Problem SK90 4.30

Tool Bounds

Execution Time	60.032352ms
Answer	TIMEOUT
Input	SK90 4.30

stdout:

TIMEOUT

We consider the following Problem:

  Strict Trs:
    {  g(.(x, .(y, z))) -> g(.(.(x, y), z))
     , g(.(x, nil())) -> .(g(x), nil())
     , g(nil()) -> nil()
     , f(.(.(x, y), z)) -> f(.(x, .(y, z)))
     , f(.(nil(), y)) -> .(nil(), f(y))
     , f(nil()) -> nil()}
  StartTerms: all
  Strategy: none

Certificate: TIMEOUT

Proof:
  Computation stopped due to timeout after 60.0 seconds.

Arrrr..

Tool CDI

Execution Time	5.519196ms
Answer	MAYBE
Input	SK90 4.30

stdout:

MAYBE

Statistics:
Number of monomials: 826
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	23.113283ms
Answer	YES(?,O(n^3))
Input	SK90 4.30

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  g(.(x, .(y, z))) -> g(.(.(x, y), z))
     , g(.(x, nil())) -> .(g(x), nil())
     , g(nil()) -> nil()
     , f(.(.(x, y), z)) -> f(.(x, .(y, z)))
     , f(.(nil(), y)) -> .(nil(), f(y))
     , f(nil()) -> nil()}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^3))

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   nil() = [0]
           [2]
           [0]
   f(x1) = [1 1 1] x1 + [0]
           [0 0 0]      [3]
           [0 1 1]      [1]
   .(x1, x2) = [1 0 1] x1 + [1 0 0] x2 + [0]
               [0 1 1]      [0 0 0]      [1]
               [0 0 0]      [0 1 1]      [1]
   g(x1) = [1 1 2] x1 + [1]
           [0 1 1]      [1]
           [0 0 0]      [3]

Hurray, we answered YES(?,O(n^3))

Tool IDA

Execution Time	60.034782ms
Answer	TIMEOUT
Input	SK90 4.30

stdout:

Tool TRI

Execution Time	3.2073832ms
Answer	YES(?,O(n^4))
Input	SK90 4.30

stdout:

YES(?,O(n^4))

We consider the following Problem:

  Strict Trs:
    {  g(.(x, .(y, z))) -> g(.(.(x, y), z))
     , g(.(x, nil())) -> .(g(x), nil())
     , g(nil()) -> nil()
     , f(.(.(x, y), z)) -> f(.(x, .(y, z)))
     , f(.(nil(), y)) -> .(nil(), f(y))
     , f(nil()) -> nil()}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^4))

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   nil() = [0]
           [1]
           [0]
           [1]
   f(x1) = [1 0 1 0] x1 + [2]
           [0 1 2 0]      [0]
           [0 0 1 0]      [0]
           [0 0 0 1]      [0]
   .(x1, x2) = [1 0 0 0] x1 + [1 0 0 0] x2 + [0]
               [0 1 0 0]      [0 0 0 1]      [0]
               [0 0 0 2]      [0 0 1 0]      [0]
               [0 0 0 1]      [0 0 0 1]      [1]
   g(x1) = [1 2 0 0] x1 + [0]
           [0 0 0 1]      [0]
           [0 0 0 2]      [0]
           [0 0 0 1]      [1]

Hurray, we answered YES(?,O(n^4))

Tool TRI2

Execution Time	0.32430005ms
Answer	MAYBE
Input	SK90 4.30

stdout:

MAYBE

We consider the following Problem:

  Strict Trs:
    {  g(.(x, .(y, z))) -> g(.(.(x, y), z))
     , g(.(x, nil())) -> .(g(x), nil())
     , g(nil()) -> nil()
     , f(.(.(x, y), z)) -> f(.(x, .(y, z)))
     , f(.(nil(), y)) -> .(nil(), f(y))
     , f(nil()) -> nil()}
  StartTerms: all
  Strategy: none

Certificate: MAYBE

Proof:
  The input cannot be shown compatible

Arrrr..