MAYBE

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

Strict Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , times(s(x), y) -> plus(y, times(x, y))
  , times(0(), y) -> 0()
  , log(x, s(0())) -> baseError()
  , log(x, 0()) -> baseError()
  , log(s(x), s(s(b))) -> loop(s(x), s(s(b)), s(0()), 0())
  , log(0(), s(s(b))) -> logZeroError()
  , loop(x, s(s(b)), s(y), z) -> if(le(x, s(y)), x, s(s(b)), s(y), z)
  , if(false(), x, b, y, z) -> loop(x, b, times(b, y), s(z))
  , if(true(), x, b, y, z) -> z }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { le^#(s(x), s(y)) -> c_1(le^#(x, y))
  , le^#(s(x), 0()) -> c_2()
  , le^#(0(), y) -> c_3()
  , plus^#(s(x), y) -> c_4(plus^#(x, y))
  , plus^#(0(), y) -> c_5()
  , times^#(s(x), y) -> c_6(plus^#(y, times(x, y)), times^#(x, y))
  , times^#(0(), y) -> c_7()
  , log^#(x, s(0())) -> c_8()
  , log^#(x, 0()) -> c_9()
  , log^#(s(x), s(s(b))) -> c_10(loop^#(s(x), s(s(b)), s(0()), 0()))
  , log^#(0(), s(s(b))) -> c_11()
  , loop^#(x, s(s(b)), s(y), z) ->
    c_12(if^#(le(x, s(y)), x, s(s(b)), s(y), z), le^#(x, s(y)))
  , if^#(false(), x, b, y, z) ->
    c_13(loop^#(x, b, times(b, y), s(z)), times^#(b, y))
  , if^#(true(), x, b, y, z) -> c_14() }

and mark the set of starting terms.

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))
  , le^#(s(x), 0()) -> c_2()
  , le^#(0(), y) -> c_3()
  , plus^#(s(x), y) -> c_4(plus^#(x, y))
  , plus^#(0(), y) -> c_5()
  , times^#(s(x), y) -> c_6(plus^#(y, times(x, y)), times^#(x, y))
  , times^#(0(), y) -> c_7()
  , log^#(x, s(0())) -> c_8()
  , log^#(x, 0()) -> c_9()
  , log^#(s(x), s(s(b))) -> c_10(loop^#(s(x), s(s(b)), s(0()), 0()))
  , log^#(0(), s(s(b))) -> c_11()
  , loop^#(x, s(s(b)), s(y), z) ->
    c_12(if^#(le(x, s(y)), x, s(s(b)), s(y), z), le^#(x, s(y)))
  , if^#(false(), x, b, y, z) ->
    c_13(loop^#(x, b, times(b, y), s(z)), times^#(b, y))
  , if^#(true(), x, b, y, z) -> c_14() }
Weak Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , times(s(x), y) -> plus(y, times(x, y))
  , times(0(), y) -> 0()
  , log(x, s(0())) -> baseError()
  , log(x, 0()) -> baseError()
  , log(s(x), s(s(b))) -> loop(s(x), s(s(b)), s(0()), 0())
  , log(0(), s(s(b))) -> logZeroError()
  , loop(x, s(s(b)), s(y), z) -> if(le(x, s(y)), x, s(s(b)), s(y), z)
  , if(false(), x, b, y, z) -> loop(x, b, times(b, y), s(z))
  , if(true(), x, b, y, z) -> z }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

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

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))
  , plus^#(s(x), y) -> c_4(plus^#(x, y))
  , times^#(s(x), y) -> c_6(plus^#(y, times(x, y)), times^#(x, y))
  , log^#(s(x), s(s(b))) -> c_10(loop^#(s(x), s(s(b)), s(0()), 0()))
  , loop^#(x, s(s(b)), s(y), z) ->
    c_12(if^#(le(x, s(y)), x, s(s(b)), s(y), z), le^#(x, s(y)))
  , if^#(false(), x, b, y, z) ->
    c_13(loop^#(x, b, times(b, y), s(z)), times^#(b, y)) }
Weak DPs:
  { le^#(s(x), 0()) -> c_2()
  , le^#(0(), y) -> c_3()
  , plus^#(0(), y) -> c_5()
  , times^#(0(), y) -> c_7()
  , log^#(x, s(0())) -> c_8()
  , log^#(x, 0()) -> c_9()
  , log^#(0(), s(s(b))) -> c_11()
  , if^#(true(), x, b, y, z) -> c_14() }
Weak Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , times(s(x), y) -> plus(y, times(x, y))
  , times(0(), y) -> 0()
  , log(x, s(0())) -> baseError()
  , log(x, 0()) -> baseError()
  , log(s(x), s(s(b))) -> loop(s(x), s(s(b)), s(0()), 0())
  , log(0(), s(s(b))) -> logZeroError()
  , loop(x, s(s(b)), s(y), z) -> if(le(x, s(y)), x, s(s(b)), s(y), z)
  , if(false(), x, b, y, z) -> loop(x, b, times(b, y), s(z))
  , if(true(), x, b, y, z) -> z }
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^#(s(x), 0()) -> c_2()
, le^#(0(), y) -> c_3()
, plus^#(0(), y) -> c_5()
, times^#(0(), y) -> c_7()
, log^#(x, s(0())) -> c_8()
, log^#(x, 0()) -> c_9()
, log^#(0(), s(s(b))) -> c_11()
, if^#(true(), x, b, y, z) -> c_14() }

