MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { check(0()) -> zero()
  , check(s(0())) -> odd()
  , check(s(s(0()))) -> even()
  , check(s(s(s(x)))) -> check(s(x))
  , half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , times(x, y) -> timesIter(x, y, 0())
  , timesIter(x, y, z) -> if(check(x), x, y, z, plus(z, y))
  , if(zero(), x, y, z, u) -> z
  , if(odd(), x, y, z, u) -> timesIter(p(x), y, u)
  , if(even(), x, y, z, u) ->
    plus(timesIter(half(x), y, half(z)),
         timesIter(half(x), y, half(s(z))))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { check^#(0()) -> c_1()
  , check^#(s(0())) -> c_2()
  , check^#(s(s(0()))) -> c_3()
  , check^#(s(s(s(x)))) -> c_4(check^#(s(x)))
  , half^#(0()) -> c_5()
  , half^#(s(0())) -> c_6()
  , half^#(s(s(x))) -> c_7(half^#(x))
  , plus^#(0(), y) -> c_8()
  , plus^#(s(x), y) -> c_9(plus^#(x, y))
  , times^#(x, y) -> c_10(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_11(if^#(check(x), x, y, z, plus(z, y)), check^#(x), plus^#(z, y))
  , if^#(zero(), x, y, z, u) -> c_12()
  , if^#(odd(), x, y, z, u) -> c_13(timesIter^#(p(x), y, u), p^#(x))
  , if^#(even(), x, y, z, u) ->
    c_14(plus^#(timesIter(half(x), y, half(z)),
                timesIter(half(x), y, half(s(z)))),
         timesIter^#(half(x), y, half(z)),
         half^#(x),
         half^#(z),
         timesIter^#(half(x), y, half(s(z))),
         half^#(x),
         half^#(s(z)))
  , p^#(0()) -> c_15()
  , p^#(s(x)) -> c_16() }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { check^#(0()) -> c_1()
  , check^#(s(0())) -> c_2()
  , check^#(s(s(0()))) -> c_3()
  , check^#(s(s(s(x)))) -> c_4(check^#(s(x)))
  , half^#(0()) -> c_5()
  , half^#(s(0())) -> c_6()
  , half^#(s(s(x))) -> c_7(half^#(x))
  , plus^#(0(), y) -> c_8()
  , plus^#(s(x), y) -> c_9(plus^#(x, y))
  , times^#(x, y) -> c_10(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_11(if^#(check(x), x, y, z, plus(z, y)), check^#(x), plus^#(z, y))
  , if^#(zero(), x, y, z, u) -> c_12()
  , if^#(odd(), x, y, z, u) -> c_13(timesIter^#(p(x), y, u), p^#(x))
  , if^#(even(), x, y, z, u) ->
    c_14(plus^#(timesIter(half(x), y, half(z)),
                timesIter(half(x), y, half(s(z)))),
         timesIter^#(half(x), y, half(z)),
         half^#(x),
         half^#(z),
         timesIter^#(half(x), y, half(s(z))),
         half^#(x),
         half^#(s(z)))
  , p^#(0()) -> c_15()
  , p^#(s(x)) -> c_16() }
Weak Trs:
  { check(0()) -> zero()
  , check(s(0())) -> odd()
  , check(s(s(0()))) -> even()
  , check(s(s(s(x)))) -> check(s(x))
  , half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , times(x, y) -> timesIter(x, y, 0())
  , timesIter(x, y, z) -> if(check(x), x, y, z, plus(z, y))
  , if(zero(), x, y, z, u) -> z
  , if(odd(), x, y, z, u) -> timesIter(p(x), y, u)
  , if(even(), x, y, z, u) ->
    plus(timesIter(half(x), y, half(z)),
         timesIter(half(x), y, half(s(z))))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,2,3,5,6,8,12,15,16} by
applications of Pre({1,2,3,5,6,8,12,15,16}) = {4,7,9,11,13,14}.
Here rules are labeled as follows:

