MAYBE

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

Strict Trs:
  { plus(0(), x) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> p(minus(x, y))
  , minus(0(), x) -> 0()
  , isZero(0()) -> true()
  , isZero(s(x)) -> false()
  , facIter(x, y) -> if(isZero(x), minus(x, s(0())), y, times(y, x))
  , if(true(), x, y, z) -> y
  , if(false(), x, y, z) -> facIter(x, z)
  , factorial(x) -> facIter(x, s(0())) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { plus^#(0(), x) -> c_1()
  , plus^#(s(x), y) -> c_2(plus^#(x, y))
  , times^#(0(), y) -> c_3()
  , times^#(s(x), y) -> c_4(plus^#(y, times(x, y)), times^#(x, y))
  , p^#(0()) -> c_5()
  , p^#(s(x)) -> c_6()
  , minus^#(x, 0()) -> c_7()
  , minus^#(x, s(y)) -> c_8(p^#(minus(x, y)), minus^#(x, y))
  , minus^#(0(), x) -> c_9()
  , isZero^#(0()) -> c_10()
  , isZero^#(s(x)) -> c_11()
  , facIter^#(x, y) ->
    c_12(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
         isZero^#(x),
         minus^#(x, s(0())),
         times^#(y, x))
  , if^#(true(), x, y, z) -> c_13()
  , if^#(false(), x, y, z) -> c_14(facIter^#(x, z))
  , factorial^#(x) -> c_15(facIter^#(x, s(0()))) }

and mark the set of starting terms.

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

Strict DPs:
  { plus^#(0(), x) -> c_1()
  , plus^#(s(x), y) -> c_2(plus^#(x, y))
  , times^#(0(), y) -> c_3()
  , times^#(s(x), y) -> c_4(plus^#(y, times(x, y)), times^#(x, y))
  , p^#(0()) -> c_5()
  , p^#(s(x)) -> c_6()
  , minus^#(x, 0()) -> c_7()
  , minus^#(x, s(y)) -> c_8(p^#(minus(x, y)), minus^#(x, y))
  , minus^#(0(), x) -> c_9()
  , isZero^#(0()) -> c_10()
  , isZero^#(s(x)) -> c_11()
  , facIter^#(x, y) ->
    c_12(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
         isZero^#(x),
         minus^#(x, s(0())),
         times^#(y, x))
  , if^#(true(), x, y, z) -> c_13()
  , if^#(false(), x, y, z) -> c_14(facIter^#(x, z))
  , factorial^#(x) -> c_15(facIter^#(x, s(0()))) }
Weak Trs:
  { plus(0(), x) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> p(minus(x, y))
  , minus(0(), x) -> 0()
  , isZero(0()) -> true()
  , isZero(s(x)) -> false()
  , facIter(x, y) -> if(isZero(x), minus(x, s(0())), y, times(y, x))
  , if(true(), x, y, z) -> y
  , if(false(), x, y, z) -> facIter(x, z)
  , factorial(x) -> facIter(x, s(0())) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: plus^#(0(), x) -> c_1()
    , 2: plus^#(s(x), y) -> c_2(plus^#(x, y))
    , 3: times^#(0(), y) -> c_3()
    , 4: times^#(s(x), y) -> c_4(plus^#(y, times(x, y)), times^#(x, y))
    , 5: p^#(0()) -> c_5()
    , 6: p^#(s(x)) -> c_6()
    , 7: minus^#(x, 0()) -> c_7()
    , 8: minus^#(x, s(y)) -> c_8(p^#(minus(x, y)), minus^#(x, y))
    , 9: minus^#(0(), x) -> c_9()
    , 10: isZero^#(0()) -> c_10()
    , 11: isZero^#(s(x)) -> c_11()
    , 12: facIter^#(x, y) ->
          c_12(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
               isZero^#(x),
               minus^#(x, s(0())),
               times^#(y, x))
    , 13: if^#(true(), x, y, z) -> c_13()
    , 14: if^#(false(), x, y, z) -> c_14(facIter^#(x, z))
    , 15: factorial^#(x) -> c_15(facIter^#(x, s(0()))) }

