MAYBE

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

Strict Trs:
  { eq(0(), 0()) -> true()
  , eq(0(), s(m)) -> false()
  , eq(s(n), 0()) -> false()
  , eq(s(n), s(m)) -> eq(n, m)
  , le(0(), m) -> true()
  , le(s(n), 0()) -> false()
  , le(s(n), s(m)) -> le(n, m)
  , min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
  , min(cons(x, nil())) -> x
  , if_min(true(), cons(n, cons(m, x))) -> min(cons(n, x))
  , if_min(false(), cons(n, cons(m, x))) -> min(cons(m, x))
  , replace(n, m, cons(k, x)) ->
    if_replace(eq(n, k), n, m, cons(k, x))
  , replace(n, m, nil()) -> nil()
  , if_replace(true(), n, m, cons(k, x)) -> cons(m, x)
  , if_replace(false(), n, m, cons(k, x)) ->
    cons(k, replace(n, m, x))
  , empty(cons(n, x)) -> false()
  , empty(nil()) -> true()
  , head(cons(n, x)) -> n
  , tail(cons(n, x)) -> x
  , tail(nil()) -> nil()
  , sort(x) -> sortIter(x, nil())
  , sortIter(x, y) ->
    if(empty(x), x, y, append(y, cons(min(x), nil())))
  , if(true(), x, y, z) -> y
  , if(false(), x, y, z) ->
    sortIter(replace(min(x), head(x), tail(x)), z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { eq^#(0(), 0()) -> c_1()
  , eq^#(0(), s(m)) -> c_2()
  , eq^#(s(n), 0()) -> c_3()
  , eq^#(s(n), s(m)) -> c_4(eq^#(n, m))
  , le^#(0(), m) -> c_5()
  , le^#(s(n), 0()) -> c_6()
  , le^#(s(n), s(m)) -> c_7(le^#(n, m))
  , min^#(cons(n, cons(m, x))) ->
    c_8(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m))
  , min^#(cons(x, nil())) -> c_9()
  , if_min^#(true(), cons(n, cons(m, x))) -> c_10(min^#(cons(n, x)))
  , if_min^#(false(), cons(n, cons(m, x))) -> c_11(min^#(cons(m, x)))
  , replace^#(n, m, cons(k, x)) ->
    c_12(if_replace^#(eq(n, k), n, m, cons(k, x)), eq^#(n, k))
  , replace^#(n, m, nil()) -> c_13()
  , if_replace^#(true(), n, m, cons(k, x)) -> c_14()
  , if_replace^#(false(), n, m, cons(k, x)) ->
    c_15(replace^#(n, m, x))
  , empty^#(cons(n, x)) -> c_16()
  , empty^#(nil()) -> c_17()
  , head^#(cons(n, x)) -> c_18()
  , tail^#(cons(n, x)) -> c_19()
  , tail^#(nil()) -> c_20()
  , sort^#(x) -> c_21(sortIter^#(x, nil()))
  , sortIter^#(x, y) ->
    c_22(if^#(empty(x), x, y, append(y, cons(min(x), nil()))),
         empty^#(x),
         min^#(x))
  , if^#(true(), x, y, z) -> c_23()
  , if^#(false(), x, y, z) ->
    c_24(sortIter^#(replace(min(x), head(x), tail(x)), z),
         replace^#(min(x), head(x), tail(x)),
         min^#(x),
         head^#(x),
         tail^#(x)) }

and mark the set of starting terms.

