MAYBE

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

Strict Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), y) -> if(gt(s(x), y), x, y)
  , if(true(), x, y) -> s(minus(x, y))
  , if(false(), x, y) -> 0()
  , gt(0(), y) -> false()
  , gt(s(x), 0()) -> true()
  , gt(s(x), s(y)) -> gt(x, y)
  , mod(x, 0()) -> 0()
  , mod(x, s(y)) -> if1(lt(x, s(y)), x, s(y))
  , if1(true(), x, y) -> x
  , if1(false(), x, y) -> mod(minus(x, y), y)
  , lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { minus^#(0(), y) -> c_1()
  , minus^#(s(x), y) -> c_2(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
  , if^#(true(), x, y) -> c_3(minus^#(x, y))
  , if^#(false(), x, y) -> c_4()
  , gt^#(0(), y) -> c_5()
  , gt^#(s(x), 0()) -> c_6()
  , gt^#(s(x), s(y)) -> c_7(gt^#(x, y))
  , mod^#(x, 0()) -> c_8()
  , mod^#(x, s(y)) -> c_9(if1^#(lt(x, s(y)), x, s(y)), lt^#(x, s(y)))
  , if1^#(true(), x, y) -> c_10()
  , if1^#(false(), x, y) ->
    c_11(mod^#(minus(x, y), y), minus^#(x, y))
  , lt^#(x, 0()) -> c_12()
  , lt^#(0(), s(x)) -> c_13()
  , lt^#(s(x), s(y)) -> c_14(lt^#(x, y)) }

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), y) -> c_2(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
  , if^#(true(), x, y) -> c_3(minus^#(x, y))
  , if^#(false(), x, y) -> c_4()
  , gt^#(0(), y) -> c_5()
  , gt^#(s(x), 0()) -> c_6()
  , gt^#(s(x), s(y)) -> c_7(gt^#(x, y))
  , mod^#(x, 0()) -> c_8()
  , mod^#(x, s(y)) -> c_9(if1^#(lt(x, s(y)), x, s(y)), lt^#(x, s(y)))
  , if1^#(true(), x, y) -> c_10()
  , if1^#(false(), x, y) ->
    c_11(mod^#(minus(x, y), y), minus^#(x, y))
  , lt^#(x, 0()) -> c_12()
  , lt^#(0(), s(x)) -> c_13()
  , lt^#(s(x), s(y)) -> c_14(lt^#(x, y)) }
Weak Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), y) -> if(gt(s(x), y), x, y)
  , if(true(), x, y) -> s(minus(x, y))
  , if(false(), x, y) -> 0()
  , gt(0(), y) -> false()
  , gt(s(x), 0()) -> true()
  , gt(s(x), s(y)) -> gt(x, y)
  , mod(x, 0()) -> 0()
  , mod(x, s(y)) -> if1(lt(x, s(y)), x, s(y))
  , if1(true(), x, y) -> x
  , if1(false(), x, y) -> mod(minus(x, y), y)
  , lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: minus^#(0(), y) -> c_1()
    , 2: minus^#(s(x), y) ->
         c_2(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
    , 3: if^#(true(), x, y) -> c_3(minus^#(x, y))
    , 4: if^#(false(), x, y) -> c_4()
    , 5: gt^#(0(), y) -> c_5()
    , 6: gt^#(s(x), 0()) -> c_6()
    , 7: gt^#(s(x), s(y)) -> c_7(gt^#(x, y))
    , 8: mod^#(x, 0()) -> c_8()
    , 9: mod^#(x, s(y)) ->
         c_9(if1^#(lt(x, s(y)), x, s(y)), lt^#(x, s(y)))
    , 10: if1^#(true(), x, y) -> c_10()
    , 11: if1^#(false(), x, y) ->
          c_11(mod^#(minus(x, y), y), minus^#(x, y))
    , 12: lt^#(x, 0()) -> c_12()
    , 13: lt^#(0(), s(x)) -> c_13()
    , 14: lt^#(s(x), s(y)) -> c_14(lt^#(x, y)) }

