MAYBE

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

Strict Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), 0()) -> s(x)
  , minus(s(x), s(y)) -> minus(x, y)
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , if(true(), x, y) -> x
  , if(false(), x, y) -> y
  , perfectp(0()) -> false()
  , perfectp(s(x)) -> f(x, s(0()), s(x), s(x))
  , f(0(), y, 0(), u) -> true()
  , f(0(), y, s(z), u) -> false()
  , f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u)
  , f(s(x), s(y), z, u) ->
    if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { minus^#(0(), y) -> c_1()
  , minus^#(s(x), 0()) -> c_2()
  , minus^#(s(x), s(y)) -> c_3(minus^#(x, y))
  , le^#(0(), y) -> c_4()
  , le^#(s(x), 0()) -> c_5()
  , le^#(s(x), s(y)) -> c_6(le^#(x, y))
  , if^#(true(), x, y) -> c_7()
  , if^#(false(), x, y) -> c_8()
  , perfectp^#(0()) -> c_9()
  , perfectp^#(s(x)) -> c_10(f^#(x, s(0()), s(x), s(x)))
  , f^#(0(), y, 0(), u) -> c_11()
  , f^#(0(), y, s(z), u) -> c_12()
  , f^#(s(x), 0(), z, u) ->
    c_13(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x)))
  , f^#(s(x), s(y), z, u) ->
    c_14(if^#(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)),
         le^#(x, y),
         f^#(s(x), minus(y, x), z, u),
         minus^#(y, x),
         f^#(x, u, z, u)) }

and mark the set of starting terms.

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

Strict DPs:
  { minus^#(0(), y) -> c_1()
  , minus^#(s(x), 0()) -> c_2()
  , minus^#(s(x), s(y)) -> c_3(minus^#(x, y))
  , le^#(0(), y) -> c_4()
  , le^#(s(x), 0()) -> c_5()
  , le^#(s(x), s(y)) -> c_6(le^#(x, y))
  , if^#(true(), x, y) -> c_7()
  , if^#(false(), x, y) -> c_8()
  , perfectp^#(0()) -> c_9()
  , perfectp^#(s(x)) -> c_10(f^#(x, s(0()), s(x), s(x)))
  , f^#(0(), y, 0(), u) -> c_11()
  , f^#(0(), y, s(z), u) -> c_12()
  , f^#(s(x), 0(), z, u) ->
    c_13(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x)))
  , f^#(s(x), s(y), z, u) ->
    c_14(if^#(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)),
         le^#(x, y),
         f^#(s(x), minus(y, x), z, u),
         minus^#(y, x),
         f^#(x, u, z, u)) }
Weak Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), 0()) -> s(x)
  , minus(s(x), s(y)) -> minus(x, y)
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , if(true(), x, y) -> x
  , if(false(), x, y) -> y
  , perfectp(0()) -> false()
  , perfectp(s(x)) -> f(x, s(0()), s(x), s(x))
  , f(0(), y, 0(), u) -> true()
  , f(0(), y, s(z), u) -> false()
  , f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u)
  , f(s(x), s(y), z, u) ->
    if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,2,4,5,7,8,9,11,12} by
applications of Pre({1,2,4,5,7,8,9,11,12}) = {3,6,10,13,14}. Here
rules are labeled as follows:

  DPs:
    { 1: minus^#(0(), y) -> c_1()
    , 2: minus^#(s(x), 0()) -> c_2()
    , 3: minus^#(s(x), s(y)) -> c_3(minus^#(x, y))
    , 4: le^#(0(), y) -> c_4()
    , 5: le^#(s(x), 0()) -> c_5()
    , 6: le^#(s(x), s(y)) -> c_6(le^#(x, y))
    , 7: if^#(true(), x, y) -> c_7()
    , 8: if^#(false(), x, y) -> c_8()
    , 9: perfectp^#(0()) -> c_9()
    , 10: perfectp^#(s(x)) -> c_10(f^#(x, s(0()), s(x), s(x)))
    , 11: f^#(0(), y, 0(), u) -> c_11()
    , 12: f^#(0(), y, s(z), u) -> c_12()
    , 13: f^#(s(x), 0(), z, u) ->
          c_13(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x)))
    , 14: f^#(s(x), s(y), z, u) ->
          c_14(if^#(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)),
               le^#(x, y),
               f^#(s(x), minus(y, x), z, u),
               minus^#(y, x),
               f^#(x, u, z, u)) }