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

Strict DPs:
  { eq^#(0(), 0()) -> c_1()
  , eq^#(0(), s(m)) -> c_2()
  , eq^#(s(n), 0()) -> c_3()
  , eq^#(s(n), s(m)) -> c_4(eq^#(n, m))
  , le^#(0(), m) -> c_5()
  , le^#(s(n), 0()) -> c_6()
  , le^#(s(n), s(m)) -> c_7(le^#(n, m))
  , min^#(cons(n, cons(m, x))) ->
    c_8(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m))
  , min^#(cons(x, nil())) -> c_9()
  , if_min^#(true(), cons(n, cons(m, x))) -> c_10(min^#(cons(n, x)))
  , if_min^#(false(), cons(n, cons(m, x))) -> c_11(min^#(cons(m, x)))
  , replace^#(n, m, cons(k, x)) ->
    c_12(if_replace^#(eq(n, k), n, m, cons(k, x)), eq^#(n, k))
  , replace^#(n, m, nil()) -> c_13()
  , if_replace^#(true(), n, m, cons(k, x)) -> c_14()
  , if_replace^#(false(), n, m, cons(k, x)) ->
    c_15(replace^#(n, m, x))
  , empty^#(cons(n, x)) -> c_16()
  , empty^#(nil()) -> c_17()
  , head^#(cons(n, x)) -> c_18()
  , tail^#(cons(n, x)) -> c_19()
  , tail^#(nil()) -> c_20()
  , sort^#(x) -> c_21(sortIter^#(x, nil()))
  , sortIter^#(x, y) ->
    c_22(if^#(empty(x), x, y, append(y, cons(min(x), nil()))),
         empty^#(x),
         min^#(x))
  , if^#(true(), x, y, z) -> c_23()
  , if^#(false(), x, y, z) ->
    c_24(sortIter^#(replace(min(x), head(x), tail(x)), z),
         replace^#(min(x), head(x), tail(x)),
         min^#(x),
         head^#(x),
         tail^#(x)) }
Weak Trs:
  { eq(0(), 0()) -> true()
  , eq(0(), s(m)) -> false()
  , eq(s(n), 0()) -> false()
  , eq(s(n), s(m)) -> eq(n, m)
  , le(0(), m) -> true()
  , le(s(n), 0()) -> false()
  , le(s(n), s(m)) -> le(n, m)
  , min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
  , min(cons(x, nil())) -> x
  , if_min(true(), cons(n, cons(m, x))) -> min(cons(n, x))
  , if_min(false(), cons(n, cons(m, x))) -> min(cons(m, x))
  , replace(n, m, cons(k, x)) ->
    if_replace(eq(n, k), n, m, cons(k, x))
  , replace(n, m, nil()) -> nil()
  , if_replace(true(), n, m, cons(k, x)) -> cons(m, x)
  , if_replace(false(), n, m, cons(k, x)) ->
    cons(k, replace(n, m, x))
  , empty(cons(n, x)) -> false()
  , empty(nil()) -> true()
  , head(cons(n, x)) -> n
  , tail(cons(n, x)) -> x
  , tail(nil()) -> nil()
  , sort(x) -> sortIter(x, nil())
  , sortIter(x, y) ->
    if(empty(x), x, y, append(y, cons(min(x), nil())))
  , if(true(), x, y, z) -> y
  , if(false(), x, y, z) ->
    sortIter(replace(min(x), head(x), tail(x)), z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of
{1,2,3,5,6,9,13,14,16,17,18,19,20,23} by applications of
Pre({1,2,3,5,6,9,13,14,16,17,18,19,20,23}) =
{4,7,8,10,11,12,15,22,24}. Here rules are labeled as follows:

