MAYBE

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

Strict Trs:
  { minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(0(), x) -> 0()
  , 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)
  , quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))
  , if_quot(x, y, true(), z) -> divByZeroError()
  , if_quot(x, y, false(), true()) -> s(quot(x, y))
  , if_quot(x, y, false(), false()) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { minus^#(x, x) -> c_1()
  , minus^#(x, 0()) -> c_2()
  , minus^#(0(), x) -> c_3()
  , minus^#(s(x), s(y)) -> c_4(minus^#(x, y))
  , le^#(0(), y) -> c_5()
  , le^#(s(x), 0()) -> c_6()
  , le^#(s(x), s(y)) -> c_7(le^#(x, y))
  , quot^#(x, y) ->
    c_8(if_quot^#(minus(x, y), y, le(y, 0()), le(y, x)),
        minus^#(x, y),
        le^#(y, 0()),
        le^#(y, x))
  , if_quot^#(x, y, true(), z) -> c_9()
  , if_quot^#(x, y, false(), true()) -> c_10(quot^#(x, y))
  , if_quot^#(x, y, false(), false()) -> c_11() }

and mark the set of starting terms.

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

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

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

  DPs:
    { 1: minus^#(x, x) -> c_1()
    , 2: minus^#(x, 0()) -> c_2()
    , 3: minus^#(0(), x) -> c_3()
    , 4: minus^#(s(x), s(y)) -> c_4(minus^#(x, y))
    , 5: le^#(0(), y) -> c_5()
    , 6: le^#(s(x), 0()) -> c_6()
    , 7: le^#(s(x), s(y)) -> c_7(le^#(x, y))
    , 8: quot^#(x, y) ->
         c_8(if_quot^#(minus(x, y), y, le(y, 0()), le(y, x)),
             minus^#(x, y),
             le^#(y, 0()),
             le^#(y, x))
    , 9: if_quot^#(x, y, true(), z) -> c_9()
    , 10: if_quot^#(x, y, false(), true()) -> c_10(quot^#(x, y))
    , 11: if_quot^#(x, y, false(), false()) -> c_11() }

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

Strict DPs:
  { minus^#(s(x), s(y)) -> c_4(minus^#(x, y))
  , le^#(s(x), s(y)) -> c_7(le^#(x, y))
  , quot^#(x, y) ->
    c_8(if_quot^#(minus(x, y), y, le(y, 0()), le(y, x)),
        minus^#(x, y),
        le^#(y, 0()),
        le^#(y, x))
  , if_quot^#(x, y, false(), true()) -> c_10(quot^#(x, y)) }
Weak DPs:
  { minus^#(x, x) -> c_1()
  , minus^#(x, 0()) -> c_2()
  , minus^#(0(), x) -> c_3()
  , le^#(0(), y) -> c_5()
  , le^#(s(x), 0()) -> c_6()
  , if_quot^#(x, y, true(), z) -> c_9()
  , if_quot^#(x, y, false(), false()) -> c_11() }
Weak Trs:
  { minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(0(), x) -> 0()
  , 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)
  , quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))
  , if_quot(x, y, true(), z) -> divByZeroError()
  , if_quot(x, y, false(), true()) -> s(quot(x, y))
  , if_quot(x, y, false(), false()) -> 0() }
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^#(x, x) -> c_1()
, minus^#(x, 0()) -> c_2()
, minus^#(0(), x) -> c_3()
, le^#(0(), y) -> c_5()
, le^#(s(x), 0()) -> c_6()
, if_quot^#(x, y, true(), z) -> c_9()
, if_quot^#(x, y, false(), false()) -> c_11() }

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

Strict DPs:
  { minus^#(s(x), s(y)) -> c_4(minus^#(x, y))
  , le^#(s(x), s(y)) -> c_7(le^#(x, y))
  , quot^#(x, y) ->
    c_8(if_quot^#(minus(x, y), y, le(y, 0()), le(y, x)),
        minus^#(x, y),
        le^#(y, 0()),
        le^#(y, x))
  , if_quot^#(x, y, false(), true()) -> c_10(quot^#(x, y)) }
Weak Trs:
  { minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(0(), x) -> 0()
  , 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)
  , quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))
  , if_quot(x, y, true(), z) -> divByZeroError()
  , if_quot(x, y, false(), true()) -> s(quot(x, y))
  , if_quot(x, y, false(), false()) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { quot^#(x, y) ->
    c_8(if_quot^#(minus(x, y), y, le(y, 0()), le(y, x)),
        minus^#(x, y),
        le^#(y, 0()),
        le^#(y, 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))
  , quot^#(x, y) ->
    c_3(if_quot^#(minus(x, y), y, le(y, 0()), le(y, x)),
        minus^#(x, y),
        le^#(y, x))
  , if_quot^#(x, y, false(), true()) -> c_4(quot^#(x, y)) }
Weak Trs:
  { minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(0(), x) -> 0()
  , 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)
  , quot(x, y) -> if_quot(minus(x, y), y, le(y, 0()), le(y, x))
  , if_quot(x, y, true(), z) -> divByZeroError()
  , if_quot(x, y, false(), true()) -> s(quot(x, y))
  , if_quot(x, y, false(), false()) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

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

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))
  , quot^#(x, y) ->
    c_3(if_quot^#(minus(x, y), y, le(y, 0()), le(y, x)),
        minus^#(x, y),
        le^#(y, x))
  , if_quot^#(x, y, false(), true()) -> c_4(quot^#(x, y)) }
Weak Trs:
  { minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(0(), x) -> 0()
  , 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) }
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..