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

Strict DPs:
  { minus^#(s(x), s(y)) -> c_3(minus^#(x, y))
  , le^#(s(x), s(y)) -> c_6(le^#(x, y))
  , perfectp^#(s(x)) -> c_10(f^#(x, s(0()), s(x), s(x)))
  , f^#(s(x), 0(), z, u) ->
    c_13(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x)))
  , f^#(s(x), s(y), z, u) ->
    c_14(if^#(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)),
         le^#(x, y),
         f^#(s(x), minus(y, x), z, u),
         minus^#(y, x),
         f^#(x, u, z, u)) }
Weak DPs:
  { minus^#(0(), y) -> c_1()
  , minus^#(s(x), 0()) -> c_2()
  , le^#(0(), y) -> c_4()
  , le^#(s(x), 0()) -> c_5()
  , if^#(true(), x, y) -> c_7()
  , if^#(false(), x, y) -> c_8()
  , perfectp^#(0()) -> c_9()
  , f^#(0(), y, 0(), u) -> c_11()
  , f^#(0(), y, s(z), u) -> c_12() }
Weak Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), 0()) -> s(x)
  , minus(s(x), s(y)) -> minus(x, y)
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , if(true(), x, y) -> x
  , if(false(), x, y) -> y
  , perfectp(0()) -> false()
  , perfectp(s(x)) -> f(x, s(0()), s(x), s(x))
  , f(0(), y, 0(), u) -> true()
  , f(0(), y, s(z), u) -> false()
  , f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u)
  , f(s(x), s(y), z, u) ->
    if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) }
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.

{ minus^#(0(), y) -> c_1()
, minus^#(s(x), 0()) -> c_2()
, le^#(0(), y) -> c_4()
, le^#(s(x), 0()) -> c_5()
, if^#(true(), x, y) -> c_7()
, if^#(false(), x, y) -> c_8()
, perfectp^#(0()) -> c_9()
, f^#(0(), y, 0(), u) -> c_11()
, f^#(0(), y, s(z), u) -> c_12() }

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

Strict DPs:
  { minus^#(s(x), s(y)) -> c_3(minus^#(x, y))
  , le^#(s(x), s(y)) -> c_6(le^#(x, y))
  , perfectp^#(s(x)) -> c_10(f^#(x, s(0()), s(x), s(x)))
  , f^#(s(x), 0(), z, u) ->
    c_13(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x)))
  , f^#(s(x), s(y), z, u) ->
    c_14(if^#(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)),
         le^#(x, y),
         f^#(s(x), minus(y, x), z, u),
         minus^#(y, x),
         f^#(x, u, z, u)) }
Weak Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), 0()) -> s(x)
  , minus(s(x), s(y)) -> minus(x, y)
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , if(true(), x, y) -> x
  , if(false(), x, y) -> y
  , perfectp(0()) -> false()
  , perfectp(s(x)) -> f(x, s(0()), s(x), s(x))
  , f(0(), y, 0(), u) -> true()
  , f(0(), y, s(z), u) -> false()
  , f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u)
  , f(s(x), s(y), z, u) ->
    if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { f^#(s(x), s(y), z, u) ->
    c_14(if^#(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)),
         le^#(x, y),
         f^#(s(x), minus(y, x), z, u),
         minus^#(y, x),
         f^#(x, u, z, u)) }