  DPs:
    { 1: eq^#(0(), 0()) -> c_1()
    , 2: eq^#(0(), s(m)) -> c_2()
    , 3: eq^#(s(n), 0()) -> c_3()
    , 4: eq^#(s(n), s(m)) -> c_4(eq^#(n, m))
    , 5: le^#(0(), m) -> c_5()
    , 6: le^#(s(n), 0()) -> c_6()
    , 7: le^#(s(n), s(m)) -> c_7(le^#(n, m))
    , 8: min^#(cons(n, cons(m, x))) ->
         c_8(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m))
    , 9: min^#(cons(x, nil())) -> c_9()
    , 10: if_min^#(true(), cons(n, cons(m, x))) ->
          c_10(min^#(cons(n, x)))
    , 11: if_min^#(false(), cons(n, cons(m, x))) ->
          c_11(min^#(cons(m, x)))
    , 12: replace^#(n, m, cons(k, x)) ->
          c_12(if_replace^#(eq(n, k), n, m, cons(k, x)), eq^#(n, k))
    , 13: replace^#(n, m, nil()) -> c_13()
    , 14: if_replace^#(true(), n, m, cons(k, x)) -> c_14()
    , 15: if_replace^#(false(), n, m, cons(k, x)) ->
          c_15(replace^#(n, m, x))
    , 16: empty^#(cons(n, x)) -> c_16()
    , 17: empty^#(nil()) -> c_17()
    , 18: head^#(cons(n, x)) -> c_18()
    , 19: tail^#(cons(n, x)) -> c_19()
    , 20: tail^#(nil()) -> c_20()
    , 21: sort^#(x) -> c_21(sortIter^#(x, nil()))
    , 22: sortIter^#(x, y) ->
          c_22(if^#(empty(x), x, y, append(y, cons(min(x), nil()))),
               empty^#(x),
               min^#(x))
    , 23: if^#(true(), x, y, z) -> c_23()
    , 24: if^#(false(), x, y, z) ->
          c_24(sortIter^#(replace(min(x), head(x), tail(x)), z),
               replace^#(min(x), head(x), tail(x)),
               min^#(x),
               head^#(x),
               tail^#(x)) }

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

Strict DPs:
  { eq^#(s(n), s(m)) -> c_4(eq^#(n, m))
  , le^#(s(n), s(m)) -> c_7(le^#(n, m))
  , min^#(cons(n, cons(m, x))) ->
    c_8(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m))
  , if_min^#(true(), cons(n, cons(m, x))) -> c_10(min^#(cons(n, x)))
  , if_min^#(false(), cons(n, cons(m, x))) -> c_11(min^#(cons(m, x)))
  , replace^#(n, m, cons(k, x)) ->
    c_12(if_replace^#(eq(n, k), n, m, cons(k, x)), eq^#(n, k))
  , if_replace^#(false(), n, m, cons(k, x)) ->
    c_15(replace^#(n, m, x))
  , sort^#(x) -> c_21(sortIter^#(x, nil()))
  , sortIter^#(x, y) ->
    c_22(if^#(empty(x), x, y, append(y, cons(min(x), nil()))),
         empty^#(x),
         min^#(x))
  , if^#(false(), x, y, z) ->
    c_24(sortIter^#(replace(min(x), head(x), tail(x)), z),
         replace^#(min(x), head(x), tail(x)),
         min^#(x),
         head^#(x),
         tail^#(x)) }
Weak DPs:
  { eq^#(0(), 0()) -> c_1()
  , eq^#(0(), s(m)) -> c_2()
  , eq^#(s(n), 0()) -> c_3()
  , le^#(0(), m) -> c_5()
  , le^#(s(n), 0()) -> c_6()
  , min^#(cons(x, nil())) -> c_9()
  , replace^#(n, m, nil()) -> c_13()
  , if_replace^#(true(), n, m, cons(k, x)) -> c_14()
  , empty^#(cons(n, x)) -> c_16()
  , empty^#(nil()) -> c_17()
  , head^#(cons(n, x)) -> c_18()
  , tail^#(cons(n, x)) -> c_19()
  , tail^#(nil()) -> c_20()
  , if^#(true(), x, y, z) -> c_23() }
Weak Trs:
  { eq(0(), 0()) -> true()
  , eq(0(), s(m)) -> false()
  , eq(s(n), 0()) -> false()
  , eq(s(n), s(m)) -> eq(n, m)
  , le(0(), m) -> true()
  , le(s(n), 0()) -> false()
  , le(s(n), s(m)) -> le(n, m)
  , min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
  , min(cons(x, nil())) -> x
  , if_min(true(), cons(n, cons(m, x))) -> min(cons(n, x))
  , if_min(false(), cons(n, cons(m, x))) -> min(cons(m, x))
  , replace(n, m, cons(k, x)) ->
    if_replace(eq(n, k), n, m, cons(k, x))
  , replace(n, m, nil()) -> nil()
  , if_replace(true(), n, m, cons(k, x)) -> cons(m, x)
  , if_replace(false(), n, m, cons(k, x)) ->
    cons(k, replace(n, m, x))
  , empty(cons(n, x)) -> false()
  , empty(nil()) -> true()
  , head(cons(n, x)) -> n
  , tail(cons(n, x)) -> x
  , tail(nil()) -> nil()
  , sort(x) -> sortIter(x, nil())
  , sortIter(x, y) ->
    if(empty(x), x, y, append(y, cons(min(x), nil())))
  , if(true(), x, y, z) -> y
  , if(false(), x, y, z) ->
    sortIter(replace(min(x), head(x), tail(x)), 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.