  DPs:
    { 1: check^#(0()) -> c_1()
    , 2: check^#(s(0())) -> c_2()
    , 3: check^#(s(s(0()))) -> c_3()
    , 4: check^#(s(s(s(x)))) -> c_4(check^#(s(x)))
    , 5: half^#(0()) -> c_5()
    , 6: half^#(s(0())) -> c_6()
    , 7: half^#(s(s(x))) -> c_7(half^#(x))
    , 8: plus^#(0(), y) -> c_8()
    , 9: plus^#(s(x), y) -> c_9(plus^#(x, y))
    , 10: times^#(x, y) -> c_10(timesIter^#(x, y, 0()))
    , 11: timesIter^#(x, y, z) ->
          c_11(if^#(check(x), x, y, z, plus(z, y)), check^#(x), plus^#(z, y))
    , 12: if^#(zero(), x, y, z, u) -> c_12()
    , 13: if^#(odd(), x, y, z, u) ->
          c_13(timesIter^#(p(x), y, u), p^#(x))
    , 14: if^#(even(), x, y, z, u) ->
          c_14(plus^#(timesIter(half(x), y, half(z)),
                      timesIter(half(x), y, half(s(z)))),
               timesIter^#(half(x), y, half(z)),
               half^#(x),
               half^#(z),
               timesIter^#(half(x), y, half(s(z))),
               half^#(x),
               half^#(s(z)))
    , 15: p^#(0()) -> c_15()
    , 16: p^#(s(x)) -> c_16() }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { check^#(s(s(s(x)))) -> c_4(check^#(s(x)))
  , half^#(s(s(x))) -> c_7(half^#(x))
  , plus^#(s(x), y) -> c_9(plus^#(x, y))
  , times^#(x, y) -> c_10(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_11(if^#(check(x), x, y, z, plus(z, y)), check^#(x), plus^#(z, y))
  , if^#(odd(), x, y, z, u) -> c_13(timesIter^#(p(x), y, u), p^#(x))
  , if^#(even(), x, y, z, u) ->
    c_14(plus^#(timesIter(half(x), y, half(z)),
                timesIter(half(x), y, half(s(z)))),
         timesIter^#(half(x), y, half(z)),
         half^#(x),
         half^#(z),
         timesIter^#(half(x), y, half(s(z))),
         half^#(x),
         half^#(s(z))) }
Weak DPs:
  { check^#(0()) -> c_1()
  , check^#(s(0())) -> c_2()
  , check^#(s(s(0()))) -> c_3()
  , half^#(0()) -> c_5()
  , half^#(s(0())) -> c_6()
  , plus^#(0(), y) -> c_8()
  , if^#(zero(), x, y, z, u) -> c_12()
  , p^#(0()) -> c_15()
  , p^#(s(x)) -> c_16() }
Weak Trs:
  { check(0()) -> zero()
  , check(s(0())) -> odd()
  , check(s(s(0()))) -> even()
  , check(s(s(s(x)))) -> check(s(x))
  , half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , times(x, y) -> timesIter(x, y, 0())
  , timesIter(x, y, z) -> if(check(x), x, y, z, plus(z, y))
  , if(zero(), x, y, z, u) -> z
  , if(odd(), x, y, z, u) -> timesIter(p(x), y, u)
  , if(even(), x, y, z, u) ->
    plus(timesIter(half(x), y, half(z)),
         timesIter(half(x), y, half(s(z))))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ check^#(0()) -> c_1()
, check^#(s(0())) -> c_2()
, check^#(s(s(0()))) -> c_3()
, half^#(0()) -> c_5()
, half^#(s(0())) -> c_6()
, plus^#(0(), y) -> c_8()
, if^#(zero(), x, y, z, u) -> c_12()
, p^#(0()) -> c_15()
, p^#(s(x)) -> c_16() }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { check^#(s(s(s(x)))) -> c_4(check^#(s(x)))
  , half^#(s(s(x))) -> c_7(half^#(x))
  , plus^#(s(x), y) -> c_9(plus^#(x, y))
  , times^#(x, y) -> c_10(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_11(if^#(check(x), x, y, z, plus(z, y)), check^#(x), plus^#(z, y))
  , if^#(odd(), x, y, z, u) -> c_13(timesIter^#(p(x), y, u), p^#(x))
  , if^#(even(), x, y, z, u) ->
    c_14(plus^#(timesIter(half(x), y, half(z)),
                timesIter(half(x), y, half(s(z)))),
         timesIter^#(half(x), y, half(z)),
         half^#(x),
         half^#(z),
         timesIter^#(half(x), y, half(s(z))),
         half^#(x),
         half^#(s(z))) }
Weak Trs:
  { check(0()) -> zero()
  , check(s(0())) -> odd()
  , check(s(s(0()))) -> even()
