MAYBE

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

Strict Trs:
  { lcm(x, y) -> lcmIter(x, y, 0(), times(x, y))
  , lcmIter(x, y, z, u) -> if(or(ge(0(), x), ge(z, u)), x, y, z, u)
  , times(x, y) -> ifTimes(ge(0(), x), x, y)
  , if(true(), x, y, z, u) -> z
  , if(false(), x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
  , or(true(), y) -> true()
  , or(false(), y) -> y
  , ge(x, 0()) -> true()
  , ge(0(), s(y)) -> false()
  , ge(s(x), s(y)) -> ge(x, y)
  , if2(true(), x, y, z, u) -> z
  , if2(false(), x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
  , divisible(0(), s(y)) -> true()
  , divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
  , plus(0(), y) -> y
  , plus(s(x), y) -> s(plus(x, y))
  , ifTimes(true(), x, y) -> 0()
  , ifTimes(false(), x, y) -> plus(y, times(y, p(x)))
  , p(0()) -> s(s(0()))
  , p(s(x)) -> x
  , div(x, y, 0()) -> divisible(x, y)
  , div(0(), y, s(z)) -> false()
  , div(s(x), y, s(z)) -> div(x, y, z)
  , a() -> b()
  , a() -> c() }
Obligation:
  runtime complexity
Answer:
  MAYBE

None of the processors succeeded.

Details of failed attempt(s):
-----------------------------
1) 'WithProblem (timeout of 60 seconds)' failed due to the
   following reason:
   
   Computation stopped due to timeout after 60.0 seconds.

2) 'Best' failed due to the following reason:
   
   None of the processors succeeded.
   
   Details of failed attempt(s):
   -----------------------------
   1) 'WithProblem (timeout of 30 seconds) (timeout of 60 seconds)'
      failed due to the following reason:
      
      Computation stopped due to timeout after 30.0 seconds.
   
   2) 'Fastest (timeout of 5 seconds) (timeout of 60 seconds)' failed
      due to the following reason:
      
      None of the processors succeeded.
      
      Details of failed attempt(s):
      -----------------------------
      1) 'Bounds with perSymbol-enrichment and initial automaton 'match''
         failed due to the following reason:
         
         match-boundness of the problem could not be verified.
      
      2) 'Bounds with minimal-enrichment and initial automaton 'match''
         failed due to the following reason:
         
         match-boundness of the problem could not be verified.
      
   
   3) 'Best' failed due to the following reason:
      
      None of the processors succeeded.
      
      Details of failed attempt(s):
      -----------------------------
      1) 'Polynomial Path Order (PS) (timeout of 60 seconds)' failed due
         to the following reason:
         
         The processor is inapplicable, reason:
           Processor only applicable for innermost runtime complexity analysis
      
      2) 'bsearch-popstar (timeout of 60 seconds)' failed due to the
         following reason:
         
         The processor is inapplicable, reason:
           Processor only applicable for innermost runtime complexity analysis
      
   

3) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed
   due to the following reason:
   
   We add the following weak dependency pairs:
   
   Strict DPs:
     { lcm^#(x, y) -> c_1(lcmIter^#(x, y, 0(), times(x, y)))
     , lcmIter^#(x, y, z, u) ->
       c_2(if^#(or(ge(0(), x), ge(z, u)), x, y, z, u))
     , if^#(true(), x, y, z, u) -> c_4(z)
     , if^#(false(), x, y, z, u) ->
       c_5(if2^#(divisible(z, y), x, y, z, u))
     , times^#(x, y) -> c_3(ifTimes^#(ge(0(), x), x, y))
     , ifTimes^#(true(), x, y) -> c_17()
     , ifTimes^#(false(), x, y) -> c_18(plus^#(y, times(y, p(x))))
     , if2^#(true(), x, y, z, u) -> c_11(z)
     , if2^#(false(), x, y, z, u) ->
       c_12(lcmIter^#(x, y, plus(x, z), u))
     , or^#(true(), y) -> c_6()
     , or^#(false(), y) -> c_7(y)
     , ge^#(x, 0()) -> c_8()
     , ge^#(0(), s(y)) -> c_9()
     , ge^#(s(x), s(y)) -> c_10(ge^#(x, y))
     , divisible^#(0(), s(y)) -> c_13()
     , divisible^#(s(x), s(y)) -> c_14(div^#(s(x), s(y), s(y)))
     , div^#(x, y, 0()) -> c_21(divisible^#(x, y))
     , div^#(0(), y, s(z)) -> c_22()
     , div^#(s(x), y, s(z)) -> c_23(div^#(x, y, z))
     , plus^#(0(), y) -> c_15(y)
     , plus^#(s(x), y) -> c_16(plus^#(x, y))
     , p^#(0()) -> c_19()
     , p^#(s(x)) -> c_20(x)
     , a^#() -> c_24()
     , a^#() -> c_25() }
   