{ eq^#(0(), 0()) -> c_1()
, eq^#(0(), s(m)) -> c_2()
, eq^#(s(n), 0()) -> c_3()
, le^#(0(), m) -> c_5()
, le^#(s(n), 0()) -> c_6()
, min^#(cons(x, nil())) -> c_9()
, replace^#(n, m, nil()) -> c_13()
, if_replace^#(true(), n, m, cons(k, x)) -> c_14()
, empty^#(cons(n, x)) -> c_16()
, empty^#(nil()) -> c_17()
, head^#(cons(n, x)) -> c_18()
, tail^#(cons(n, x)) -> c_19()
, tail^#(nil()) -> c_20()
, if^#(true(), x, y, z) -> c_23() }

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

Strict DPs:
  { eq^#(s(n), s(m)) -> c_4(eq^#(n, m))
  , le^#(s(n), s(m)) -> c_7(le^#(n, m))
  , min^#(cons(n, cons(m, x))) ->
    c_8(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m))
  , if_min^#(true(), cons(n, cons(m, x))) -> c_10(min^#(cons(n, x)))
  , if_min^#(false(), cons(n, cons(m, x))) -> c_11(min^#(cons(m, x)))
  , replace^#(n, m, cons(k, x)) ->
    c_12(if_replace^#(eq(n, k), n, m, cons(k, x)), eq^#(n, k))
  , if_replace^#(false(), n, m, cons(k, x)) ->
    c_15(replace^#(n, m, x))
  , sort^#(x) -> c_21(sortIter^#(x, nil()))
  , sortIter^#(x, y) ->
    c_22(if^#(empty(x), x, y, append(y, cons(min(x), nil()))),
         empty^#(x),
         min^#(x))
  , if^#(false(), x, y, z) ->
    c_24(sortIter^#(replace(min(x), head(x), tail(x)), z),
         replace^#(min(x), head(x), tail(x)),
         min^#(x),
         head^#(x),
         tail^#(x)) }
Weak Trs:
  { eq(0(), 0()) -> true()
  , eq(0(), s(m)) -> false()
  , eq(s(n), 0()) -> false()
  , eq(s(n), s(m)) -> eq(n, m)
  , le(0(), m) -> true()
  , le(s(n), 0()) -> false()
  , le(s(n), s(m)) -> le(n, m)
  , min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
  , min(cons(x, nil())) -> x
  , if_min(true(), cons(n, cons(m, x))) -> min(cons(n, x))
  , if_min(false(), cons(n, cons(m, x))) -> min(cons(m, x))
  , replace(n, m, cons(k, x)) ->
    if_replace(eq(n, k), n, m, cons(k, x))
  , replace(n, m, nil()) -> nil()
  , if_replace(true(), n, m, cons(k, x)) -> cons(m, x)
  , if_replace(false(), n, m, cons(k, x)) ->
    cons(k, replace(n, m, x))
  , empty(cons(n, x)) -> false()
  , empty(nil()) -> true()
  , head(cons(n, x)) -> n
  , tail(cons(n, x)) -> x
  , tail(nil()) -> nil()
  , sort(x) -> sortIter(x, nil())
  , sortIter(x, y) ->
    if(empty(x), x, y, append(y, cons(min(x), nil())))
  , if(true(), x, y, z) -> y
  , if(false(), x, y, z) ->
    sortIter(replace(min(x), head(x), tail(x)), z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

