MAYBE

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

Strict Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , mod(x, 0()) -> modZeroErro()
  , mod(x, s(y)) -> modIter(x, s(y), 0(), 0())
  , modIter(x, s(y), z, u) -> if(le(x, z), x, s(y), z, u)
  , if(true(), x, y, z, u) -> u
  , if(false(), x, y, z, u) -> if2(le(y, s(u)), x, y, s(z), s(u))
  , if2(true(), x, y, z, u) -> modIter(x, y, z, 0())
  , if2(false(), x, y, z, u) -> modIter(x, y, z, u) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { le^#(0(), y) -> c_1()
  , le^#(s(x), 0()) -> c_2()
  , le^#(s(x), s(y)) -> c_3(le^#(x, y))
  , mod^#(x, 0()) -> c_4()
  , mod^#(x, s(y)) -> c_5(modIter^#(x, s(y), 0(), 0()))
  , modIter^#(x, s(y), z, u) ->
    c_6(if^#(le(x, z), x, s(y), z, u), le^#(x, z))
  , if^#(true(), x, y, z, u) -> c_7()
  , if^#(false(), x, y, z, u) ->
    c_8(if2^#(le(y, s(u)), x, y, s(z), s(u)), le^#(y, s(u)))
  , if2^#(true(), x, y, z, u) -> c_9(modIter^#(x, y, z, 0()))
  , if2^#(false(), x, y, z, u) -> c_10(modIter^#(x, y, z, u)) }

and mark the set of starting terms.

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

Strict DPs:
  { le^#(0(), y) -> c_1()
  , le^#(s(x), 0()) -> c_2()
  , le^#(s(x), s(y)) -> c_3(le^#(x, y))
  , mod^#(x, 0()) -> c_4()
  , mod^#(x, s(y)) -> c_5(modIter^#(x, s(y), 0(), 0()))
  , modIter^#(x, s(y), z, u) ->
    c_6(if^#(le(x, z), x, s(y), z, u), le^#(x, z))
  , if^#(true(), x, y, z, u) -> c_7()
  , if^#(false(), x, y, z, u) ->
    c_8(if2^#(le(y, s(u)), x, y, s(z), s(u)), le^#(y, s(u)))
  , if2^#(true(), x, y, z, u) -> c_9(modIter^#(x, y, z, 0()))
  , if2^#(false(), x, y, z, u) -> c_10(modIter^#(x, y, z, u)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , mod(x, 0()) -> modZeroErro()
  , mod(x, s(y)) -> modIter(x, s(y), 0(), 0())
  , modIter(x, s(y), z, u) -> if(le(x, z), x, s(y), z, u)
  , if(true(), x, y, z, u) -> u
  , if(false(), x, y, z, u) -> if2(le(y, s(u)), x, y, s(z), s(u))
  , if2(true(), x, y, z, u) -> modIter(x, y, z, 0())
  , if2(false(), x, y, z, u) -> modIter(x, y, z, u) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: le^#(0(), y) -> c_1()
    , 2: le^#(s(x), 0()) -> c_2()
    , 3: le^#(s(x), s(y)) -> c_3(le^#(x, y))
    , 4: mod^#(x, 0()) -> c_4()
    , 5: mod^#(x, s(y)) -> c_5(modIter^#(x, s(y), 0(), 0()))
    , 6: modIter^#(x, s(y), z, u) ->
         c_6(if^#(le(x, z), x, s(y), z, u), le^#(x, z))
    , 7: if^#(true(), x, y, z, u) -> c_7()
    , 8: if^#(false(), x, y, z, u) ->
         c_8(if2^#(le(y, s(u)), x, y, s(z), s(u)), le^#(y, s(u)))
    , 9: if2^#(true(), x, y, z, u) -> c_9(modIter^#(x, y, z, 0()))
    , 10: if2^#(false(), x, y, z, u) -> c_10(modIter^#(x, y, z, u)) }

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_3(le^#(x, y))
  , mod^#(x, s(y)) -> c_5(modIter^#(x, s(y), 0(), 0()))
  , modIter^#(x, s(y), z, u) ->
    c_6(if^#(le(x, z), x, s(y), z, u), le^#(x, z))
  , if^#(false(), x, y, z, u) ->
    c_8(if2^#(le(y, s(u)), x, y, s(z), s(u)), le^#(y, s(u)))
  , if2^#(true(), x, y, z, u) -> c_9(modIter^#(x, y, z, 0()))
  , if2^#(false(), x, y, z, u) -> c_10(modIter^#(x, y, z, u)) }
Weak DPs:
  { le^#(0(), y) -> c_1()
  , le^#(s(x), 0()) -> c_2()
  , mod^#(x, 0()) -> c_4()
  , if^#(true(), x, y, z, u) -> c_7() }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , mod(x, 0()) -> modZeroErro()
  , mod(x, s(y)) -> modIter(x, s(y), 0(), 0())
  , modIter(x, s(y), z, u) -> if(le(x, z), x, s(y), z, u)
  , if(true(), x, y, z, u) -> u
  , if(false(), x, y, z, u) -> if2(le(y, s(u)), x, y, s(z), s(u))
  , if2(true(), x, y, z, u) -> modIter(x, y, z, 0())
  , if2(false(), x, y, z, u) -> modIter(x, y, 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.

