MAYBE

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

Strict Trs:
  { inc(s(x)) -> s(inc(x))
  , inc(0()) -> s(0())
  , plus(x, y) -> ifPlus(eq(x, 0()), minus(x, s(0())), x, inc(x))
  , ifPlus(false(), x, y, z) -> plus(x, z)
  , ifPlus(true(), x, y, z) -> y
  , eq(x, x) -> true()
  , eq(s(x), s(y)) -> eq(x, y)
  , eq(s(x), 0()) -> false()
  , eq(0(), s(x)) -> false()
  , eq(0(), 0()) -> true()
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , minus(0(), x) -> 0()
  , times(x, y) -> timesIter(x, y, 0())
  , timesIter(x, y, z) ->
    ifTimes(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z))
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, u)
  , ifTimes(true(), x, y, z, u) -> z
  , f() -> g()
  , f() -> h() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { inc^#(s(x)) -> c_1(inc^#(x))
  , inc^#(0()) -> c_2()
  , plus^#(x, y) ->
    c_3(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
        eq^#(x, 0()),
        minus^#(x, s(0())),
        inc^#(x))
  , ifPlus^#(false(), x, y, z) -> c_4(plus^#(x, z))
  , ifPlus^#(true(), x, y, z) -> c_5()
  , eq^#(x, x) -> c_6()
  , eq^#(s(x), s(y)) -> c_7(eq^#(x, y))
  , eq^#(s(x), 0()) -> c_8()
  , eq^#(0(), s(x)) -> c_9()
  , eq^#(0(), 0()) -> c_10()
  , minus^#(x, x) -> c_11()
  , minus^#(x, 0()) -> c_12()
  , minus^#(s(x), s(y)) -> c_13(minus^#(x, y))
  , minus^#(0(), x) -> c_14()
  , times^#(x, y) -> c_15(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_16(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
         eq^#(x, 0()),
         minus^#(x, s(0())),
         plus^#(y, z))
  , ifTimes^#(false(), x, y, z, u) -> c_17(timesIter^#(x, y, u))
  , ifTimes^#(true(), x, y, z, u) -> c_18()
  , f^#() -> c_19()
  , f^#() -> c_20() }

and mark the set of starting terms.

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

Strict DPs:
  { inc^#(s(x)) -> c_1(inc^#(x))
  , inc^#(0()) -> c_2()
  , plus^#(x, y) ->
    c_3(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
        eq^#(x, 0()),
        minus^#(x, s(0())),
        inc^#(x))
  , ifPlus^#(false(), x, y, z) -> c_4(plus^#(x, z))
  , ifPlus^#(true(), x, y, z) -> c_5()
  , eq^#(x, x) -> c_6()
  , eq^#(s(x), s(y)) -> c_7(eq^#(x, y))
  , eq^#(s(x), 0()) -> c_8()
  , eq^#(0(), s(x)) -> c_9()
  , eq^#(0(), 0()) -> c_10()
  , minus^#(x, x) -> c_11()
  , minus^#(x, 0()) -> c_12()
  , minus^#(s(x), s(y)) -> c_13(minus^#(x, y))
  , minus^#(0(), x) -> c_14()
  , times^#(x, y) -> c_15(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_16(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
         eq^#(x, 0()),
         minus^#(x, s(0())),
         plus^#(y, z))
  , ifTimes^#(false(), x, y, z, u) -> c_17(timesIter^#(x, y, u))
  , ifTimes^#(true(), x, y, z, u) -> c_18()
  , f^#() -> c_19()
  , f^#() -> c_20() }
Weak Trs:
  { inc(s(x)) -> s(inc(x))
  , inc(0()) -> s(0())
  , plus(x, y) -> ifPlus(eq(x, 0()), minus(x, s(0())), x, inc(x))
  , ifPlus(false(), x, y, z) -> plus(x, z)
  , ifPlus(true(), x, y, z) -> y
  , eq(x, x) -> true()
  , eq(s(x), s(y)) -> eq(x, y)
  , eq(s(x), 0()) -> false()
  , eq(0(), s(x)) -> false()
  , eq(0(), 0()) -> true()
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , minus(0(), x) -> 0()
  , times(x, y) -> timesIter(x, y, 0())
  , timesIter(x, y, z) ->
    ifTimes(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z))
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, u)
  , ifTimes(true(), x, y, z, u) -> z
  , f() -> g()
  , f() -> h() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of
{2,5,6,8,9,10,11,12,14,18,19,20} by applications of
Pre({2,5,6,8,9,10,11,12,14,18,19,20}) = {1,3,7,13,16}. Here rules
are labeled as follows:

  DPs:
    { 1: inc^#(s(x)) -> c_1(inc^#(x))
    , 2: inc^#(0()) -> c_2()
    , 3: plus^#(x, y) ->
         c_3(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
             eq^#(x, 0()),
             minus^#(x, s(0())),
             inc^#(x))
    , 4: ifPlus^#(false(), x, y, z) -> c_4(plus^#(x, z))
    , 5: ifPlus^#(true(), x, y, z) -> c_5()
    , 6: eq^#(x, x) -> c_6()
    , 7: eq^#(s(x), s(y)) -> c_7(eq^#(x, y))
    , 8: eq^#(s(x), 0()) -> c_8()
    , 9: eq^#(0(), s(x)) -> c_9()
    , 10: eq^#(0(), 0()) -> c_10()
    , 11: minus^#(x, x) -> c_11()
    , 12: minus^#(x, 0()) -> c_12()
    , 13: minus^#(s(x), s(y)) -> c_13(minus^#(x, y))
    , 14: minus^#(0(), x) -> c_14()
    , 15: times^#(x, y) -> c_15(timesIter^#(x, y, 0()))
    , 16: timesIter^#(x, y, z) ->
          c_16(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
               eq^#(x, 0()),
               minus^#(x, s(0())),
               plus^#(y, z))
    , 17: ifTimes^#(false(), x, y, z, u) -> c_17(timesIter^#(x, y, u))
    , 18: ifTimes^#(true(), x, y, z, u) -> c_18()
    , 19: f^#() -> c_19()
    , 20: f^#() -> c_20() }

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

Strict DPs:
  { inc^#(s(x)) -> c_1(inc^#(x))
  , plus^#(x, y) ->
    c_3(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
        eq^#(x, 0()),
        minus^#(x, s(0())),
        inc^#(x))
  , ifPlus^#(false(), x, y, z) -> c_4(plus^#(x, z))
  , eq^#(s(x), s(y)) -> c_7(eq^#(x, y))
  , minus^#(s(x), s(y)) -> c_13(minus^#(x, y))
  , times^#(x, y) -> c_15(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_16(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
         eq^#(x, 0()),
         minus^#(x, s(0())),
         plus^#(y, z))
  , ifTimes^#(false(), x, y, z, u) -> c_17(timesIter^#(x, y, u)) }
Weak DPs:
  { inc^#(0()) -> c_2()
  , ifPlus^#(true(), x, y, z) -> c_5()
  , eq^#(x, x) -> c_6()
  , eq^#(s(x), 0()) -> c_8()
  , eq^#(0(), s(x)) -> c_9()
  , eq^#(0(), 0()) -> c_10()
  , minus^#(x, x) -> c_11()
  , minus^#(x, 0()) -> c_12()
  , minus^#(0(), x) -> c_14()
  , ifTimes^#(true(), x, y, z, u) -> c_18()
  , f^#() -> c_19()
  , f^#() -> c_20() }
Weak Trs:
  { inc(s(x)) -> s(inc(x))
  , inc(0()) -> s(0())
  , plus(x, y) -> ifPlus(eq(x, 0()), minus(x, s(0())), x, inc(x))
  , ifPlus(false(), x, y, z) -> plus(x, z)
  , ifPlus(true(), x, y, z) -> y
  , eq(x, x) -> true()
  , eq(s(x), s(y)) -> eq(x, y)
  , eq(s(x), 0()) -> false()
  , eq(0(), s(x)) -> false()
  , eq(0(), 0()) -> true()
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , minus(0(), x) -> 0()
  , times(x, y) -> timesIter(x, y, 0())
  , timesIter(x, y, z) ->
    ifTimes(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z))
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, u)
  , ifTimes(true(), x, y, z, u) -> z
  , f() -> g()
  , f() -> h() }
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.

{ inc^#(0()) -> c_2()
, ifPlus^#(true(), x, y, z) -> c_5()
, eq^#(x, x) -> c_6()
, eq^#(s(x), 0()) -> c_8()
, eq^#(0(), s(x)) -> c_9()
, eq^#(0(), 0()) -> c_10()
, minus^#(x, x) -> c_11()
, minus^#(x, 0()) -> c_12()
, minus^#(0(), x) -> c_14()
, ifTimes^#(true(), x, y, z, u) -> c_18()
, f^#() -> c_19()
, f^#() -> c_20() }

