*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        a__f(X) -> f(X)
        a__f(f(a())) -> a__f(g(f(a())))
        mark(a()) -> a()
        mark(f(X)) -> a__f(mark(X))
        mark(g(X)) -> g(X)
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {a__f/1,mark/1} / {a/0,f/1,g/1}
      Obligation:
        Full
        basic terms: {a__f,mark}/{a,f,g}
    Applied Processor:
      ToInnermost
    Proof:
      switch to innermost, as the system is overlay and right linear and does not contain weak rules
*** 1.1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        a__f(X) -> f(X)
        a__f(f(a())) -> a__f(g(f(a())))
        mark(a()) -> a()
        mark(f(X)) -> a__f(mark(X))
        mark(g(X)) -> g(X)
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {a__f/1,mark/1} / {a/0,f/1,g/1}
      Obligation:
        Innermost
        basic terms: {a__f,mark}/{a,f,g}
    Applied Processor:
      Bounds {initialAutomaton = minimal, enrichment = match}
    Proof:
      The problem is match-bounded by 3.
      The enriched problem is compatible with follwoing automaton.
        a_0() -> 2
        a_1() -> 1
        a_1() -> 3
        a_1() -> 5
        a_2() -> 8
        a__f_0(2) -> 1
        a__f_1(3) -> 1
        a__f_1(3) -> 3
        a__f_2(6) -> 1
        a__f_2(6) -> 3
        f_0(2) -> 2
        f_1(2) -> 1
        f_1(5) -> 4
        f_2(3) -> 1
        f_2(3) -> 3
        f_2(8) -> 7
        f_3(6) -> 1
        f_3(6) -> 3
        g_0(2) -> 2
        g_1(2) -> 1
        g_1(2) -> 3
        g_1(4) -> 3
        g_2(7) -> 6
        mark_0(2) -> 1
        mark_1(2) -> 3
*** 1.1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        a__f(X) -> f(X)
        a__f(f(a())) -> a__f(g(f(a())))
        mark(a()) -> a()
        mark(f(X)) -> a__f(mark(X))
        mark(g(X)) -> g(X)
      Signature:
        {a__f/1,mark/1} / {a/0,f/1,g/1}
      Obligation:
        Innermost
        basic terms: {a__f,mark}/{a,f,g}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).