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

Strict DPs:
  { plus^#(s(x), y) -> c_2(plus^#(x, y))
  , times^#(s(x), y) -> c_4(plus^#(y, times(x, y)), times^#(x, y))
  , minus^#(x, s(y)) -> c_8(p^#(minus(x, y)), minus^#(x, y))
  , facIter^#(x, y) ->
    c_12(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
         isZero^#(x),
         minus^#(x, s(0())),
         times^#(y, x))
  , if^#(false(), x, y, z) -> c_14(facIter^#(x, z))
  , factorial^#(x) -> c_15(facIter^#(x, s(0()))) }
Weak DPs:
  { plus^#(0(), x) -> c_1()
  , times^#(0(), y) -> c_3()
  , p^#(0()) -> c_5()
  , p^#(s(x)) -> c_6()
  , minus^#(x, 0()) -> c_7()
  , minus^#(0(), x) -> c_9()
  , isZero^#(0()) -> c_10()
  , isZero^#(s(x)) -> c_11()
  , if^#(true(), x, y, z) -> c_13() }
Weak Trs:
  { plus(0(), x) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> p(minus(x, y))
  , minus(0(), x) -> 0()
  , isZero(0()) -> true()
  , isZero(s(x)) -> false()
  , facIter(x, y) -> if(isZero(x), minus(x, s(0())), y, times(y, x))
  , if(true(), x, y, z) -> y
  , if(false(), x, y, z) -> facIter(x, z)
  , factorial(x) -> facIter(x, s(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.

{ plus^#(0(), x) -> c_1()
, times^#(0(), y) -> c_3()
, p^#(0()) -> c_5()
, p^#(s(x)) -> c_6()
, minus^#(x, 0()) -> c_7()
, minus^#(0(), x) -> c_9()
, isZero^#(0()) -> c_10()
, isZero^#(s(x)) -> c_11()
, if^#(true(), x, y, z) -> c_13() }

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

Strict DPs:
  { plus^#(s(x), y) -> c_2(plus^#(x, y))
  , times^#(s(x), y) -> c_4(plus^#(y, times(x, y)), times^#(x, y))
  , minus^#(x, s(y)) -> c_8(p^#(minus(x, y)), minus^#(x, y))
  , facIter^#(x, y) ->
    c_12(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
         isZero^#(x),
         minus^#(x, s(0())),
         times^#(y, x))
  , if^#(false(), x, y, z) -> c_14(facIter^#(x, z))
  , factorial^#(x) -> c_15(facIter^#(x, s(0()))) }
Weak Trs:
  { plus(0(), x) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> p(minus(x, y))
  , minus(0(), x) -> 0()
  , isZero(0()) -> true()
  , isZero(s(x)) -> false()
  , facIter(x, y) -> if(isZero(x), minus(x, s(0())), y, times(y, x))
  , if(true(), x, y, z) -> y
  , if(false(), x, y, z) -> facIter(x, z)
  , factorial(x) -> facIter(x, s(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:

  { minus^#(x, s(y)) -> c_8(p^#(minus(x, y)), minus^#(x, y))
  , facIter^#(x, y) ->
    c_12(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
         isZero^#(x),
         minus^#(x, s(0())),
         times^#(y, x)) }