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

Strict DPs:
  { inc^#(s(x)) -> c_1(inc^#(x))
  , plus^#(x, y) ->
    c_3(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
        eq^#(x, 0()),
        minus^#(x, s(0())),
        inc^#(x))
  , ifPlus^#(false(), x, y, z) -> c_4(plus^#(x, z))
  , eq^#(s(x), s(y)) -> c_7(eq^#(x, y))
  , minus^#(s(x), s(y)) -> c_13(minus^#(x, y))
  , times^#(x, y) -> c_15(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_16(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
         eq^#(x, 0()),
         minus^#(x, s(0())),
         plus^#(y, z))
  , ifTimes^#(false(), x, y, z, u) -> c_17(timesIter^#(x, y, u)) }
Weak Trs:
  { inc(s(x)) -> s(inc(x))
  , inc(0()) -> s(0())
  , plus(x, y) -> ifPlus(eq(x, 0()), minus(x, s(0())), x, inc(x))
  , ifPlus(false(), x, y, z) -> plus(x, z)
  , ifPlus(true(), x, y, z) -> y
  , eq(x, x) -> true()
  , eq(s(x), s(y)) -> eq(x, y)
  , eq(s(x), 0()) -> false()
  , eq(0(), s(x)) -> false()
  , eq(0(), 0()) -> true()
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , minus(0(), x) -> 0()
  , times(x, y) -> timesIter(x, y, 0())
  , timesIter(x, y, z) ->
    ifTimes(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z))
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, u)
  , ifTimes(true(), x, y, z, u) -> z
  , f() -> g()
  , f() -> h() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { plus^#(x, y) ->
    c_3(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
        eq^#(x, 0()),
        minus^#(x, s(0())),
        inc^#(x))
  , timesIter^#(x, y, z) ->
    c_16(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
         eq^#(x, 0()),
         minus^#(x, s(0())),
         plus^#(y, z)) }

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

Strict DPs:
  { inc^#(s(x)) -> c_1(inc^#(x))
  , plus^#(x, y) ->
    c_2(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
        minus^#(x, s(0())),
        inc^#(x))
  , ifPlus^#(false(), x, y, z) -> c_3(plus^#(x, z))
  , eq^#(s(x), s(y)) -> c_4(eq^#(x, y))
  , minus^#(s(x), s(y)) -> c_5(minus^#(x, y))
  , times^#(x, y) -> c_6(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_7(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
        minus^#(x, s(0())),
        plus^#(y, z))
  , ifTimes^#(false(), x, y, z, u) -> c_8(timesIter^#(x, y, u)) }
Weak Trs:
  { inc(s(x)) -> s(inc(x))
  , inc(0()) -> s(0())
  , plus(x, y) -> ifPlus(eq(x, 0()), minus(x, s(0())), x, inc(x))
  , ifPlus(false(), x, y, z) -> plus(x, z)
  , ifPlus(true(), x, y, z) -> y
  , eq(x, x) -> true()
  , eq(s(x), s(y)) -> eq(x, y)
  , eq(s(x), 0()) -> false()
  , eq(0(), s(x)) -> false()
  , eq(0(), 0()) -> true()
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , minus(0(), x) -> 0()
  , times(x, y) -> timesIter(x, y, 0())
  , timesIter(x, y, z) ->
    ifTimes(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z))
  , ifTimes(false(), x, y, z, u) -> timesIter(x, y, u)
  , ifTimes(true(), x, y, z, u) -> z
  , f() -> g()
  , f() -> h() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { inc(s(x)) -> s(inc(x))
    , inc(0()) -> s(0())
    , plus(x, y) -> ifPlus(eq(x, 0()), minus(x, s(0())), x, inc(x))
    , ifPlus(false(), x, y, z) -> plus(x, z)
    , ifPlus(true(), x, y, z) -> y
    , eq(x, x) -> true()
    , eq(s(x), s(y)) -> eq(x, y)
    , eq(s(x), 0()) -> false()
    , eq(0(), s(x)) -> false()
    , eq(0(), 0()) -> true()
    , minus(x, x) -> 0()
    , minus(x, 0()) -> x
    , minus(s(x), s(y)) -> minus(x, y)
    , minus(0(), x) -> 0() }