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))
  , plus^#(s(x), y) -> c_4(plus^#(x, y))
  , times^#(s(x), y) -> c_6(plus^#(y, times(x, y)), times^#(x, y))
  , log^#(s(x), s(s(b))) -> c_10(loop^#(s(x), s(s(b)), s(0()), 0()))
  , loop^#(x, s(s(b)), s(y), z) ->
    c_12(if^#(le(x, s(y)), x, s(s(b)), s(y), z), le^#(x, s(y)))
  , if^#(false(), x, b, y, z) ->
    c_13(loop^#(x, b, times(b, y), s(z)), times^#(b, y)) }
Weak Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , times(s(x), y) -> plus(y, times(x, y))
  , times(0(), y) -> 0()
  , log(x, s(0())) -> baseError()
  , log(x, 0()) -> baseError()
  , log(s(x), s(s(b))) -> loop(s(x), s(s(b)), s(0()), 0())
  , log(0(), s(s(b))) -> logZeroError()
  , loop(x, s(s(b)), s(y), z) -> if(le(x, s(y)), x, s(s(b)), s(y), z)
  , if(false(), x, b, y, z) -> loop(x, b, times(b, y), s(z))
  , if(true(), x, b, y, z) -> z }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: le^#(s(x), s(y)) -> c_1(le^#(x, y))
     -->_1 le^#(s(x), s(y)) -> c_1(le^#(x, y)) :1
  
  2: plus^#(s(x), y) -> c_4(plus^#(x, y))
     -->_1 plus^#(s(x), y) -> c_4(plus^#(x, y)) :2
  
  3: times^#(s(x), y) -> c_6(plus^#(y, times(x, y)), times^#(x, y))
     -->_2 times^#(s(x), y) ->
           c_6(plus^#(y, times(x, y)), times^#(x, y)) :3
     -->_1 plus^#(s(x), y) -> c_4(plus^#(x, y)) :2
  
  4: log^#(s(x), s(s(b))) -> c_10(loop^#(s(x), s(s(b)), s(0()), 0()))
     -->_1 loop^#(x, s(s(b)), s(y), z) ->
           c_12(if^#(le(x, s(y)), x, s(s(b)), s(y), z), le^#(x, s(y))) :5
  
  5: loop^#(x, s(s(b)), s(y), z) ->
     c_12(if^#(le(x, s(y)), x, s(s(b)), s(y), z), le^#(x, s(y)))
     -->_1 if^#(false(), x, b, y, z) ->
           c_13(loop^#(x, b, times(b, y), s(z)), times^#(b, y)) :6
     -->_2 le^#(s(x), s(y)) -> c_1(le^#(x, y)) :1
  
  6: if^#(false(), x, b, y, z) ->
     c_13(loop^#(x, b, times(b, y), s(z)), times^#(b, y))
     -->_1 loop^#(x, s(s(b)), s(y), z) ->
           c_12(if^#(le(x, s(y)), x, s(s(b)), s(y), z), le^#(x, s(y))) :5
     -->_2 times^#(s(x), y) ->
           c_6(plus^#(y, times(x, y)), times^#(x, y)) :3
  

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).

  { log^#(s(x), s(s(b))) ->
    c_10(loop^#(s(x), s(s(b)), s(0()), 0())) }


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))
  , plus^#(s(x), y) -> c_4(plus^#(x, y))
  , times^#(s(x), y) -> c_6(plus^#(y, times(x, y)), times^#(x, y))
  , loop^#(x, s(s(b)), s(y), z) ->
    c_12(if^#(le(x, s(y)), x, s(s(b)), s(y), z), le^#(x, s(y)))
  , if^#(false(), x, b, y, z) ->
    c_13(loop^#(x, b, times(b, y), s(z)), times^#(b, y)) }
Weak Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , times(s(x), y) -> plus(y, times(x, y))
  , times(0(), y) -> 0()
  , log(x, s(0())) -> baseError()
  , log(x, 0()) -> baseError()
  , log(s(x), s(s(b))) -> loop(s(x), s(s(b)), s(0()), 0())
  , log(0(), s(s(b))) -> logZeroError()
  , loop(x, s(s(b)), s(y), z) -> if(le(x, s(y)), x, s(s(b)), s(y), z)
  , if(false(), x, b, y, z) -> loop(x, b, times(b, y), s(z))
  , if(true(), x, b, y, z) -> z }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

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

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))
  , plus^#(s(x), y) -> c_4(plus^#(x, y))
  , times^#(s(x), y) -> c_6(plus^#(y, times(x, y)), times^#(x, y))
  , loop^#(x, s(s(b)), s(y), z) ->
    c_12(if^#(le(x, s(y)), x, s(s(b)), s(y), z), le^#(x, s(y)))
  , if^#(false(), x, b, y, z) ->
    c_13(loop^#(x, b, times(b, y), s(z)), times^#(b, y)) }
Weak Trs:
  { le(s(x), s(y)) -> le(x, y)
  , le(s(x), 0()) -> false()
  , le(0(), y) -> true()
  , plus(s(x), y) -> s(plus(x, y))
  , plus(0(), y) -> y
  , times(s(x), y) -> plus(y, times(x, y))
  , times(0(), y) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..