Problem Secret 06 TRS 9

Tool Bounds

Execution Time	4.3447018e-2ms
Answer	YES(?,O(n^1))
Input	Secret 06 TRS 9

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  b(y, 0()) -> y
     , c(b(y, c(x))) -> c(c(b(a(0(), 0()), y)))
     , a(x, y) -> b(x, b(0(), c(y)))}
  StartTerms: all
  Strategy: none

Certificate: YES(?,O(n^1))

Proof:
  The problem is match-bounded by 3.
  The enriched problem is compatible with the following automaton:
  {  a_0(1, 1) -> 1
   , a_1(5, 6) -> 3
   , a_1(5, 6) -> 4
   , a_2(10, 10) -> 15
   , 0_0() -> 1
   , 0_1() -> 5
   , 0_1() -> 6
   , 0_2() -> 10
   , 0_3() -> 17
   , c_0(1) -> 1
   , c_1(1) -> 8
   , c_1(2) -> 1
   , c_1(2) -> 8
   , c_1(3) -> 2
   , c_1(12) -> 3
   , c_2(6) -> 11
   , c_2(13) -> 2
   , c_2(14) -> 13
   , c_3(10) -> 18
   , b_0(1, 1) -> 1
   , b_1(1, 7) -> 1
   , b_1(4, 1) -> 3
   , b_1(4, 4) -> 12
   , b_1(6, 8) -> 7
   , b_2(5, 9) -> 3
   , b_2(5, 9) -> 4
   , b_2(10, 11) -> 9
   , b_2(15, 4) -> 14
   , b_3(10, 16) -> 15
   , b_3(17, 18) -> 16}

Hurray, we answered YES(?,O(n^1))

Tool CDI

Execution Time	60.05463ms
Answer	TIMEOUT
Input	Secret 06 TRS 9

stdout:

TIMEOUT

Statistics:
Number of monomials: 5702
Last formula building started for bound 3
Last SAT solving started for bound 0

Tool EDA

Execution Time	10.04581ms
Answer	YES(?,O(n^3))
Input	Secret 06 TRS 9

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  b(y, 0()) -> y
     , c(b(y, c(x))) -> c(c(b(a(0(), 0()), y)))
     , a(x, y) -> b(x, b(0(), c(y)))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   a(x1, x2) = [1 0 0] x1 + [1 0 3] x2 + [2]
               [0 1 0]      [0 0 0]      [0]
               [0 1 1]      [0 0 0]      [0]
   0() = [1]
         [0]
         [0]
   c(x1) = [1 0 3] x1 + [0]
           [0 0 0]      [3]
           [0 0 0]      [0]
   b(x1, x2) = [1 0 0] x1 + [1 0 0] x2 + [0]
               [0 1 0]      [0 0 0]      [0]
               [0 1 1]      [0 1 0]      [0]

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

Tool IDA

Execution Time	14.892351ms
Answer	YES(?,O(n^2))
Input	Secret 06 TRS 9

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  b(y, 0()) -> y
     , c(b(y, c(x))) -> c(c(b(a(0(), 0()), y)))
     , a(x, y) -> b(x, b(0(), c(y)))}
  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:
   a(x1, x2) = [1 0 0] x1 + [1 2 0] x2 + [3]
               [0 1 2]      [0 0 0]      [0]
               [0 0 1]      [0 0 0]      [0]
   0() = [0]
         [0]
         [0]
   c(x1) = [1 2 0] x1 + [0]
           [0 0 0]      [0]
           [0 0 0]      [2]
   b(x1, x2) = [1 0 0] x1 + [1 0 0] x2 + [1]
               [0 1 2]      [0 0 2]      [0]
               [0 0 1]      [0 0 0]      [0]

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

Tool TRI

Execution Time	0.9347811ms
Answer	YES(?,O(n^3))
Input	Secret 06 TRS 9

stdout:

YES(?,O(n^3))

We consider the following Problem:

  Strict Trs:
    {  b(y, 0()) -> y
     , c(b(y, c(x))) -> c(c(b(a(0(), 0()), y)))
     , a(x, y) -> b(x, b(0(), c(y)))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   a(x1, x2) = [1 3 1] x1 + [1 2 2] x2 + [3]
               [0 1 2]      [0 0 0]      [0]
               [0 0 1]      [0 0 0]      [0]
   0() = [0]
         [0]
         [0]
   c(x1) = [1 2 0] x1 + [0]
           [0 0 0]      [0]
           [0 0 0]      [2]
   b(x1, x2) = [1 2 0] x1 + [1 0 0] x2 + [1]
               [0 1 2]      [0 0 2]      [0]
               [0 0 1]      [0 0 0]      [0]

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

Tool TRI2

Execution Time	0.20159698ms
Answer	MAYBE
Input	Secret 06 TRS 9

stdout:

MAYBE

We consider the following Problem:

  Strict Trs:
    {  b(y, 0()) -> y
     , c(b(y, c(x))) -> c(c(b(a(0(), 0()), y)))
     , a(x, y) -> b(x, b(0(), c(y)))}
  StartTerms: all
  Strategy: none

Certificate: MAYBE

Proof:
  The input cannot be shown compatible

Arrrr..