*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        check(no(x)) -> no(x)
        check(no(x)) -> no(check(x))
        check(rec(x)) -> rec(check(x))
        check(sent(x)) -> sent(check(x))
        check(up(x)) -> up(check(x))
        no(up(x)) -> up(no(x))
        rec(bot()) -> up(sent(bot()))
        rec(no(x)) -> sent(rec(x))
        rec(rec(x)) -> sent(rec(x))
        rec(sent(x)) -> sent(rec(x))
        rec(up(x)) -> up(rec(x))
        sent(up(x)) -> up(sent(x))
        top(no(up(x))) -> top(check(rec(x)))
        top(rec(up(x))) -> top(check(rec(x)))
        top(sent(up(x))) -> top(check(rec(x)))
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {check/1,no/1,rec/1,sent/1,top/1} / {bot/0,up/1}
      Obligation:
        Full
        basic terms: {check,no,rec,sent,top}/{bot,up}
    Applied Processor:
      Bounds {initialAutomaton = minimal, enrichment = match}
    Proof:
      The problem is match-bounded by 1.
      The enriched problem is compatible with follwoing automaton.
        bot_0() -> 2
        bot_1() -> 4
        check_0(2) -> 1
        check_1(2) -> 3
        no_0(2) -> 1
        no_1(2) -> 3
        rec_0(2) -> 1
        rec_1(2) -> 3
        sent_0(2) -> 1
        sent_1(2) -> 3
        sent_1(4) -> 3
        top_0(2) -> 1
        up_0(2) -> 2
        up_1(3) -> 1
        up_1(3) -> 3
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        check(no(x)) -> no(x)
        check(no(x)) -> no(check(x))
        check(rec(x)) -> rec(check(x))
        check(sent(x)) -> sent(check(x))
        check(up(x)) -> up(check(x))
        no(up(x)) -> up(no(x))
        rec(bot()) -> up(sent(bot()))
        rec(no(x)) -> sent(rec(x))
        rec(rec(x)) -> sent(rec(x))
        rec(sent(x)) -> sent(rec(x))
        rec(up(x)) -> up(rec(x))
        sent(up(x)) -> up(sent(x))
        top(no(up(x))) -> top(check(rec(x)))
        top(rec(up(x))) -> top(check(rec(x)))
        top(sent(up(x))) -> top(check(rec(x)))
      Signature:
        {check/1,no/1,rec/1,sent/1,top/1} / {bot/0,up/1}
      Obligation:
        Full
        basic terms: {check,no,rec,sent,top}/{bot,up}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).