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)
  , pred(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> pred(minus(x, y))
  , gcd(0(), y) -> y
  , gcd(s(x), 0()) -> s(x)
  , gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
  , if_gcd(true(), x, y) -> gcd(minus(x, y), y)
  , if_gcd(false(), x, y) -> gcd(minus(y, x), x) }
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))
  , pred^#(s(x)) -> c_4()
  , minus^#(x, 0()) -> c_5()
  , minus^#(x, s(y)) -> c_6(pred^#(minus(x, y)), minus^#(x, y))
  , gcd^#(0(), y) -> c_7()
  , gcd^#(s(x), 0()) -> c_8()
  , gcd^#(s(x), s(y)) ->
    c_9(if_gcd^#(le(y, x), s(x), s(y)), le^#(y, x))
  , if_gcd^#(true(), x, y) ->
    c_10(gcd^#(minus(x, y), y), minus^#(x, y))
  , if_gcd^#(false(), x, y) ->
    c_11(gcd^#(minus(y, x), x), minus^#(y, x)) }

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))
  , pred^#(s(x)) -> c_4()
  , minus^#(x, 0()) -> c_5()
  , minus^#(x, s(y)) -> c_6(pred^#(minus(x, y)), minus^#(x, y))
  , gcd^#(0(), y) -> c_7()
  , gcd^#(s(x), 0()) -> c_8()
  , gcd^#(s(x), s(y)) ->
    c_9(if_gcd^#(le(y, x), s(x), s(y)), le^#(y, x))
  , if_gcd^#(true(), x, y) ->
    c_10(gcd^#(minus(x, y), y), minus^#(x, y))
  , if_gcd^#(false(), x, y) ->
    c_11(gcd^#(minus(y, x), x), minus^#(y, x)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , pred(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> pred(minus(x, y))
  , gcd(0(), y) -> y
  , gcd(s(x), 0()) -> s(x)
  , gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
  , if_gcd(true(), x, y) -> gcd(minus(x, y), y)
  , if_gcd(false(), x, y) -> gcd(minus(y, x), x) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,2,4,5,7,8} by
applications of Pre({1,2,4,5,7,8}) = {3,6,9,10,11}. 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: pred^#(s(x)) -> c_4()
    , 5: minus^#(x, 0()) -> c_5()
    , 6: minus^#(x, s(y)) -> c_6(pred^#(minus(x, y)), minus^#(x, y))
    , 7: gcd^#(0(), y) -> c_7()
    , 8: gcd^#(s(x), 0()) -> c_8()
    , 9: gcd^#(s(x), s(y)) ->
         c_9(if_gcd^#(le(y, x), s(x), s(y)), le^#(y, x))
    , 10: if_gcd^#(true(), x, y) ->
          c_10(gcd^#(minus(x, y), y), minus^#(x, y))
    , 11: if_gcd^#(false(), x, y) ->
          c_11(gcd^#(minus(y, x), x), minus^#(y, x)) }

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))
  , minus^#(x, s(y)) -> c_6(pred^#(minus(x, y)), minus^#(x, y))
  , gcd^#(s(x), s(y)) ->
    c_9(if_gcd^#(le(y, x), s(x), s(y)), le^#(y, x))
  , if_gcd^#(true(), x, y) ->
    c_10(gcd^#(minus(x, y), y), minus^#(x, y))
  , if_gcd^#(false(), x, y) ->
    c_11(gcd^#(minus(y, x), x), minus^#(y, x)) }
Weak DPs:
  { le^#(0(), y) -> c_1()
  , le^#(s(x), 0()) -> c_2()
  , pred^#(s(x)) -> c_4()
  , minus^#(x, 0()) -> c_5()
  , gcd^#(0(), y) -> c_7()
  , gcd^#(s(x), 0()) -> c_8() }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , pred(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> pred(minus(x, y))
  , gcd(0(), y) -> y
  , gcd(s(x), 0()) -> s(x)
  , gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
  , if_gcd(true(), x, y) -> gcd(minus(x, y), y)
  , if_gcd(false(), x, y) -> gcd(minus(y, x), x) }
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()
, pred^#(s(x)) -> c_4()
, minus^#(x, 0()) -> c_5()
, gcd^#(0(), y) -> c_7()
, gcd^#(s(x), 0()) -> c_8() }

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))
  , minus^#(x, s(y)) -> c_6(pred^#(minus(x, y)), minus^#(x, y))
  , gcd^#(s(x), s(y)) ->
    c_9(if_gcd^#(le(y, x), s(x), s(y)), le^#(y, x))
  , if_gcd^#(true(), x, y) ->
    c_10(gcd^#(minus(x, y), y), minus^#(x, y))
  , if_gcd^#(false(), x, y) ->
    c_11(gcd^#(minus(y, x), x), minus^#(y, x)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , pred(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> pred(minus(x, y))
  , gcd(0(), y) -> y
  , gcd(s(x), 0()) -> s(x)
  , gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
  , if_gcd(true(), x, y) -> gcd(minus(x, y), y)
  , if_gcd(false(), x, y) -> gcd(minus(y, x), x) }
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_6(pred^#(minus(x, y)), minus^#(x, y)) }

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_1(le^#(x, y))
  , minus^#(x, s(y)) -> c_2(minus^#(x, y))
  , gcd^#(s(x), s(y)) ->
    c_3(if_gcd^#(le(y, x), s(x), s(y)), le^#(y, x))
  , if_gcd^#(true(), x, y) ->
    c_4(gcd^#(minus(x, y), y), minus^#(x, y))
  , if_gcd^#(false(), x, y) ->
    c_5(gcd^#(minus(y, x), x), minus^#(y, x)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , pred(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> pred(minus(x, y))
  , gcd(0(), y) -> y
  , gcd(s(x), 0()) -> s(x)
  , gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
  , if_gcd(true(), x, y) -> gcd(minus(x, y), y)
  , if_gcd(false(), x, y) -> gcd(minus(y, x), x) }
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)
    , pred(s(x)) -> x
    , minus(x, 0()) -> x
    , minus(x, s(y)) -> pred(minus(x, y)) }

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

Strict DPs:
  { le^#(s(x), s(y)) -> c_1(le^#(x, y))
  , minus^#(x, s(y)) -> c_2(minus^#(x, y))
  , gcd^#(s(x), s(y)) ->
    c_3(if_gcd^#(le(y, x), s(x), s(y)), le^#(y, x))
  , if_gcd^#(true(), x, y) ->
    c_4(gcd^#(minus(x, y), y), minus^#(x, y))
  , if_gcd^#(false(), x, y) ->
    c_5(gcd^#(minus(y, x), x), minus^#(y, x)) }
Weak Trs:
  { le(0(), y) -> true()
  , le(s(x), 0()) -> false()
  , le(s(x), s(y)) -> le(x, y)
  , pred(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> pred(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..