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

Strict DPs:
  { minus^#(s(x), y) -> c_2(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
  , if^#(true(), x, y) -> c_3(minus^#(x, y))
  , gt^#(s(x), s(y)) -> c_7(gt^#(x, y))
  , mod^#(x, s(y)) -> c_9(if1^#(lt(x, s(y)), x, s(y)), lt^#(x, s(y)))
  , if1^#(false(), x, y) ->
    c_11(mod^#(minus(x, y), y), minus^#(x, y))
  , lt^#(s(x), s(y)) -> c_14(lt^#(x, y)) }
Weak DPs:
  { minus^#(0(), y) -> c_1()
  , if^#(false(), x, y) -> c_4()
  , gt^#(0(), y) -> c_5()
  , gt^#(s(x), 0()) -> c_6()
  , mod^#(x, 0()) -> c_8()
  , if1^#(true(), x, y) -> c_10()
  , lt^#(x, 0()) -> c_12()
  , lt^#(0(), s(x)) -> c_13() }
Weak Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), y) -> if(gt(s(x), y), x, y)
  , if(true(), x, y) -> s(minus(x, y))
  , if(false(), x, y) -> 0()
  , gt(0(), y) -> false()
  , gt(s(x), 0()) -> true()
  , gt(s(x), s(y)) -> gt(x, y)
  , mod(x, 0()) -> 0()
  , mod(x, s(y)) -> if1(lt(x, s(y)), x, s(y))
  , if1(true(), x, y) -> x
  , if1(false(), x, y) -> mod(minus(x, y), y)
  , lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y) }
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()
, if^#(false(), x, y) -> c_4()
, gt^#(0(), y) -> c_5()
, gt^#(s(x), 0()) -> c_6()
, mod^#(x, 0()) -> c_8()
, if1^#(true(), x, y) -> c_10()
, lt^#(x, 0()) -> c_12()
, lt^#(0(), s(x)) -> c_13() }

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

Strict DPs:
  { minus^#(s(x), y) -> c_2(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
  , if^#(true(), x, y) -> c_3(minus^#(x, y))
  , gt^#(s(x), s(y)) -> c_7(gt^#(x, y))
  , mod^#(x, s(y)) -> c_9(if1^#(lt(x, s(y)), x, s(y)), lt^#(x, s(y)))
  , if1^#(false(), x, y) ->
    c_11(mod^#(minus(x, y), y), minus^#(x, y))
  , lt^#(s(x), s(y)) -> c_14(lt^#(x, y)) }
Weak Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), y) -> if(gt(s(x), y), x, y)
  , if(true(), x, y) -> s(minus(x, y))
  , if(false(), x, y) -> 0()
  , gt(0(), y) -> false()
  , gt(s(x), 0()) -> true()
  , gt(s(x), s(y)) -> gt(x, y)
  , mod(x, 0()) -> 0()
  , mod(x, s(y)) -> if1(lt(x, s(y)), x, s(y))
  , if1(true(), x, y) -> x
  , if1(false(), x, y) -> mod(minus(x, y), y)
  , lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(x, y) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { minus(0(), y) -> 0()
    , minus(s(x), y) -> if(gt(s(x), y), x, y)
    , if(true(), x, y) -> s(minus(x, y))
    , if(false(), x, y) -> 0()
    , gt(0(), y) -> false()
    , gt(s(x), 0()) -> true()
    , gt(s(x), s(y)) -> gt(x, y)
    , lt(x, 0()) -> false()
    , lt(0(), s(x)) -> true()
    , lt(s(x), s(y)) -> lt(x, y) }

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

Strict DPs:
  { minus^#(s(x), y) -> c_2(if^#(gt(s(x), y), x, y), gt^#(s(x), y))
  , if^#(true(), x, y) -> c_3(minus^#(x, y))
  , gt^#(s(x), s(y)) -> c_7(gt^#(x, y))
  , mod^#(x, s(y)) -> c_9(if1^#(lt(x, s(y)), x, s(y)), lt^#(x, s(y)))
  , if1^#(false(), x, y) ->
    c_11(mod^#(minus(x, y), y), minus^#(x, y))
  , lt^#(s(x), s(y)) -> c_14(lt^#(x, y)) }
Weak Trs:
  { minus(0(), y) -> 0()
  , minus(s(x), y) -> if(gt(s(x), y), x, y)
  , if(true(), x, y) -> s(minus(x, y))
  , if(false(), x, y) -> 0()
  , gt(0(), y) -> false()
  , gt(s(x), 0()) -> true()
  , gt(s(x), s(y)) -> gt(x, y)
  , lt(x, 0()) -> false()
  , lt(0(), s(x)) -> true()
  , lt(s(x), s(y)) -> lt(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..