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

Strict DPs:
  { minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
  , le^#(s(x), s(y)) -> c_2(le^#(x, y))
  , perfectp^#(s(x)) -> c_3(f^#(x, s(0()), s(x), s(x)))
  , f^#(s(x), 0(), z, u) ->
    c_4(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x)))
  , f^#(s(x), s(y), z, u) ->
    c_5(le^#(x, y),
        f^#(s(x), minus(y, x), z, u),
        minus^#(y, x),
        f^#(x, u, z, u)) }
Weak Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), 0()) -> s(x)
  , minus(s(x), s(y)) -> minus(x, y)
  , le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , if(true(), x, y) -> x
  , if(false(), x, y) -> y
  , perfectp(0()) -> false()
  , perfectp(s(x)) -> f(x, s(0()), s(x), s(x))
  , f(0(), y, 0(), u) -> true()
  , f(0(), y, s(z), u) -> false()
  , f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u)
  , f(s(x), s(y), z, u) ->
    if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { minus(0(), y) -> 0()
    , minus(s(x), 0()) -> s(x)
    , minus(s(x), s(y)) -> minus(x, y) }

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

Strict DPs:
  { minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
  , le^#(s(x), s(y)) -> c_2(le^#(x, y))
  , perfectp^#(s(x)) -> c_3(f^#(x, s(0()), s(x), s(x)))
  , f^#(s(x), 0(), z, u) ->
    c_4(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x)))
  , f^#(s(x), s(y), z, u) ->
    c_5(le^#(x, y),
        f^#(s(x), minus(y, x), z, u),
        minus^#(y, x),
        f^#(x, u, z, u)) }
Weak Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), 0()) -> s(x)
  , minus(s(x), s(y)) -> minus(x, y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
     -->_1 minus^#(s(x), s(y)) -> c_1(minus^#(x, y)) :1
  
  2: le^#(s(x), s(y)) -> c_2(le^#(x, y))
     -->_1 le^#(s(x), s(y)) -> c_2(le^#(x, y)) :2
  
  3: perfectp^#(s(x)) -> c_3(f^#(x, s(0()), s(x), s(x)))
     -->_1 f^#(s(x), s(y), z, u) ->
           c_5(le^#(x, y),
               f^#(s(x), minus(y, x), z, u),
               minus^#(y, x),
               f^#(x, u, z, u)) :5
  
  4: f^#(s(x), 0(), z, u) ->
     c_4(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x)))
     -->_1 f^#(s(x), s(y), z, u) ->
           c_5(le^#(x, y),
               f^#(s(x), minus(y, x), z, u),
               minus^#(y, x),
               f^#(x, u, z, u)) :5
     -->_1 f^#(s(x), 0(), z, u) ->
           c_4(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x))) :4
     -->_2 minus^#(s(x), s(y)) -> c_1(minus^#(x, y)) :1
  
  5: f^#(s(x), s(y), z, u) ->
     c_5(le^#(x, y),
         f^#(s(x), minus(y, x), z, u),
         minus^#(y, x),
         f^#(x, u, z, u))
     -->_4 f^#(s(x), s(y), z, u) ->
           c_5(le^#(x, y),
               f^#(s(x), minus(y, x), z, u),
               minus^#(y, x),
               f^#(x, u, z, u)) :5
     -->_2 f^#(s(x), s(y), z, u) ->
           c_5(le^#(x, y),
               f^#(s(x), minus(y, x), z, u),
               minus^#(y, x),
               f^#(x, u, z, u)) :5
     -->_4 f^#(s(x), 0(), z, u) ->
           c_4(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x))) :4
     -->_2 f^#(s(x), 0(), z, u) ->
           c_4(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x))) :4
     -->_1 le^#(s(x), s(y)) -> c_2(le^#(x, y)) :2
     -->_3 minus^#(s(x), s(y)) -> c_1(minus^#(x, y)) :1
  

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).

  { perfectp^#(s(x)) -> c_3(f^#(x, s(0()), s(x), s(x))) }


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

Strict DPs:
  { minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
  , le^#(s(x), s(y)) -> c_2(le^#(x, y))
  , f^#(s(x), 0(), z, u) ->
    c_4(f^#(x, u, minus(z, s(x)), u), minus^#(z, s(x)))
  , f^#(s(x), s(y), z, u) ->
    c_5(le^#(x, y),
        f^#(s(x), minus(y, x), z, u),
        minus^#(y, x),
        f^#(x, u, z, u)) }
Weak Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), 0()) -> s(x)
  , minus(s(x), s(y)) -> minus(x, y) }
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:
      
      The input cannot be shown compatible
   
   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..