Problem SK90 2.31

Tool Bounds

Execution Time	60.033443ms
Answer	TIMEOUT
Input	SK90 2.31

stdout:

TIMEOUT

We consider the following Problem:

  Strict Trs:
    {  +(s(x), y) -> s(+(x, y))
     , +(x, s(y)) -> s(+(x, y))
     , +(x, 0()) -> x
     , odd(s(x)) -> not(odd(x))
     , odd(0()) -> false()
     , not(false()) -> true()
     , not(true()) -> false()}
  StartTerms: all
  Strategy: none

Certificate: TIMEOUT

Proof:
  Computation stopped due to timeout after 60.0 seconds.

Arrrr..

Tool CDI

Execution Time	0.25201893ms
Answer	MAYBE
Input	SK90 2.31

stdout:

MAYBE

Statistics:
Number of monomials: 248
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.3857081ms
Answer	YES(?,O(n^2))
Input	SK90 2.31

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  +(s(x), y) -> s(+(x, y))
     , +(x, s(y)) -> s(+(x, y))
     , +(x, 0()) -> x
     , odd(s(x)) -> not(odd(x))
     , odd(0()) -> false()
     , not(false()) -> true()
     , not(true()) -> false()}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^2))

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   true() = [0]
            [1]
   not(x1) = [1 1] x1 + [0]
             [0 0]      [1]
   false() = [0]
             [1]
   0() = [1]
         [0]
   odd(x1) = [1 1] x1 + [0]
             [0 0]      [1]
   s(x1) = [1 0] x1 + [3]
           [0 1]      [1]
   +(x1, x2) = [1 1] x1 + [1 1] x2 + [0]
               [0 1]      [0 1]      [0]

Hurray, we answered YES(?,O(n^2))

Tool IDA

Execution Time	0.63167ms
Answer	YES(?,O(n^2))
Input	SK90 2.31

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  +(s(x), y) -> s(+(x, y))
     , +(x, s(y)) -> s(+(x, y))
     , +(x, 0()) -> x
     , odd(s(x)) -> not(odd(x))
     , odd(0()) -> false()
     , not(false()) -> true()
     , not(true()) -> false()}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^2))

Proof:
  We have the following EDA-non-satisfying and IDA(2)-non-satisfying matrix interpretation:
  Interpretation Functions:
   true() = [0]
            [1]
   not(x1) = [1 1] x1 + [0]
             [0 0]      [1]
   false() = [0]
             [1]
   0() = [1]
         [3]
   odd(x1) = [1 1] x1 + [0]
             [0 0]      [1]
   s(x1) = [1 0] x1 + [3]
           [0 1]      [1]
   +(x1, x2) = [1 1] x1 + [1 1] x2 + [0]
               [0 1]      [0 1]      [0]

Hurray, we answered YES(?,O(n^2))

Tool TRI

Execution Time	0.1866448ms
Answer	YES(?,O(n^2))
Input	SK90 2.31

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  +(s(x), y) -> s(+(x, y))
     , +(x, s(y)) -> s(+(x, y))
     , +(x, 0()) -> x
     , odd(s(x)) -> not(odd(x))
     , odd(0()) -> false()
     , not(false()) -> true()
     , not(true()) -> false()}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^2))

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   true() = [2]
            [1]
   not(x1) = [1 2] x1 + [0]
             [0 0]      [1]
   false() = [3]
             [0]
   0() = [3]
         [1]
   odd(x1) = [1 1] x1 + [0]
             [0 0]      [1]
   s(x1) = [1 0] x1 + [3]
           [0 1]      [1]
   +(x1, x2) = [1 1] x1 + [1 1] x2 + [0]
               [0 1]      [0 1]      [0]

Hurray, we answered YES(?,O(n^2))

Tool TRI2

Execution Time	0.13722396ms
Answer	YES(?,O(n^2))
Input	SK90 2.31

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  +(s(x), y) -> s(+(x, y))
     , +(x, s(y)) -> s(+(x, y))
     , +(x, 0()) -> x
     , odd(s(x)) -> not(odd(x))
     , odd(0()) -> false()
     , not(false()) -> true()
     , not(true()) -> false()}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^2))

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   true() = [2]
            [1]
   not(x1) = [1 2] x1 + [0]
             [0 0]      [1]
   false() = [3]
             [0]
   0() = [3]
         [1]
   odd(x1) = [1 1] x1 + [0]
             [0 0]      [1]
   s(x1) = [1 0] x1 + [3]
           [0 1]      [1]
   +(x1, x2) = [1 1] x1 + [1 1] x2 + [0]
               [0 1]      [0 1]      [0]

Hurray, we answered YES(?,O(n^2))