MAYBE

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

Strict Trs:
  { plus(x, y) -> ifPlus(isZero(x), x, inc(y))
  , ifPlus(true(), x, y) -> p(y)
  , ifPlus(false(), x, y) -> plus(p(x), y)
  , isZero(0()) -> true()
  , isZero(s(0())) -> false()
  , isZero(s(s(x))) -> isZero(s(x))
  , inc(x) -> s(x)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , p(s(s(x))) -> s(p(s(x)))
  , times(x, y) -> timesIter(0(), x, y, 0())
  , timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
  , ifTimes(true(), i, x, y, z) -> z
  , ifTimes(false(), i, x, y, z) ->
    timesIter(inc(i), x, y, plus(z, y))
  , ge(x, 0()) -> true()
  , ge(0(), s(y)) -> false()
  , ge(s(x), s(y)) -> ge(x, y)
  , f0(x, y, z) -> d()
  , f0(0(), y, x) -> f1(x, y, x)
  , f1(x, y, z) -> f2(x, y, z)
  , f1(x, y, z) -> c()
  , f2(x, 1(), z) -> f0(x, z, z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { plus^#(x, y) ->
    c_1(ifPlus^#(isZero(x), x, inc(y)), isZero^#(x), inc^#(y))
  , ifPlus^#(true(), x, y) -> c_2(p^#(y))
  , ifPlus^#(false(), x, y) -> c_3(plus^#(p(x), y), p^#(x))
  , isZero^#(0()) -> c_4()
  , isZero^#(s(0())) -> c_5()
  , isZero^#(s(s(x))) -> c_6(isZero^#(s(x)))
  , inc^#(x) -> c_7()
  , inc^#(0()) -> c_8()
  , inc^#(s(x)) -> c_9(inc^#(x))
  , p^#(0()) -> c_10()
  , p^#(s(x)) -> c_11()
  , p^#(s(s(x))) -> c_12(p^#(s(x)))
  , times^#(x, y) -> c_13(timesIter^#(0(), x, y, 0()))
  , timesIter^#(i, x, y, z) ->
    c_14(ifTimes^#(ge(i, x), i, x, y, z), ge^#(i, x))
  , ifTimes^#(true(), i, x, y, z) -> c_15()
  , ifTimes^#(false(), i, x, y, z) ->
    c_16(timesIter^#(inc(i), x, y, plus(z, y)), inc^#(i), plus^#(z, y))
  , ge^#(x, 0()) -> c_17()
  , ge^#(0(), s(y)) -> c_18()
  , ge^#(s(x), s(y)) -> c_19(ge^#(x, y))
  , f0^#(x, y, z) -> c_20()
  , f0^#(0(), y, x) -> c_21(f1^#(x, y, x))
  , f1^#(x, y, z) -> c_22(f2^#(x, y, z))
  , f1^#(x, y, z) -> c_23()
  , f2^#(x, 1(), z) -> c_24(f0^#(x, z, z)) }

and mark the set of starting terms.

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

Strict DPs:
  { plus^#(x, y) ->
    c_1(ifPlus^#(isZero(x), x, inc(y)), isZero^#(x), inc^#(y))
  , ifPlus^#(true(), x, y) -> c_2(p^#(y))
  , ifPlus^#(false(), x, y) -> c_3(plus^#(p(x), y), p^#(x))
  , isZero^#(0()) -> c_4()
  , isZero^#(s(0())) -> c_5()
  , isZero^#(s(s(x))) -> c_6(isZero^#(s(x)))
  , inc^#(x) -> c_7()
  , inc^#(0()) -> c_8()
  , inc^#(s(x)) -> c_9(inc^#(x))
  , p^#(0()) -> c_10()
  , p^#(s(x)) -> c_11()
  , p^#(s(s(x))) -> c_12(p^#(s(x)))
  , times^#(x, y) -> c_13(timesIter^#(0(), x, y, 0()))
  , timesIter^#(i, x, y, z) ->
    c_14(ifTimes^#(ge(i, x), i, x, y, z), ge^#(i, x))
  , ifTimes^#(true(), i, x, y, z) -> c_15()
  , ifTimes^#(false(), i, x, y, z) ->
    c_16(timesIter^#(inc(i), x, y, plus(z, y)), inc^#(i), plus^#(z, y))
  , ge^#(x, 0()) -> c_17()
  , ge^#(0(), s(y)) -> c_18()
  , ge^#(s(x), s(y)) -> c_19(ge^#(x, y))
  , f0^#(x, y, z) -> c_20()
  , f0^#(0(), y, x) -> c_21(f1^#(x, y, x))
  , f1^#(x, y, z) -> c_22(f2^#(x, y, z))
  , f1^#(x, y, z) -> c_23()
  , f2^#(x, 1(), z) -> c_24(f0^#(x, z, z)) }
Weak Trs:
  { plus(x, y) -> ifPlus(isZero(x), x, inc(y))
  , ifPlus(true(), x, y) -> p(y)
  , ifPlus(false(), x, y) -> plus(p(x), y)
  , isZero(0()) -> true()
  , isZero(s(0())) -> false()
  , isZero(s(s(x))) -> isZero(s(x))
  , inc(x) -> s(x)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , p(s(s(x))) -> s(p(s(x)))
  , times(x, y) -> timesIter(0(), x, y, 0())
  , timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
  , ifTimes(true(), i, x, y, z) -> z
  , ifTimes(false(), i, x, y, z) ->
    timesIter(inc(i), x, y, plus(z, y))
  , ge(x, 0()) -> true()
  , ge(0(), s(y)) -> false()
  , ge(s(x), s(y)) -> ge(x, y)
  , f0(x, y, z) -> d()
  , f0(0(), y, x) -> f1(x, y, x)
  , f1(x, y, z) -> f2(x, y, z)
  , f1(x, y, z) -> c()
  , f2(x, 1(), z) -> f0(x, z, z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of
{4,5,7,8,10,11,15,17,18,20,23} by applications of
Pre({4,5,7,8,10,11,15,17,18,20,23}) =
{1,2,3,6,9,12,14,16,19,21,24}. Here rules are labeled as follows:

  DPs:
    { 1: plus^#(x, y) ->
         c_1(ifPlus^#(isZero(x), x, inc(y)), isZero^#(x), inc^#(y))
    , 2: ifPlus^#(true(), x, y) -> c_2(p^#(y))
    , 3: ifPlus^#(false(), x, y) -> c_3(plus^#(p(x), y), p^#(x))
    , 4: isZero^#(0()) -> c_4()
    , 5: isZero^#(s(0())) -> c_5()
    , 6: isZero^#(s(s(x))) -> c_6(isZero^#(s(x)))
    , 7: inc^#(x) -> c_7()
    , 8: inc^#(0()) -> c_8()
    , 9: inc^#(s(x)) -> c_9(inc^#(x))
    , 10: p^#(0()) -> c_10()
    , 11: p^#(s(x)) -> c_11()
    , 12: p^#(s(s(x))) -> c_12(p^#(s(x)))
    , 13: times^#(x, y) -> c_13(timesIter^#(0(), x, y, 0()))
    , 14: timesIter^#(i, x, y, z) ->
          c_14(ifTimes^#(ge(i, x), i, x, y, z), ge^#(i, x))
    , 15: ifTimes^#(true(), i, x, y, z) -> c_15()
    , 16: ifTimes^#(false(), i, x, y, z) ->
          c_16(timesIter^#(inc(i), x, y, plus(z, y)), inc^#(i), plus^#(z, y))
    , 17: ge^#(x, 0()) -> c_17()
    , 18: ge^#(0(), s(y)) -> c_18()
    , 19: ge^#(s(x), s(y)) -> c_19(ge^#(x, y))
    , 20: f0^#(x, y, z) -> c_20()
    , 21: f0^#(0(), y, x) -> c_21(f1^#(x, y, x))
    , 22: f1^#(x, y, z) -> c_22(f2^#(x, y, z))
    , 23: f1^#(x, y, z) -> c_23()
    , 24: f2^#(x, 1(), z) -> c_24(f0^#(x, z, z)) }