  { sortIter^#(x, y) ->
    c_22(if^#(empty(x), x, y, append(y, cons(min(x), nil()))),
         empty^#(x),
         min^#(x))
  , if^#(false(), x, y, z) ->
    c_24(sortIter^#(replace(min(x), head(x), tail(x)), z),
         replace^#(min(x), head(x), tail(x)),
         min^#(x),
         head^#(x),
         tail^#(x)) }

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

Strict DPs:
  { eq^#(s(n), s(m)) -> c_1(eq^#(n, m))
  , le^#(s(n), s(m)) -> c_2(le^#(n, m))
  , min^#(cons(n, cons(m, x))) ->
    c_3(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m))
  , if_min^#(true(), cons(n, cons(m, x))) -> c_4(min^#(cons(n, x)))
  , if_min^#(false(), cons(n, cons(m, x))) -> c_5(min^#(cons(m, x)))
  , replace^#(n, m, cons(k, x)) ->
    c_6(if_replace^#(eq(n, k), n, m, cons(k, x)), eq^#(n, k))
  , if_replace^#(false(), n, m, cons(k, x)) ->
    c_7(replace^#(n, m, x))
  , sort^#(x) -> c_8(sortIter^#(x, nil()))
  , sortIter^#(x, y) ->
    c_9(if^#(empty(x), x, y, append(y, cons(min(x), nil()))), min^#(x))
  , if^#(false(), x, y, z) ->
    c_10(sortIter^#(replace(min(x), head(x), tail(x)), z),
         replace^#(min(x), head(x), tail(x)),
         min^#(x)) }
Weak Trs:
  { eq(0(), 0()) -> true()
  , eq(0(), s(m)) -> false()
  , eq(s(n), 0()) -> false()
  , eq(s(n), s(m)) -> eq(n, m)
  , le(0(), m) -> true()
  , le(s(n), 0()) -> false()
  , le(s(n), s(m)) -> le(n, m)
  , min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
  , min(cons(x, nil())) -> x
  , if_min(true(), cons(n, cons(m, x))) -> min(cons(n, x))
  , if_min(false(), cons(n, cons(m, x))) -> min(cons(m, x))
  , replace(n, m, cons(k, x)) ->
    if_replace(eq(n, k), n, m, cons(k, x))
  , replace(n, m, nil()) -> nil()
  , if_replace(true(), n, m, cons(k, x)) -> cons(m, x)
  , if_replace(false(), n, m, cons(k, x)) ->
    cons(k, replace(n, m, x))
  , empty(cons(n, x)) -> false()
  , empty(nil()) -> true()
  , head(cons(n, x)) -> n
  , tail(cons(n, x)) -> x
  , tail(nil()) -> nil()
  , sort(x) -> sortIter(x, nil())
  , sortIter(x, y) ->
    if(empty(x), x, y, append(y, cons(min(x), nil())))
  , if(true(), x, y, z) -> y
  , if(false(), x, y, z) ->
    sortIter(replace(min(x), head(x), tail(x)), z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { eq(0(), 0()) -> true()
    , eq(0(), s(m)) -> false()
    , eq(s(n), 0()) -> false()
    , eq(s(n), s(m)) -> eq(n, m)
    , le(0(), m) -> true()
    , le(s(n), 0()) -> false()
    , le(s(n), s(m)) -> le(n, m)
    , min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
    , min(cons(x, nil())) -> x
    , if_min(true(), cons(n, cons(m, x))) -> min(cons(n, x))
    , if_min(false(), cons(n, cons(m, x))) -> min(cons(m, x))
    , replace(n, m, cons(k, x)) ->
      if_replace(eq(n, k), n, m, cons(k, x))
    , replace(n, m, nil()) -> nil()
    , if_replace(true(), n, m, cons(k, x)) -> cons(m, x)
    , if_replace(false(), n, m, cons(k, x)) ->
      cons(k, replace(n, m, x))
    , empty(cons(n, x)) -> false()
    , empty(nil()) -> true()
    , head(cons(n, x)) -> n
    , tail(cons(n, x)) -> x
    , tail(nil()) -> nil() }

