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] mark(h(a)) -> mark(h(a))
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 = mark(h(a)) and theta2(theta1(l)) = theta1(r|p).
Hence, the term theta1(l) = mark(h(a)) 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^#(h(_0)) -> mark^#(g(_0,_0)), mark^#(h(_0)) -> active^#(h(mark(_0))), active^#(g(a,_0)) -> mark^#(f(b,_0)), mark^#(g(_0,_1)) -> active^#(g(mark(_0),_1)), active^#(f(_0,_0)) -> mark^#(h(a)), mark^#(f(_0,_1)) -> active^#(f(mark(_0),_1)), mark^#(h(_0)) -> mark^#(_0), mark^#(g(_0,_1)) -> mark^#(_0), mark^#(f(_0,_1)) -> mark^#(_0)]
  TRS = {active(h(_0)) -> mark(g(_0,_0)), active(g(a,_0)) -> mark(f(b,_0)), active(f(_0,_0)) -> mark(h(a)), active(a) -> mark(b), mark(h(_0)) -> active(h(mark(_0))), mark(g(_0,_1)) -> active(g(mark(_0),_1)), mark(a) -> active(a), mark(f(_0,_1)) -> active(f(mark(_0),_1)), mark(b) -> active(b), h(mark(_0)) -> h(_0), h(active(_0)) -> h(_0), g(mark(_0),_1) -> g(_0,_1), g(_0,mark(_1)) -> g(_0,_1), g(active(_0),_1) -> g(_0,_1), g(_0,active(_1)) -> g(_0,_1), f(mark(_0),_1) -> f(_0,_1), f(_0,mark(_1)) -> f(_0,_1), f(active(_0),_1) -> f(_0,_1), f(_0,active(_1)) -> f(_0,_1)}
    ## Trying with homeomorphic embeddings... Failed!
    ## Trying with polynomial interpretations... Too many coefficients (17)! 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, 9 unfolded rules generated.
        # Iteration 1: nontermination not detected, 47 unfolded rules generated.
        # Iteration 2: nontermination not detected, 103 unfolded rules generated.
        # Iteration 3: nontermination not detected, 223 unfolded rules generated.
        # Iteration 4: nontermination not detected, 624 unfolded rules generated.
        # Iteration 5: nontermination not detected, 1333 unfolded rules generated.
        # Iteration 6: nontermination not detected, 2326 unfolded rules generated.
        # Iteration 7: nontermination not detected, 4119 unfolded rules generated.
        # Iteration 8: nontermination not detected, 6996 unfolded rules generated.
        # Iteration 9: nontermination detected, 315 unfolded rules generated.
        Here is the successful unfolding. Let IR be the TRS under analysis.
        L0 = mark^#(h(_0)) -> active^#(h(mark(_0))) [trans] is in U_IR^0.
        D = active^#(h(_0)) -> mark^#(g(_0,_0)) is a dependency pair of IR.
        We build a composed triple from L0 and D.
        ==> L1 = [mark^#(h(_0)) -> active^#(h(mark(_0))), active^#(h(_1)) -> mark^#(g(_1,_1))] [comp] is in U_IR^1.
        Let p1 = [0, 0].
        We unfold the first rule of L1 forwards at position p1
        with the rule mark(a) -> active(a).
        ==> L2 = mark^#(h(a)) -> mark^#(g(active(a),active(a))) [trans] is in U_IR^2.
        D = mark^#(g(_0,_1)) -> active^#(g(mark(_0),_1)) is a dependency pair of IR.
        We build a composed triple from L2 and D.
        ==> L3 = [mark^#(h(a)) -> mark^#(g(active(a),active(a))), mark^#(g(_0,_1)) -> active^#(g(mark(_0),_1))] [comp] is in U_IR^3.
        Let p3 = [0].
        We unfold the first rule of L3 forwards at position p3
        with the rule g(active(_0),_1) -> g(_0,_1).
        ==> L4 = mark^#(h(a)) -> active^#(g(mark(a),active(a))) [trans] is in U_IR^4.
        D = active^#(g(a,_0)) -> mark^#(f(b,_0)) is a dependency pair of IR.
        We build a composed triple from L4 and D.
        ==> L5 = [mark^#(h(a)) -> active^#(g(mark(a),active(a))), active^#(g(a,_0)) -> mark^#(f(b,_0))] [comp] is in U_IR^5.
        Let p5 = [0].
        We unfold the first rule of L5 forwards at position p5
        with the rule g(mark(_0),_1) -> g(_0,_1).
        ==> L6 = [mark^#(h(a)) -> active^#(g(a,active(a))), active^#(g(a,_0)) -> mark^#(f(b,_0))] [comp] is in U_IR^6.
        Let p6 = [0, 1].
        We unfold the first rule of L6 forwards at position p6
        with the rule active(a) -> mark(b).
        ==> L7 = mark^#(h(a)) -> mark^#(f(b,mark(b))) [trans] is in U_IR^7.
        D = mark^#(f(_0,_1)) -> active^#(f(mark(_0),_1)) is a dependency pair of IR.
        We build a composed triple from L7 and D.
        ==> L8 = mark^#(h(a)) -> active^#(f(mark(b),mark(b))) [trans] is in U_IR^8.
        D = active^#(f(_0,_0)) -> mark^#(h(a)) is a dependency pair of IR.
        We build a composed triple from L8 and D.
        ==> L9 = mark^#(h(a)) -> mark^#(h(a)) [trans] 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 = 16095
Number of unfolded rules generated by all the parallel proofs = 83578