  , check(s(s(s(x)))) -> check(s(x))
  , half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , times(x, y) -> timesIter(x, y, 0())
  , timesIter(x, y, z) -> if(check(x), x, y, z, plus(z, y))
  , if(zero(), x, y, z, u) -> z
  , if(odd(), x, y, z, u) -> timesIter(p(x), y, u)
  , if(even(), x, y, z, u) ->
    plus(timesIter(half(x), y, half(z)),
         timesIter(half(x), y, half(s(z))))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { if^#(odd(), x, y, z, u) ->
    c_13(timesIter^#(p(x), y, u), p^#(x)) }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { check^#(s(s(s(x)))) -> c_1(check^#(s(x)))
  , half^#(s(s(x))) -> c_2(half^#(x))
  , plus^#(s(x), y) -> c_3(plus^#(x, y))
  , times^#(x, y) -> c_4(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_5(if^#(check(x), x, y, z, plus(z, y)), check^#(x), plus^#(z, y))
  , if^#(odd(), x, y, z, u) -> c_6(timesIter^#(p(x), y, u))
  , if^#(even(), x, y, z, u) ->
    c_7(plus^#(timesIter(half(x), y, half(z)),
               timesIter(half(x), y, half(s(z)))),
        timesIter^#(half(x), y, half(z)),
        half^#(x),
        half^#(z),
        timesIter^#(half(x), y, half(s(z))),
        half^#(x),
        half^#(s(z))) }
Weak Trs:
  { check(0()) -> zero()
  , check(s(0())) -> odd()
  , check(s(s(0()))) -> even()
  , check(s(s(s(x)))) -> check(s(x))
  , half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , times(x, y) -> timesIter(x, y, 0())
  , timesIter(x, y, z) -> if(check(x), x, y, z, plus(z, y))
  , if(zero(), x, y, z, u) -> z
  , if(odd(), x, y, z, u) -> timesIter(p(x), y, u)
  , if(even(), x, y, z, u) ->
    plus(timesIter(half(x), y, half(z)),
         timesIter(half(x), y, half(s(z))))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { check(0()) -> zero()
    , check(s(0())) -> odd()
    , check(s(s(0()))) -> even()
    , check(s(s(s(x)))) -> check(s(x))
    , half(0()) -> 0()
    , half(s(0())) -> 0()
    , half(s(s(x))) -> s(half(x))
    , plus(0(), y) -> y
    , plus(s(x), y) -> s(plus(x, y))
    , timesIter(x, y, z) -> if(check(x), x, y, z, plus(z, y))
    , if(zero(), x, y, z, u) -> z
    , if(odd(), x, y, z, u) -> timesIter(p(x), y, u)
    , if(even(), x, y, z, u) ->
      plus(timesIter(half(x), y, half(z)),
           timesIter(half(x), y, half(s(z))))
    , p(0()) -> 0()
    , p(s(x)) -> x }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { check^#(s(s(s(x)))) -> c_1(check^#(s(x)))
  , half^#(s(s(x))) -> c_2(half^#(x))
  , plus^#(s(x), y) -> c_3(plus^#(x, y))
  , times^#(x, y) -> c_4(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_5(if^#(check(x), x, y, z, plus(z, y)), check^#(x), plus^#(z, y))
  , if^#(odd(), x, y, z, u) -> c_6(timesIter^#(p(x), y, u))
  , if^#(even(), x, y, z, u) ->
    c_7(plus^#(timesIter(half(x), y, half(z)),
               timesIter(half(x), y, half(s(z)))),
        timesIter^#(half(x), y, half(z)),
        half^#(x),
        half^#(z),
        timesIter^#(half(x), y, half(s(z))),
        half^#(x),
        half^#(s(z))) }
Weak Trs:
  { check(0()) -> zero()
  , check(s(0())) -> odd()
  , check(s(s(0()))) -> even()
  , check(s(s(s(x)))) -> check(s(x))
  , half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , timesIter(x, y, z) -> if(check(x), x, y, z, plus(z, y))
  , if(zero(), x, y, z, u) -> z
  , if(odd(), x, y, z, u) -> timesIter(p(x), y, u)
  , if(even(), x, y, z, u) ->
    plus(timesIter(half(x), y, half(z)),
         timesIter(half(x), y, half(s(z))))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: check^#(s(s(s(x)))) -> c_1(check^#(s(x)))
     -->_1 check^#(s(s(s(x)))) -> c_1(check^#(s(x))) :1
  
