* Step 1: Bounds WORST_CASE(?,O(n^1))
    + Considered Problem:
        - Strict TRS:
            +(x,0()) -> x
            +(x,s(y)) -> s(+(x,y))
            +(s(x),y) -> s(+(x,y))
            not(false()) -> true()
            not(true()) -> false()
            odd(0()) -> false()
            odd(s(x)) -> not(odd(x))
        - Signature:
            {+/2,not/1,odd/1} / {0/0,false/0,s/1,true/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {+,not,odd} and constructors {0,false,s,true}
    + Applied Processor:
        Bounds {initialAutomaton = minimal, enrichment = match}
    + Details:
        The problem is match-bounded by 2.
        The enriched problem is compatible with follwoing automaton.
          +_0(2,2) -> 1
          +_1(2,2) -> 3
          0_0() -> 1
          0_0() -> 2
          0_0() -> 3
          false_0() -> 1
          false_0() -> 2
          false_0() -> 3
          false_1() -> 1
          false_1() -> 4
          false_2() -> 1
          false_2() -> 4
          not_0(2) -> 1
          not_1(4) -> 1
          not_1(4) -> 4
          odd_0(2) -> 1
          odd_1(2) -> 4
          s_0(2) -> 1
          s_0(2) -> 2
          s_0(2) -> 3
          s_1(3) -> 1
          s_1(3) -> 3
          true_0() -> 1
          true_0() -> 2
          true_0() -> 3
          true_1() -> 1
          true_2() -> 1
          true_2() -> 4
          2 -> 1
          2 -> 3
* Step 2: EmptyProcessor WORST_CASE(?,O(1))
    + Considered Problem:
        - Weak TRS:
            +(x,0()) -> x
            +(x,s(y)) -> s(+(x,y))
            +(s(x),y) -> s(+(x,y))
            not(false()) -> true()
            not(true()) -> false()
            odd(0()) -> false()
            odd(s(x)) -> not(odd(x))
        - Signature:
            {+/2,not/1,odd/1} / {0/0,false/0,s/1,true/0}
        - Obligation:
            innermost runtime complexity wrt. defined symbols {+,not,odd} and constructors {0,false,s,true}
    + Applied Processor:
        EmptyProcessor
    + Details:
        The problem is already closed. The intended complexity is O(1).

WORST_CASE(?,O(n^1))