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

Strict DPs:
  { eq^#(s(n), s(m)) -> c_1(eq^#(n, m))
  , le^#(s(n), s(m)) -> c_2(le^#(n, m))
  , min^#(cons(n, cons(m, x))) ->
    c_3(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m))
  , if_min^#(true(), cons(n, cons(m, x))) -> c_4(min^#(cons(n, x)))
  , if_min^#(false(), cons(n, cons(m, x))) -> c_5(min^#(cons(m, x)))
  , replace^#(n, m, cons(k, x)) ->
    c_6(if_replace^#(eq(n, k), n, m, cons(k, x)), eq^#(n, k))
  , if_replace^#(false(), n, m, cons(k, x)) ->
    c_7(replace^#(n, m, x))
  , sort^#(x) -> c_8(sortIter^#(x, nil()))
  , sortIter^#(x, y) ->
    c_9(if^#(empty(x), x, y, append(y, cons(min(x), nil()))), min^#(x))
  , if^#(false(), x, y, z) ->
    c_10(sortIter^#(replace(min(x), head(x), tail(x)), z),
         replace^#(min(x), head(x), tail(x)),
         min^#(x)) }
Weak Trs:
  { eq(0(), 0()) -> true()
  , eq(0(), s(m)) -> false()
  , eq(s(n), 0()) -> false()
  , eq(s(n), s(m)) -> eq(n, m)
  , le(0(), m) -> true()
  , le(s(n), 0()) -> false()
  , le(s(n), s(m)) -> le(n, m)
  , min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
  , min(cons(x, nil())) -> x
  , if_min(true(), cons(n, cons(m, x))) -> min(cons(n, x))
  , if_min(false(), cons(n, cons(m, x))) -> min(cons(m, x))
  , replace(n, m, cons(k, x)) ->
    if_replace(eq(n, k), n, m, cons(k, x))
  , replace(n, m, nil()) -> nil()
  , if_replace(true(), n, m, cons(k, x)) -> cons(m, x)
  , if_replace(false(), n, m, cons(k, x)) ->
    cons(k, replace(n, m, x))
  , empty(cons(n, x)) -> false()
  , empty(nil()) -> true()
  , head(cons(n, x)) -> n
  , tail(cons(n, x)) -> x
  , tail(nil()) -> nil() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: eq^#(s(n), s(m)) -> c_1(eq^#(n, m))
     -->_1 eq^#(s(n), s(m)) -> c_1(eq^#(n, m)) :1
  
  2: le^#(s(n), s(m)) -> c_2(le^#(n, m))
     -->_1 le^#(s(n), s(m)) -> c_2(le^#(n, m)) :2
  
  3: min^#(cons(n, cons(m, x))) ->
     c_3(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m))
     -->_1 if_min^#(false(), cons(n, cons(m, x))) ->
           c_5(min^#(cons(m, x))) :5
     -->_1 if_min^#(true(), cons(n, cons(m, x))) ->
           c_4(min^#(cons(n, x))) :4
     -->_2 le^#(s(n), s(m)) -> c_2(le^#(n, m)) :2
  
  4: if_min^#(true(), cons(n, cons(m, x))) -> c_4(min^#(cons(n, x)))
     -->_1 min^#(cons(n, cons(m, x))) ->
           c_3(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m)) :3
  
  5: if_min^#(false(), cons(n, cons(m, x))) -> c_5(min^#(cons(m, x)))
     -->_1 min^#(cons(n, cons(m, x))) ->
           c_3(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m)) :3
  
  6: replace^#(n, m, cons(k, x)) ->
     c_6(if_replace^#(eq(n, k), n, m, cons(k, x)), eq^#(n, k))
     -->_1 if_replace^#(false(), n, m, cons(k, x)) ->
           c_7(replace^#(n, m, x)) :7
     -->_2 eq^#(s(n), s(m)) -> c_1(eq^#(n, m)) :1
  