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

Strict DPs:
  { plus^#(s(x), y) -> c_1(plus^#(x, y))
  , times^#(s(x), y) -> c_2(plus^#(y, times(x, y)), times^#(x, y))
  , minus^#(x, s(y)) -> c_3(minus^#(x, y))
  , facIter^#(x, y) ->
    c_4(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
        minus^#(x, s(0())),
        times^#(y, x))
  , if^#(false(), x, y, z) -> c_5(facIter^#(x, z))
  , factorial^#(x) -> c_6(facIter^#(x, s(0()))) }
Weak Trs:
  { plus(0(), x) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> p(minus(x, y))
  , minus(0(), x) -> 0()
  , isZero(0()) -> true()
  , isZero(s(x)) -> false()
  , facIter(x, y) -> if(isZero(x), minus(x, s(0())), y, times(y, x))
  , if(true(), x, y, z) -> y
  , if(false(), x, y, z) -> facIter(x, z)
  , factorial(x) -> facIter(x, s(0())) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { plus(0(), x) -> x
    , plus(s(x), y) -> s(plus(x, y))
    , times(0(), y) -> 0()
    , times(s(x), y) -> plus(y, times(x, y))
    , p(0()) -> 0()
    , p(s(x)) -> x
    , minus(x, 0()) -> x
    , minus(x, s(y)) -> p(minus(x, y))
    , minus(0(), x) -> 0()
    , isZero(0()) -> true()
    , isZero(s(x)) -> false() }

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

Strict DPs:
  { plus^#(s(x), y) -> c_1(plus^#(x, y))
  , times^#(s(x), y) -> c_2(plus^#(y, times(x, y)), times^#(x, y))
  , minus^#(x, s(y)) -> c_3(minus^#(x, y))
  , facIter^#(x, y) ->
    c_4(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
        minus^#(x, s(0())),
        times^#(y, x))
  , if^#(false(), x, y, z) -> c_5(facIter^#(x, z))
  , factorial^#(x) -> c_6(facIter^#(x, s(0()))) }
Weak Trs:
  { plus(0(), x) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> p(minus(x, y))
  , minus(0(), x) -> 0()
  , isZero(0()) -> true()
  , isZero(s(x)) -> false() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: plus^#(s(x), y) -> c_1(plus^#(x, y))
     -->_1 plus^#(s(x), y) -> c_1(plus^#(x, y)) :1
  
  2: times^#(s(x), y) -> c_2(plus^#(y, times(x, y)), times^#(x, y))
     -->_2 times^#(s(x), y) ->
           c_2(plus^#(y, times(x, y)), times^#(x, y)) :2
     -->_1 plus^#(s(x), y) -> c_1(plus^#(x, y)) :1
  
  3: minus^#(x, s(y)) -> c_3(minus^#(x, y))
     -->_1 minus^#(x, s(y)) -> c_3(minus^#(x, y)) :3
  
  4: facIter^#(x, y) ->
     c_4(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
         minus^#(x, s(0())),
         times^#(y, x))
     -->_1 if^#(false(), x, y, z) -> c_5(facIter^#(x, z)) :5
     -->_2 minus^#(x, s(y)) -> c_3(minus^#(x, y)) :3
     -->_3 times^#(s(x), y) ->
           c_2(plus^#(y, times(x, y)), times^#(x, y)) :2
  
  5: if^#(false(), x, y, z) -> c_5(facIter^#(x, z))
     -->_1 facIter^#(x, y) ->
           c_4(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
               minus^#(x, s(0())),
               times^#(y, x)) :4
  
  6: factorial^#(x) -> c_6(facIter^#(x, s(0())))
     -->_1 facIter^#(x, y) ->
           c_4(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
               minus^#(x, s(0())),
               times^#(y, x)) :4
  

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).

  { factorial^#(x) -> c_6(facIter^#(x, s(0()))) }


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

Strict DPs:
  { plus^#(s(x), y) -> c_1(plus^#(x, y))
  , times^#(s(x), y) -> c_2(plus^#(y, times(x, y)), times^#(x, y))
  , minus^#(x, s(y)) -> c_3(minus^#(x, y))
  , facIter^#(x, y) ->
    c_4(if^#(isZero(x), minus(x, s(0())), y, times(y, x)),
        minus^#(x, s(0())),
        times^#(y, x))
  , if^#(false(), x, y, z) -> c_5(facIter^#(x, z)) }
Weak Trs:
  { plus(0(), x) -> x
  , plus(s(x), y) -> s(plus(x, y))
  , times(0(), y) -> 0()
  , times(s(x), y) -> plus(y, times(x, y))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> p(minus(x, y))
  , minus(0(), x) -> 0()
  , isZero(0()) -> true()
  , isZero(s(x)) -> false() }
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..