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

Strict DPs:
  { plus^#(x, y) ->
    c_1(ifPlus^#(isZero(x), x, inc(y)), isZero^#(x), inc^#(y))
  , ifPlus^#(true(), x, y) -> c_2(p^#(y))
  , ifPlus^#(false(), x, y) -> c_3(plus^#(p(x), y), p^#(x))
  , isZero^#(s(s(x))) -> c_6(isZero^#(s(x)))
  , inc^#(s(x)) -> c_9(inc^#(x))
  , p^#(s(s(x))) -> c_12(p^#(s(x)))
  , times^#(x, y) -> c_13(timesIter^#(0(), x, y, 0()))
  , timesIter^#(i, x, y, z) ->
    c_14(ifTimes^#(ge(i, x), i, x, y, z), ge^#(i, x))
  , ifTimes^#(false(), i, x, y, z) ->
    c_16(timesIter^#(inc(i), x, y, plus(z, y)), inc^#(i), plus^#(z, y))
  , ge^#(s(x), s(y)) -> c_19(ge^#(x, y))
  , f0^#(0(), y, x) -> c_21(f1^#(x, y, x))
  , f1^#(x, y, z) -> c_22(f2^#(x, y, z))
  , f2^#(x, 1(), z) -> c_24(f0^#(x, z, z)) }
Weak DPs:
  { isZero^#(0()) -> c_4()
  , isZero^#(s(0())) -> c_5()
  , inc^#(x) -> c_7()
  , inc^#(0()) -> c_8()
  , p^#(0()) -> c_10()
  , p^#(s(x)) -> c_11()
  , ifTimes^#(true(), i, x, y, z) -> c_15()
  , ge^#(x, 0()) -> c_17()
  , ge^#(0(), s(y)) -> c_18()
  , f0^#(x, y, z) -> c_20()
  , f1^#(x, y, z) -> c_23() }
Weak Trs:
  { plus(x, y) -> ifPlus(isZero(x), x, inc(y))
  , ifPlus(true(), x, y) -> p(y)
  , ifPlus(false(), x, y) -> plus(p(x), y)
  , isZero(0()) -> true()
  , isZero(s(0())) -> false()
  , isZero(s(s(x))) -> isZero(s(x))
  , inc(x) -> s(x)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , p(s(s(x))) -> s(p(s(x)))
  , times(x, y) -> timesIter(0(), x, y, 0())
  , timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
  , ifTimes(true(), i, x, y, z) -> z
  , ifTimes(false(), i, x, y, z) ->
    timesIter(inc(i), x, y, plus(z, y))
  , ge(x, 0()) -> true()
  , ge(0(), s(y)) -> false()
  , ge(s(x), s(y)) -> ge(x, y)
  , f0(x, y, z) -> d()
  , f0(0(), y, x) -> f1(x, y, x)
  , f1(x, y, z) -> f2(x, y, z)
  , f1(x, y, z) -> c()
  , f2(x, 1(), z) -> f0(x, 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.

{ isZero^#(0()) -> c_4()
, isZero^#(s(0())) -> c_5()
, inc^#(x) -> c_7()
, inc^#(0()) -> c_8()
, p^#(0()) -> c_10()
, p^#(s(x)) -> c_11()
, ifTimes^#(true(), i, x, y, z) -> c_15()
, ge^#(x, 0()) -> c_17()
, ge^#(0(), s(y)) -> c_18()
, f0^#(x, y, z) -> c_20()
, f1^#(x, y, z) -> c_23() }

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

Strict DPs:
  { plus^#(x, y) ->
    c_1(ifPlus^#(isZero(x), x, inc(y)), isZero^#(x), inc^#(y))
  , ifPlus^#(true(), x, y) -> c_2(p^#(y))
  , ifPlus^#(false(), x, y) -> c_3(plus^#(p(x), y), p^#(x))
  , isZero^#(s(s(x))) -> c_6(isZero^#(s(x)))
  , inc^#(s(x)) -> c_9(inc^#(x))
  , p^#(s(s(x))) -> c_12(p^#(s(x)))
  , times^#(x, y) -> c_13(timesIter^#(0(), x, y, 0()))
  , timesIter^#(i, x, y, z) ->
    c_14(ifTimes^#(ge(i, x), i, x, y, z), ge^#(i, x))
  , ifTimes^#(false(), i, x, y, z) ->
    c_16(timesIter^#(inc(i), x, y, plus(z, y)), inc^#(i), plus^#(z, y))
  , ge^#(s(x), s(y)) -> c_19(ge^#(x, y))
  , f0^#(0(), y, x) -> c_21(f1^#(x, y, x))
  , f1^#(x, y, z) -> c_22(f2^#(x, y, z))
  , f2^#(x, 1(), z) -> c_24(f0^#(x, z, z)) }
Weak Trs:
  { plus(x, y) -> ifPlus(isZero(x), x, inc(y))
  , ifPlus(true(), x, y) -> p(y)
  , ifPlus(false(), x, y) -> plus(p(x), y)
  , isZero(0()) -> true()
  , isZero(s(0())) -> false()
  , isZero(s(s(x))) -> isZero(s(x))
  , inc(x) -> s(x)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , p(s(s(x))) -> s(p(s(x)))
  , times(x, y) -> timesIter(0(), x, y, 0())
  , timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
  , ifTimes(true(), i, x, y, z) -> z
  , ifTimes(false(), i, x, y, z) ->
    timesIter(inc(i), x, y, plus(z, y))
  , ge(x, 0()) -> true()
  , ge(0(), s(y)) -> false()
  , ge(s(x), s(y)) -> ge(x, y)
  , f0(x, y, z) -> d()
  , f0(0(), y, x) -> f1(x, y, x)
  , f1(x, y, z) -> f2(x, y, z)
  , f1(x, y, z) -> c()
  , f2(x, 1(), z) -> f0(x, z, z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Consider the dependency graph

