first rule of L4 forwards at position p4
        with the rule f(mark(_0),_1,_2) -> f(_0,_1,_2).
        ==> L5 = active^#(f(b,mark(c),c)) -> active^#(f(b,mark(c),c)) [trans] 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 = 498
Number of unfolded rules generated by all the parallel proofs = 2247