   and mark the set of starting terms.
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { lcm^#(x, y) -> c_1(lcmIter^#(x, y, 0(), times(x, y)))
     , lcmIter^#(x, y, z, u) ->
       c_2(if^#(or(ge(0(), x), ge(z, u)), x, y, z, u))
     , if^#(true(), x, y, z, u) -> c_4(z)
     , if^#(false(), x, y, z, u) ->
       c_5(if2^#(divisible(z, y), x, y, z, u))
     , times^#(x, y) -> c_3(ifTimes^#(ge(0(), x), x, y))
     , ifTimes^#(true(), x, y) -> c_17()
     , ifTimes^#(false(), x, y) -> c_18(plus^#(y, times(y, p(x))))
     , if2^#(true(), x, y, z, u) -> c_11(z)
     , if2^#(false(), x, y, z, u) ->
       c_12(lcmIter^#(x, y, plus(x, z), u))
     , or^#(true(), y) -> c_6()
     , or^#(false(), y) -> c_7(y)
     , ge^#(x, 0()) -> c_8()
     , ge^#(0(), s(y)) -> c_9()
     , ge^#(s(x), s(y)) -> c_10(ge^#(x, y))
     , divisible^#(0(), s(y)) -> c_13()
     , divisible^#(s(x), s(y)) -> c_14(div^#(s(x), s(y), s(y)))
     , div^#(x, y, 0()) -> c_21(divisible^#(x, y))
     , div^#(0(), y, s(z)) -> c_22()
     , div^#(s(x), y, s(z)) -> c_23(div^#(x, y, z))
     , plus^#(0(), y) -> c_15(y)
     , plus^#(s(x), y) -> c_16(plus^#(x, y))
     , p^#(0()) -> c_19()
     , p^#(s(x)) -> c_20(x)
     , a^#() -> c_24()
     , a^#() -> c_25() }
   Strict Trs:
     { lcm(x, y) -> lcmIter(x, y, 0(), times(x, y))
     , lcmIter(x, y, z, u) -> if(or(ge(0(), x), ge(z, u)), x, y, z, u)
     , times(x, y) -> ifTimes(ge(0(), x), x, y)
     , if(true(), x, y, z, u) -> z
     , if(false(), x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
     , or(true(), y) -> true()
     , or(false(), y) -> y
     , ge(x, 0()) -> true()
     , ge(0(), s(y)) -> false()
     , ge(s(x), s(y)) -> ge(x, y)
     , if2(true(), x, y, z, u) -> z
     , if2(false(), x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
     , divisible(0(), s(y)) -> true()
     , divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
     , plus(0(), y) -> y
     , plus(s(x), y) -> s(plus(x, y))
     , ifTimes(true(), x, y) -> 0()
     , ifTimes(false(), x, y) -> plus(y, times(y, p(x)))
     , p(0()) -> s(s(0()))
     , p(s(x)) -> x
     , div(x, y, 0()) -> divisible(x, y)
     , div(0(), y, s(z)) -> false()
     , div(s(x), y, s(z)) -> div(x, y, z)
     , a() -> b()
     , a() -> c() }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   We estimate the number of application of
   {6,10,12,13,15,18,22,24,25} by applications of
   Pre({6,10,12,13,15,18,22,24,25}) = {3,5,8,11,14,17,19,20,23}. Here
   rules are labeled as follows:
   