  1: plus^#(x, y) ->
     c_1(ifPlus^#(isZero(x), x, inc(y)), isZero^#(x), inc^#(y))
     -->_3 inc^#(s(x)) -> c_9(inc^#(x)) :5
     -->_2 isZero^#(s(s(x))) -> c_6(isZero^#(s(x))) :4
     -->_1 ifPlus^#(false(), x, y) -> c_3(plus^#(p(x), y), p^#(x)) :3
     -->_1 ifPlus^#(true(), x, y) -> c_2(p^#(y)) :2
  
  2: ifPlus^#(true(), x, y) -> c_2(p^#(y))
     -->_1 p^#(s(s(x))) -> c_12(p^#(s(x))) :6
  
  3: ifPlus^#(false(), x, y) -> c_3(plus^#(p(x), y), p^#(x))
     -->_2 p^#(s(s(x))) -> c_12(p^#(s(x))) :6
     -->_1 plus^#(x, y) ->
           c_1(ifPlus^#(isZero(x), x, inc(y)), isZero^#(x), inc^#(y)) :1
  
  4: isZero^#(s(s(x))) -> c_6(isZero^#(s(x)))
     -->_1 isZero^#(s(s(x))) -> c_6(isZero^#(s(x))) :4
  
  5: inc^#(s(x)) -> c_9(inc^#(x))
     -->_1 inc^#(s(x)) -> c_9(inc^#(x)) :5
  
  6: p^#(s(s(x))) -> c_12(p^#(s(x)))
     -->_1 p^#(s(s(x))) -> c_12(p^#(s(x))) :6
  
  7: times^#(x, y) -> c_13(timesIter^#(0(), x, y, 0()))
     -->_1 timesIter^#(i, x, y, z) ->
           c_14(ifTimes^#(ge(i, x), i, x, y, z), ge^#(i, x)) :8
  
  8: timesIter^#(i, x, y, z) ->
     c_14(ifTimes^#(ge(i, x), i, x, y, z), ge^#(i, x))
     -->_2 ge^#(s(x), s(y)) -> c_19(ge^#(x, y)) :10
     -->_1 ifTimes^#(false(), i, x, y, z) ->
           c_16(timesIter^#(inc(i), x, y, plus(z, y)),
                inc^#(i),
                plus^#(z, y)) :9
  
  9: ifTimes^#(false(), i, x, y, z) ->
     c_16(timesIter^#(inc(i), x, y, plus(z, y)), inc^#(i), plus^#(z, y))
     -->_1 timesIter^#(i, x, y, z) ->
           c_14(ifTimes^#(ge(i, x), i, x, y, z), ge^#(i, x)) :8
     -->_2 inc^#(s(x)) -> c_9(inc^#(x)) :5
     -->_3 plus^#(x, y) ->
           c_1(ifPlus^#(isZero(x), x, inc(y)), isZero^#(x), inc^#(y)) :1
  
  10: ge^#(s(x), s(y)) -> c_19(ge^#(x, y))
     -->_1 ge^#(s(x), s(y)) -> c_19(ge^#(x, y)) :10
  
  11: f0^#(0(), y, x) -> c_21(f1^#(x, y, x))
     -->_1 f1^#(x, y, z) -> c_22(f2^#(x, y, z)) :12
  
  12: f1^#(x, y, z) -> c_22(f2^#(x, y, z))
     -->_1 f2^#(x, 1(), z) -> c_24(f0^#(x, z, z)) :13
  
  13: f2^#(x, 1(), z) -> c_24(f0^#(x, z, z))
     -->_1 f0^#(0(), y, x) -> c_21(f1^#(x, y, x)) :11
  

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_13(timesIter^#(0(), x, y, 0())) }


