MAYBE

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

Strict Trs:
  { minus(X, s(Y)) -> pred(minus(X, Y))
  , minus(X, 0()) -> X
  , pred(s(X)) -> X
  , le(s(X), s(Y)) -> le(X, Y)
  , le(s(X), 0()) -> false()
  , le(0(), Y) -> true()
  , gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
  , gcd(s(X), 0()) -> s(X)
  , gcd(0(), Y) -> 0()
  , if(false(), s(X), s(Y)) -> gcd(minus(Y, X), s(X))
  , if(true(), s(X), s(Y)) -> gcd(minus(X, Y), s(Y)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { minus^#(X, s(Y)) -> c_1(pred^#(minus(X, Y)), minus^#(X, Y))
  , minus^#(X, 0()) -> c_2()
  , pred^#(s(X)) -> c_3()
  , le^#(s(X), s(Y)) -> c_4(le^#(X, Y))
  , le^#(s(X), 0()) -> c_5()
  , le^#(0(), Y) -> c_6()
  , gcd^#(s(X), s(Y)) -> c_7(if^#(le(Y, X), s(X), s(Y)), le^#(Y, X))
  , gcd^#(s(X), 0()) -> c_8()
  , gcd^#(0(), Y) -> c_9()
  , if^#(false(), s(X), s(Y)) ->
    c_10(gcd^#(minus(Y, X), s(X)), minus^#(Y, X))
  , if^#(true(), s(X), s(Y)) ->
    c_11(gcd^#(minus(X, Y), s(Y)), minus^#(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^#(X, s(Y)) -> c_1(pred^#(minus(X, Y)), minus^#(X, Y))
  , minus^#(X, 0()) -> c_2()
  , pred^#(s(X)) -> c_3()
  , le^#(s(X), s(Y)) -> c_4(le^#(X, Y))
  , le^#(s(X), 0()) -> c_5()
  , le^#(0(), Y) -> c_6()
  , gcd^#(s(X), s(Y)) -> c_7(if^#(le(Y, X), s(X), s(Y)), le^#(Y, X))
  , gcd^#(s(X), 0()) -> c_8()
  , gcd^#(0(), Y) -> c_9()
  , if^#(false(), s(X), s(Y)) ->
    c_10(gcd^#(minus(Y, X), s(X)), minus^#(Y, X))
  , if^#(true(), s(X), s(Y)) ->
    c_11(gcd^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
Weak Trs:
  { minus(X, s(Y)) -> pred(minus(X, Y))
  , minus(X, 0()) -> X
  , pred(s(X)) -> X
  , le(s(X), s(Y)) -> le(X, Y)
  , le(s(X), 0()) -> false()
  , le(0(), Y) -> true()
  , gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
  , gcd(s(X), 0()) -> s(X)
  , gcd(0(), Y) -> 0()
  , if(false(), s(X), s(Y)) -> gcd(minus(Y, X), s(X))
  , if(true(), s(X), s(Y)) -> gcd(minus(X, Y), s(Y)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  DPs:
    { 1: minus^#(X, s(Y)) -> c_1(pred^#(minus(X, Y)), minus^#(X, Y))
    , 2: minus^#(X, 0()) -> c_2()
    , 3: pred^#(s(X)) -> c_3()
    , 4: le^#(s(X), s(Y)) -> c_4(le^#(X, Y))
    , 5: le^#(s(X), 0()) -> c_5()
    , 6: le^#(0(), Y) -> c_6()
    , 7: gcd^#(s(X), s(Y)) ->
         c_7(if^#(le(Y, X), s(X), s(Y)), le^#(Y, X))
    , 8: gcd^#(s(X), 0()) -> c_8()
    , 9: gcd^#(0(), Y) -> c_9()
    , 10: if^#(false(), s(X), s(Y)) ->
          c_10(gcd^#(minus(Y, X), s(X)), minus^#(Y, X))
    , 11: if^#(true(), s(X), s(Y)) ->
          c_11(gcd^#(minus(X, Y), s(Y)), minus^#(X, Y)) }