  2: half^#(s(s(x))) -> c_2(half^#(x))
     -->_1 half^#(s(s(x))) -> c_2(half^#(x)) :2
  
  3: plus^#(s(x), y) -> c_3(plus^#(x, y))
     -->_1 plus^#(s(x), y) -> c_3(plus^#(x, y)) :3
  
  4: times^#(x, y) -> c_4(timesIter^#(x, y, 0()))
     -->_1 timesIter^#(x, y, z) ->
           c_5(if^#(check(x), x, y, z, plus(z, y)),
               check^#(x),
               plus^#(z, y)) :5
  
  5: timesIter^#(x, y, z) ->
     c_5(if^#(check(x), x, y, z, plus(z, y)), check^#(x), plus^#(z, y))
     -->_1 if^#(even(), x, y, z, u) ->
           c_7(plus^#(timesIter(half(x), y, half(z)),
                      timesIter(half(x), y, half(s(z)))),
               timesIter^#(half(x), y, half(z)),
               half^#(x),
               half^#(z),
               timesIter^#(half(x), y, half(s(z))),
               half^#(x),
               half^#(s(z))) :7
     -->_1 if^#(odd(), x, y, z, u) -> c_6(timesIter^#(p(x), y, u)) :6
     -->_3 plus^#(s(x), y) -> c_3(plus^#(x, y)) :3
     -->_2 check^#(s(s(s(x)))) -> c_1(check^#(s(x))) :1
  
  6: if^#(odd(), x, y, z, u) -> c_6(timesIter^#(p(x), y, u))
     -->_1 timesIter^#(x, y, z) ->
           c_5(if^#(check(x), x, y, z, plus(z, y)),
               check^#(x),
               plus^#(z, y)) :5
  
  7: if^#(even(), x, y, z, u) ->
     c_7(plus^#(timesIter(half(x), y, half(z)),
                timesIter(half(x), y, half(s(z)))),
         timesIter^#(half(x), y, half(z)),
         half^#(x),
         half^#(z),
         timesIter^#(half(x), y, half(s(z))),
         half^#(x),
         half^#(s(z)))
     -->_5 timesIter^#(x, y, z) ->
           c_5(if^#(check(x), x, y, z, plus(z, y)),
               check^#(x),
               plus^#(z, y)) :5
     -->_2 timesIter^#(x, y, z) ->
           c_5(if^#(check(x), x, y, z, plus(z, y)),
               check^#(x),
               plus^#(z, y)) :5
     -->_1 plus^#(s(x), y) -> c_3(plus^#(x, y)) :3
     -->_7 half^#(s(s(x))) -> c_2(half^#(x)) :2
     -->_6 half^#(s(s(x))) -> c_2(half^#(x)) :2
     -->_4 half^#(s(s(x))) -> c_2(half^#(x)) :2
     -->_3 half^#(s(s(x))) -> c_2(half^#(x)) :2
  

Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).

  { times^#(x, y) -> c_4(timesIter^#(x, y, 0())) }


We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { check^#(s(s(s(x)))) -> c_1(check^#(s(x)))
  , half^#(s(s(x))) -> c_2(half^#(x))
  , plus^#(s(x), y) -> c_3(plus^#(x, y))
  , timesIter^#(x, y, z) ->
    c_5(if^#(check(x), x, y, z, plus(z, y)), check^#(x), plus^#(z, y))
  , if^#(odd(), x, y, z, u) -> c_6(timesIter^#(p(x), y, u))
  , if^#(even(), x, y, z, u) ->
    c_7(plus^#(timesIter(half(x), y, half(z)),
               timesIter(half(x), y, half(s(z)))),
        timesIter^#(half(x), y, half(z)),
        half^#(x),
        half^#(z),
        timesIter^#(half(x), y, half(s(z))),
        half^#(x),
        half^#(s(z))) }
Weak Trs:
  { check(0()) -> zero()
  , check(s(0())) -> odd()
  , check(s(s(0()))) -> even()
  , check(s(s(s(x)))) -> check(s(x))
  , half(0()) -> 0()
  , half(s(0())) -> 0()
  , half(s(s(x))) -> s(half(x))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , timesIter(x, y, z) -> if(check(x), x, y, z, plus(z, y))
  , if(zero(), x, y, z, u) -> z
  , if(odd(), x, y, z, u) -> timesIter(p(x), y, u)
  , if(even(), x, y, z, u) ->
    plus(timesIter(half(x), y, half(z)),
         timesIter(half(x), y, half(s(z))))
  , p(0()) -> 0()
  , p(s(x)) -> x }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

None of the processors succeeded.

Details of failed attempt(s):
-----------------------------
1) 'matrices' failed due to the following reason:
   
   None of the processors succeeded.
   
   Details of failed attempt(s):
   -----------------------------
   1) 'matrix interpretation of dimension 4' failed due to the
      following reason:
      
      Following exception was raised:
        stack overflow
   
   2) 'matrix interpretation of dimension 3' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   3) 'matrix interpretation of dimension 3' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   4) 'matrix interpretation of dimension 2' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   5) 'matrix interpretation of dimension 2' failed due to the
      following reason:
      
      The input cannot be shown compatible
   
   6) 'matrix interpretation of dimension 1' failed due to the
      following reason:
      
      The input cannot be shown compatible
   

2) 'empty' failed due to the following reason:
   
   Empty strict component of the problem is NOT empty.


Arrrr..