     DPs:
       { 1: lcm^#(x, y) -> c_1(lcmIter^#(x, y, 0(), times(x, y)))
       , 2: lcmIter^#(x, y, z, u) ->
            c_2(if^#(or(ge(0(), x), ge(z, u)), x, y, z, u))
       , 3: if^#(true(), x, y, z, u) -> c_4(z)
       , 4: if^#(false(), x, y, z, u) ->
            c_5(if2^#(divisible(z, y), x, y, z, u))
       , 5: times^#(x, y) -> c_3(ifTimes^#(ge(0(), x), x, y))
       , 6: ifTimes^#(true(), x, y) -> c_17()
       , 7: ifTimes^#(false(), x, y) -> c_18(plus^#(y, times(y, p(x))))
       , 8: if2^#(true(), x, y, z, u) -> c_11(z)
       , 9: if2^#(false(), x, y, z, u) ->
            c_12(lcmIter^#(x, y, plus(x, z), u))
       , 10: or^#(true(), y) -> c_6()
       , 11: or^#(false(), y) -> c_7(y)
       , 12: ge^#(x, 0()) -> c_8()
       , 13: ge^#(0(), s(y)) -> c_9()
       , 14: ge^#(s(x), s(y)) -> c_10(ge^#(x, y))
       , 15: divisible^#(0(), s(y)) -> c_13()
       , 16: divisible^#(s(x), s(y)) -> c_14(div^#(s(x), s(y), s(y)))
       , 17: div^#(x, y, 0()) -> c_21(divisible^#(x, y))
       , 18: div^#(0(), y, s(z)) -> c_22()
       , 19: div^#(s(x), y, s(z)) -> c_23(div^#(x, y, z))
       , 20: plus^#(0(), y) -> c_15(y)
       , 21: plus^#(s(x), y) -> c_16(plus^#(x, y))
       , 22: p^#(0()) -> c_19()
       , 23: p^#(s(x)) -> c_20(x)
       , 24: a^#() -> c_24()
       , 25: a^#() -> c_25() }
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { lcm^#(x, y) -> c_1(lcmIter^#(x, y, 0(), times(x, y)))
     , lcmIter^#(x, y, z, u) ->
       c_2(if^#(or(ge(0(), x), ge(z, u)), x, y, z, u))
     , if^#(true(), x, y, z, u) -> c_4(z)
     , if^#(false(), x, y, z, u) ->
       c_5(if2^#(divisible(z, y), x, y, z, u))
     , times^#(x, y) -> c_3(ifTimes^#(ge(0(), x), x, y))
     , ifTimes^#(false(), x, y) -> c_18(plus^#(y, times(y, p(x))))
     , if2^#(true(), x, y, z, u) -> c_11(z)
     , if2^#(false(), x, y, z, u) ->
       c_12(lcmIter^#(x, y, plus(x, z), u))
     , or^#(false(), y) -> c_7(y)
     , ge^#(s(x), s(y)) -> c_10(ge^#(x, y))
     , divisible^#(s(x), s(y)) -> c_14(div^#(s(x), s(y), s(y)))
     , div^#(x, y, 0()) -> c_21(divisible^#(x, y))
     , div^#(s(x), y, s(z)) -> c_23(div^#(x, y, z))
     , plus^#(0(), y) -> c_15(y)
     , plus^#(s(x), y) -> c_16(plus^#(x, y))
     , p^#(s(x)) -> c_20(x) }
   Strict Trs:
     { lcm(x, y) -> lcmIter(x, y, 0(), times(x, y))
     , lcmIter(x, y, z, u) -> if(or(ge(0(), x), ge(z, u)), x, y, z, u)
     , times(x, y) -> ifTimes(ge(0(), x), x, y)
     , if(true(), x, y, z, u) -> z
     , if(false(), x, y, z, u) -> if2(divisible(z, y), x, y, z, u)
     , or(true(), y) -> true()
     , or(false(), y) -> y
     , ge(x, 0()) -> true()
     , ge(0(), s(y)) -> false()
     , ge(s(x), s(y)) -> ge(x, y)
     , if2(true(), x, y, z, u) -> z
     , if2(false(), x, y, z, u) -> lcmIter(x, y, plus(x, z), u)
     , divisible(0(), s(y)) -> true()
     , divisible(s(x), s(y)) -> div(s(x), s(y), s(y))
     , plus(0(), y) -> y
     , plus(s(x), y) -> s(plus(x, y))
     , ifTimes(true(), x, y) -> 0()
     , ifTimes(false(), x, y) -> plus(y, times(y, p(x)))
     , p(0()) -> s(s(0()))
     , p(s(x)) -> x
     , div(x, y, 0()) -> divisible(x, y)
     , div(0(), y, s(z)) -> false()
     , div(s(x), y, s(z)) -> div(x, y, z)
     , a() -> b()
     , a() -> c() }
   Weak DPs:
     { ifTimes^#(true(), x, y) -> c_17()
     , or^#(true(), y) -> c_6()
     , ge^#(x, 0()) -> c_8()
     , ge^#(0(), s(y)) -> c_9()
     , divisible^#(0(), s(y)) -> c_13()
     , div^#(0(), y, s(z)) -> c_22()
     , p^#(0()) -> c_19()
     , a^#() -> c_24()
     , a^#() -> c_25() }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   Empty strict component of the problem is NOT empty.


Arrrr..