  7: if_replace^#(false(), n, m, cons(k, x)) ->
     c_7(replace^#(n, m, x))
     -->_1 replace^#(n, m, cons(k, x)) ->
           c_6(if_replace^#(eq(n, k), n, m, cons(k, x)), eq^#(n, k)) :6
  
  8: sort^#(x) -> c_8(sortIter^#(x, nil()))
     -->_1 sortIter^#(x, y) ->
           c_9(if^#(empty(x), x, y, append(y, cons(min(x), nil()))),
               min^#(x)) :9
  
  9: sortIter^#(x, y) ->
     c_9(if^#(empty(x), x, y, append(y, cons(min(x), nil()))), min^#(x))
     -->_1 if^#(false(), x, y, z) ->
           c_10(sortIter^#(replace(min(x), head(x), tail(x)), z),
                replace^#(min(x), head(x), tail(x)),
                min^#(x)) :10
     -->_2 min^#(cons(n, cons(m, x))) ->
           c_3(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m)) :3
  
  10: if^#(false(), x, y, z) ->
      c_10(sortIter^#(replace(min(x), head(x), tail(x)), z),
           replace^#(min(x), head(x), tail(x)),
           min^#(x))
     -->_1 sortIter^#(x, y) ->
           c_9(if^#(empty(x), x, y, append(y, cons(min(x), nil()))),
               min^#(x)) :9
     -->_2 replace^#(n, m, cons(k, x)) ->
           c_6(if_replace^#(eq(n, k), n, m, cons(k, x)), eq^#(n, k)) :6
     -->_3 min^#(cons(n, cons(m, x))) ->
           c_3(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m)) :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).

  { sort^#(x) -> c_8(sortIter^#(x, nil())) }


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

Strict DPs:
  { eq^#(s(n), s(m)) -> c_1(eq^#(n, m))
  , le^#(s(n), s(m)) -> c_2(le^#(n, m))
  , min^#(cons(n, cons(m, x))) ->
    c_3(if_min^#(le(n, m), cons(n, cons(m, x))), le^#(n, m))
  , if_min^#(true(), cons(n, cons(m, x))) -> c_4(min^#(cons(n, x)))
  , if_min^#(false(), cons(n, cons(m, x))) -> c_5(min^#(cons(m, x)))
  , replace^#(n, m, cons(k, x)) ->
    c_6(if_replace^#(eq(n, k), n, m, cons(k, x)), eq^#(n, k))
  , if_replace^#(false(), n, m, cons(k, x)) ->
    c_7(replace^#(n, m, x))
  , sortIter^#(x, y) ->
    c_9(if^#(empty(x), x, y, append(y, cons(min(x), nil()))), min^#(x))
  , if^#(false(), x, y, z) ->
    c_10(sortIter^#(replace(min(x), head(x), tail(x)), z),
         replace^#(min(x), head(x), tail(x)),
         min^#(x)) }
Weak Trs:
  { eq(0(), 0()) -> true()
  , eq(0(), s(m)) -> false()
  , eq(s(n), 0()) -> false()
  , eq(s(n), s(m)) -> eq(n, m)
  , le(0(), m) -> true()
  , le(s(n), 0()) -> false()
  , le(s(n), s(m)) -> le(n, m)
  , min(cons(n, cons(m, x))) -> if_min(le(n, m), cons(n, cons(m, x)))
  , min(cons(x, nil())) -> x
  , if_min(true(), cons(n, cons(m, x))) -> min(cons(n, x))
  , if_min(false(), cons(n, cons(m, x))) -> min(cons(m, x))
  , replace(n, m, cons(k, x)) ->
    if_replace(eq(n, k), n, m, cons(k, x))
  , replace(n, m, nil()) -> nil()
  , if_replace(true(), n, m, cons(k, x)) -> cons(m, x)
  , if_replace(false(), n, m, cons(k, x)) ->
    cons(k, replace(n, m, x))
  , empty(cons(n, x)) -> false()
  , empty(nil()) -> true()
  , head(cons(n, x)) -> n
  , tail(cons(n, x)) -> x
  , tail(nil()) -> nil() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..