{ le^#(0(), y) -> c_1()
, le^#(s(x), 0()) -> c_2()
, mod^#(x, 0()) -> c_4()
, if^#(true(), x, y, z, u) -> c_7() }

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_3(le^#(x, y))
  , mod^#(x, s(y)) -> c_5(modIter^#(x, s(y), 0(), 0()))
  , modIter^#(x, s(y), z, u) ->
    c_6(if^#(le(x, z), x, s(y), z, u), le^#(x, z))
  , if^#(false(), x, y, z, u) ->
    c_8(if2^#(le(y, s(u)), x, y, s(z), s(u)), le^#(y, s(u)))
  , if2^#(true(), x, y, z, u) -> c_9(modIter^#(x, y, z, 0()))
  , if2^#(false(), x, y, z, u) -> c_10(modIter^#(x, y, z, u)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , mod(x, 0()) -> modZeroErro()
  , mod(x, s(y)) -> modIter(x, s(y), 0(), 0())
  , modIter(x, s(y), z, u) -> if(le(x, z), x, s(y), z, u)
  , if(true(), x, y, z, u) -> u
  , if(false(), x, y, z, u) -> if2(le(y, s(u)), x, y, s(z), s(u))
  , if2(true(), x, y, z, u) -> modIter(x, y, z, 0())
  , if2(false(), x, y, z, u) -> modIter(x, y, z, u) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: le^#(s(x), s(y)) -> c_3(le^#(x, y))
     -->_1 le^#(s(x), s(y)) -> c_3(le^#(x, y)) :1
  
  2: mod^#(x, s(y)) -> c_5(modIter^#(x, s(y), 0(), 0()))
     -->_1 modIter^#(x, s(y), z, u) ->
           c_6(if^#(le(x, z), x, s(y), z, u), le^#(x, z)) :3
  
  3: modIter^#(x, s(y), z, u) ->
     c_6(if^#(le(x, z), x, s(y), z, u), le^#(x, z))
     -->_1 if^#(false(), x, y, z, u) ->
           c_8(if2^#(le(y, s(u)), x, y, s(z), s(u)), le^#(y, s(u))) :4
     -->_2 le^#(s(x), s(y)) -> c_3(le^#(x, y)) :1
  
  4: if^#(false(), x, y, z, u) ->
     c_8(if2^#(le(y, s(u)), x, y, s(z), s(u)), le^#(y, s(u)))
     -->_1 if2^#(false(), x, y, z, u) -> c_10(modIter^#(x, y, z, u)) :6
     -->_1 if2^#(true(), x, y, z, u) -> c_9(modIter^#(x, y, z, 0())) :5
     -->_2 le^#(s(x), s(y)) -> c_3(le^#(x, y)) :1
  
  5: if2^#(true(), x, y, z, u) -> c_9(modIter^#(x, y, z, 0()))
     -->_1 modIter^#(x, s(y), z, u) ->
           c_6(if^#(le(x, z), x, s(y), z, u), le^#(x, z)) :3
  
  6: if2^#(false(), x, y, z, u) -> c_10(modIter^#(x, y, z, u))
     -->_1 modIter^#(x, s(y), z, u) ->
           c_6(if^#(le(x, z), x, s(y), z, u), le^#(x, z)) :3
  

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).

  { mod^#(x, s(y)) -> c_5(modIter^#(x, s(y), 0(), 0())) }


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

Strict DPs:
  { le^#(s(x), s(y)) -> c_3(le^#(x, y))
  , modIter^#(x, s(y), z, u) ->
    c_6(if^#(le(x, z), x, s(y), z, u), le^#(x, z))
  , if^#(false(), x, y, z, u) ->
    c_8(if2^#(le(y, s(u)), x, y, s(z), s(u)), le^#(y, s(u)))
  , if2^#(true(), x, y, z, u) -> c_9(modIter^#(x, y, z, 0()))
  , if2^#(false(), x, y, z, u) -> c_10(modIter^#(x, y, z, u)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , mod(x, 0()) -> modZeroErro()
  , mod(x, s(y)) -> modIter(x, s(y), 0(), 0())
  , modIter(x, s(y), z, u) -> if(le(x, z), x, s(y), z, u)
  , if(true(), x, y, z, u) -> u
  , if(false(), x, y, z, u) -> if2(le(y, s(u)), x, y, s(z), s(u))
  , if2(true(), x, y, z, u) -> modIter(x, y, z, 0())
  , if2(false(), x, y, z, u) -> modIter(x, y, z, u) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { le(0(), y) -> true()
    , le(s(x), 0()) -> false()
    , le(s(x), s(y)) -> le(x, y) }

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_3(le^#(x, y))
  , modIter^#(x, s(y), z, u) ->
    c_6(if^#(le(x, z), x, s(y), z, u), le^#(x, z))
  , if^#(false(), x, y, z, u) ->
    c_8(if2^#(le(y, s(u)), x, y, s(z), s(u)), le^#(y, s(u)))
  , if2^#(true(), x, y, z, u) -> c_9(modIter^#(x, y, z, 0()))
  , if2^#(false(), x, y, z, u) -> c_10(modIter^#(x, y, z, u)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..