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

Strict DPs:
  { plus^#(x, y) ->
    c_1(ifPlus^#(isZero(x), x, inc(y)), isZero^#(x), inc^#(y))
  , ifPlus^#(true(), x, y) -> c_2(p^#(y))
  , ifPlus^#(false(), x, y) -> c_3(plus^#(p(x), y), p^#(x))
  , isZero^#(s(s(x))) -> c_6(isZero^#(s(x)))
  , inc^#(s(x)) -> c_9(inc^#(x))
  , p^#(s(s(x))) -> c_12(p^#(s(x)))
  , timesIter^#(i, x, y, z) ->
    c_14(ifTimes^#(ge(i, x), i, x, y, z), ge^#(i, x))
  , ifTimes^#(false(), i, x, y, z) ->
    c_16(timesIter^#(inc(i), x, y, plus(z, y)), inc^#(i), plus^#(z, y))
  , ge^#(s(x), s(y)) -> c_19(ge^#(x, y))
  , f0^#(0(), y, x) -> c_21(f1^#(x, y, x))
  , f1^#(x, y, z) -> c_22(f2^#(x, y, z))
  , f2^#(x, 1(), z) -> c_24(f0^#(x, z, z)) }
Weak Trs:
  { plus(x, y) -> ifPlus(isZero(x), x, inc(y))
  , ifPlus(true(), x, y) -> p(y)
  , ifPlus(false(), x, y) -> plus(p(x), y)
  , isZero(0()) -> true()
  , isZero(s(0())) -> false()
  , isZero(s(s(x))) -> isZero(s(x))
  , inc(x) -> s(x)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , p(s(s(x))) -> s(p(s(x)))
  , times(x, y) -> timesIter(0(), x, y, 0())
  , timesIter(i, x, y, z) -> ifTimes(ge(i, x), i, x, y, z)
  , ifTimes(true(), i, x, y, z) -> z
  , ifTimes(false(), i, x, y, z) ->
    timesIter(inc(i), x, y, plus(z, y))
  , ge(x, 0()) -> true()
  , ge(0(), s(y)) -> false()
  , ge(s(x), s(y)) -> ge(x, y)
  , f0(x, y, z) -> d()
  , f0(0(), y, x) -> f1(x, y, x)
  , f1(x, y, z) -> f2(x, y, z)
  , f1(x, y, z) -> c()
  , f2(x, 1(), z) -> f0(x, z, z) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { plus(x, y) -> ifPlus(isZero(x), x, inc(y))
    , ifPlus(true(), x, y) -> p(y)
    , ifPlus(false(), x, y) -> plus(p(x), y)
    , isZero(0()) -> true()
    , isZero(s(0())) -> false()
    , isZero(s(s(x))) -> isZero(s(x))
    , inc(x) -> s(x)
    , inc(0()) -> s(0())
    , inc(s(x)) -> s(inc(x))
    , p(0()) -> 0()
    , p(s(x)) -> x
    , p(s(s(x))) -> s(p(s(x)))
    , ge(x, 0()) -> true()
    , ge(0(), s(y)) -> false()
    , ge(s(x), s(y)) -> ge(x, y) }

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

Strict DPs:
  { plus^#(x, y) ->
    c_1(ifPlus^#(isZero(x), x, inc(y)), isZero^#(x), inc^#(y))
  , ifPlus^#(true(), x, y) -> c_2(p^#(y))
  , ifPlus^#(false(), x, y) -> c_3(plus^#(p(x), y), p^#(x))
  , isZero^#(s(s(x))) -> c_6(isZero^#(s(x)))
  , inc^#(s(x)) -> c_9(inc^#(x))
  , p^#(s(s(x))) -> c_12(p^#(s(x)))
  , timesIter^#(i, x, y, z) ->
    c_14(ifTimes^#(ge(i, x), i, x, y, z), ge^#(i, x))
  , ifTimes^#(false(), i, x, y, z) ->
    c_16(timesIter^#(inc(i), x, y, plus(z, y)), inc^#(i), plus^#(z, y))
  , ge^#(s(x), s(y)) -> c_19(ge^#(x, y))
  , f0^#(0(), y, x) -> c_21(f1^#(x, y, x))
  , f1^#(x, y, z) -> c_22(f2^#(x, y, z))
  , f2^#(x, 1(), z) -> c_24(f0^#(x, z, z)) }
Weak Trs:
  { plus(x, y) -> ifPlus(isZero(x), x, inc(y))
  , ifPlus(true(), x, y) -> p(y)
  , ifPlus(false(), x, y) -> plus(p(x), y)
  , isZero(0()) -> true()
  , isZero(s(0())) -> false()
  , isZero(s(s(x))) -> isZero(s(x))
  , inc(x) -> s(x)
  , inc(0()) -> s(0())
  , inc(s(x)) -> s(inc(x))
  , p(0()) -> 0()
  , p(s(x)) -> x
  , p(s(s(x))) -> s(p(s(x)))
  , ge(x, 0()) -> true()
  , ge(0(), s(y)) -> false()
  , ge(s(x), s(y)) -> ge(x, y) }
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..