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

Strict DPs:
  { minus^#(X, s(Y)) -> c_1(pred^#(minus(X, Y)), minus^#(X, Y))
  , le^#(s(X), s(Y)) -> c_4(le^#(X, Y))
  , gcd^#(s(X), s(Y)) -> c_7(if^#(le(Y, X), s(X), s(Y)), le^#(Y, X))
  , if^#(false(), s(X), s(Y)) ->
    c_10(gcd^#(minus(Y, X), s(X)), minus^#(Y, X))
  , if^#(true(), s(X), s(Y)) ->
    c_11(gcd^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
Weak DPs:
  { minus^#(X, 0()) -> c_2()
  , pred^#(s(X)) -> c_3()
  , le^#(s(X), 0()) -> c_5()
  , le^#(0(), Y) -> c_6()
  , gcd^#(s(X), 0()) -> c_8()
  , gcd^#(0(), Y) -> c_9() }
Weak Trs:
  { minus(X, s(Y)) -> pred(minus(X, Y))
  , minus(X, 0()) -> X
  , pred(s(X)) -> X
  , le(s(X), s(Y)) -> le(X, Y)
  , le(s(X), 0()) -> false()
  , le(0(), Y) -> true()
  , gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
  , gcd(s(X), 0()) -> s(X)
  , gcd(0(), Y) -> 0()
  , if(false(), s(X), s(Y)) -> gcd(minus(Y, X), s(X))
  , if(true(), s(X), s(Y)) -> gcd(minus(X, Y), s(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^#(X, 0()) -> c_2()
, pred^#(s(X)) -> c_3()
, le^#(s(X), 0()) -> c_5()
, le^#(0(), Y) -> c_6()
, gcd^#(s(X), 0()) -> c_8()
, gcd^#(0(), Y) -> c_9() }

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

Strict DPs:
  { minus^#(X, s(Y)) -> c_1(pred^#(minus(X, Y)), minus^#(X, Y))
  , le^#(s(X), s(Y)) -> c_4(le^#(X, Y))
  , gcd^#(s(X), s(Y)) -> c_7(if^#(le(Y, X), s(X), s(Y)), le^#(Y, X))
  , if^#(false(), s(X), s(Y)) ->
    c_10(gcd^#(minus(Y, X), s(X)), minus^#(Y, X))
  , if^#(true(), s(X), s(Y)) ->
    c_11(gcd^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
Weak Trs:
  { minus(X, s(Y)) -> pred(minus(X, Y))
  , minus(X, 0()) -> X
  , pred(s(X)) -> X
  , le(s(X), s(Y)) -> le(X, Y)
  , le(s(X), 0()) -> false()
  , le(0(), Y) -> true()
  , gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
  , gcd(s(X), 0()) -> s(X)
  , gcd(0(), Y) -> 0()
  , if(false(), s(X), s(Y)) -> gcd(minus(Y, X), s(X))
  , if(true(), s(X), s(Y)) -> gcd(minus(X, Y), s(Y)) }
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_1(pred^#(minus(X, Y)), minus^#(X, Y)) }

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

Strict DPs:
  { minus^#(X, s(Y)) -> c_1(minus^#(X, Y))
  , le^#(s(X), s(Y)) -> c_2(le^#(X, Y))
  , gcd^#(s(X), s(Y)) -> c_3(if^#(le(Y, X), s(X), s(Y)), le^#(Y, X))
  , if^#(false(), s(X), s(Y)) ->
    c_4(gcd^#(minus(Y, X), s(X)), minus^#(Y, X))
  , if^#(true(), s(X), s(Y)) ->
    c_5(gcd^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
Weak Trs:
  { minus(X, s(Y)) -> pred(minus(X, Y))
  , minus(X, 0()) -> X
  , pred(s(X)) -> X
  , le(s(X), s(Y)) -> le(X, Y)
  , le(s(X), 0()) -> false()
  , le(0(), Y) -> true()
  , gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
  , gcd(s(X), 0()) -> s(X)
  , gcd(0(), Y) -> 0()
  , if(false(), s(X), s(Y)) -> gcd(minus(Y, X), s(X))
  , if(true(), s(X), s(Y)) -> gcd(minus(X, Y), s(Y)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { minus(X, s(Y)) -> pred(minus(X, Y))
    , minus(X, 0()) -> X
    , pred(s(X)) -> X
    , le(s(X), s(Y)) -> le(X, Y)
    , le(s(X), 0()) -> false()
    , le(0(), Y) -> true() }

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

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