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

Strict DPs:
  { inc^#(s(x)) -> c_1(inc^#(x))
  , plus^#(x, y) ->
    c_2(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
        minus^#(x, s(0())),
        inc^#(x))
  , ifPlus^#(false(), x, y, z) -> c_3(plus^#(x, z))
  , eq^#(s(x), s(y)) -> c_4(eq^#(x, y))
  , minus^#(s(x), s(y)) -> c_5(minus^#(x, y))
  , times^#(x, y) -> c_6(timesIter^#(x, y, 0()))
  , timesIter^#(x, y, z) ->
    c_7(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
        minus^#(x, s(0())),
        plus^#(y, z))
  , ifTimes^#(false(), x, y, z, u) -> c_8(timesIter^#(x, y, u)) }
Weak Trs:
  { inc(s(x)) -> s(inc(x))
  , inc(0()) -> s(0())
  , plus(x, y) -> ifPlus(eq(x, 0()), minus(x, s(0())), x, inc(x))
  , ifPlus(false(), x, y, z) -> plus(x, z)
  , ifPlus(true(), x, y, z) -> y
  , eq(x, x) -> true()
  , eq(s(x), s(y)) -> eq(x, y)
  , eq(s(x), 0()) -> false()
  , eq(0(), s(x)) -> false()
  , eq(0(), 0()) -> true()
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , minus(0(), x) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: inc^#(s(x)) -> c_1(inc^#(x))
     -->_1 inc^#(s(x)) -> c_1(inc^#(x)) :1
  
  2: plus^#(x, y) ->
     c_2(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
         minus^#(x, s(0())),
         inc^#(x))
     -->_2 minus^#(s(x), s(y)) -> c_5(minus^#(x, y)) :5
     -->_1 ifPlus^#(false(), x, y, z) -> c_3(plus^#(x, z)) :3
     -->_3 inc^#(s(x)) -> c_1(inc^#(x)) :1
  
  3: ifPlus^#(false(), x, y, z) -> c_3(plus^#(x, z))
     -->_1 plus^#(x, y) ->
           c_2(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
               minus^#(x, s(0())),
               inc^#(x)) :2
  
  4: eq^#(s(x), s(y)) -> c_4(eq^#(x, y))
     -->_1 eq^#(s(x), s(y)) -> c_4(eq^#(x, y)) :4
  
  5: minus^#(s(x), s(y)) -> c_5(minus^#(x, y))
     -->_1 minus^#(s(x), s(y)) -> c_5(minus^#(x, y)) :5
  
  6: times^#(x, y) -> c_6(timesIter^#(x, y, 0()))
     -->_1 timesIter^#(x, y, z) ->
           c_7(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
               minus^#(x, s(0())),
               plus^#(y, z)) :7
  
  7: timesIter^#(x, y, z) ->
     c_7(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
         minus^#(x, s(0())),
         plus^#(y, z))
     -->_1 ifTimes^#(false(), x, y, z, u) ->
           c_8(timesIter^#(x, y, u)) :8
     -->_2 minus^#(s(x), s(y)) -> c_5(minus^#(x, y)) :5
     -->_3 plus^#(x, y) ->
           c_2(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
               minus^#(x, s(0())),
               inc^#(x)) :2
  
  8: ifTimes^#(false(), x, y, z, u) -> c_8(timesIter^#(x, y, u))
     -->_1 timesIter^#(x, y, z) ->
           c_7(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
               minus^#(x, s(0())),
               plus^#(y, z)) :7
  

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).

  { times^#(x, y) -> c_6(timesIter^#(x, y, 0())) }


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

Strict DPs:
  { inc^#(s(x)) -> c_1(inc^#(x))
  , plus^#(x, y) ->
    c_2(ifPlus^#(eq(x, 0()), minus(x, s(0())), x, inc(x)),
        minus^#(x, s(0())),
        inc^#(x))
  , ifPlus^#(false(), x, y, z) -> c_3(plus^#(x, z))
  , eq^#(s(x), s(y)) -> c_4(eq^#(x, y))
  , minus^#(s(x), s(y)) -> c_5(minus^#(x, y))
  , timesIter^#(x, y, z) ->
    c_7(ifTimes^#(eq(x, 0()), minus(x, s(0())), y, z, plus(y, z)),
        minus^#(x, s(0())),
        plus^#(y, z))
  , ifTimes^#(false(), x, y, z, u) -> c_8(timesIter^#(x, y, u)) }
Weak Trs:
  { inc(s(x)) -> s(inc(x))
  , inc(0()) -> s(0())
  , plus(x, y) -> ifPlus(eq(x, 0()), minus(x, s(0())), x, inc(x))
  , ifPlus(false(), x, y, z) -> plus(x, z)
  , ifPlus(true(), x, y, z) -> y
  , eq(x, x) -> true()
  , eq(s(x), s(y)) -> eq(x, y)
  , eq(s(x), 0()) -> false()
  , eq(0(), s(x)) -> false()
  , eq(0(), 0()) -> true()
  , minus(x, x) -> 0()
  , minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , minus(0(), x) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..