Problem Maude 06 PALINDROME nokinds

Tool Bounds

Execution Time	60.03829ms
Answer	TIMEOUT
Input	Maude 06 PALINDROME nokinds

stdout:

TIMEOUT

We consider the following Problem:

  Strict Trs:
    {  isQid(u()) -> tt()
     , isQid(o()) -> tt()
     , isQid(i()) -> tt()
     , isQid(e()) -> tt()
     , isQid(a()) -> tt()
     , isPal(nil()) -> tt()
     , isPal(V) -> isNePal(V)
     , isNePal(__(I, __(P, I))) -> and(isQid(I), isPal(P))
     , isNePal(V) -> isQid(V)
     , isNeList(__(V1, V2)) -> and(isNeList(V1), isList(V2))
     , isNeList(__(V1, V2)) -> and(isList(V1), isNeList(V2))
     , isNeList(V) -> isQid(V)
     , isList(__(V1, V2)) -> and(isList(V1), isList(V2))
     , isList(nil()) -> tt()
     , isList(V) -> isNeList(V)
     , and(tt(), X) -> X
     , __(nil(), X) -> X
     , __(X, nil()) -> X
     , __(__(X, Y), Z) -> __(X, __(Y, Z))}
  StartTerms: all
  Strategy: none

Certificate: TIMEOUT

Proof:
  Computation stopped due to timeout after 60.0 seconds.

Arrrr..

Tool CDI

Execution Time	3.9634662ms
Answer	MAYBE
Input	Maude 06 PALINDROME nokinds

stdout:

MAYBE

Statistics:
Number of monomials: 1403
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.82873297ms
Answer	YES(?,O(n^2))
Input	Maude 06 PALINDROME nokinds

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  isQid(u()) -> tt()
     , isQid(o()) -> tt()
     , isQid(i()) -> tt()
     , isQid(e()) -> tt()
     , isQid(a()) -> tt()
     , isPal(nil()) -> tt()
     , isPal(V) -> isNePal(V)
     , isNePal(__(I, __(P, I))) -> and(isQid(I), isPal(P))
     , isNePal(V) -> isQid(V)
     , isNeList(__(V1, V2)) -> and(isNeList(V1), isList(V2))
     , isNeList(__(V1, V2)) -> and(isList(V1), isNeList(V2))
     , isNeList(V) -> isQid(V)
     , isList(__(V1, V2)) -> and(isList(V1), isList(V2))
     , isList(nil()) -> tt()
     , isList(V) -> isNeList(V)
     , and(tt(), X) -> X
     , __(nil(), X) -> X
     , __(X, nil()) -> X
     , __(__(X, Y), Z) -> __(X, __(Y, Z))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   __(x1, x2) = [1 2] x1 + [1 0] x2 + [3]
                [0 1]      [0 1]      [2]
   nil() = [0]
           [0]
   and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                 [0 0]      [0 1]      [0]
   tt() = [1]
          [0]
   isList(x1) = [1 0] x1 + [2]
                [0 0]      [0]
   isNeList(x1) = [1 0] x1 + [1]
                  [0 0]      [0]
   isQid(x1) = [1 0] x1 + [0]
               [0 0]      [0]
   isNePal(x1) = [1 0] x1 + [1]
                 [0 0]      [0]
   isPal(x1) = [1 0] x1 + [2]
               [0 0]      [0]
   a() = [2]
         [3]
   e() = [2]
         [3]
   i() = [2]
         [3]
   o() = [2]
         [3]
   u() = [2]
         [3]

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

Tool IDA

Execution Time	1.432359ms
Answer	YES(?,O(n^2))
Input	Maude 06 PALINDROME nokinds

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  isQid(u()) -> tt()
     , isQid(o()) -> tt()
     , isQid(i()) -> tt()
     , isQid(e()) -> tt()
     , isQid(a()) -> tt()
     , isPal(nil()) -> tt()
     , isPal(V) -> isNePal(V)
     , isNePal(__(I, __(P, I))) -> and(isQid(I), isPal(P))
     , isNePal(V) -> isQid(V)
     , isNeList(__(V1, V2)) -> and(isNeList(V1), isList(V2))
     , isNeList(__(V1, V2)) -> and(isList(V1), isNeList(V2))
     , isNeList(V) -> isQid(V)
     , isList(__(V1, V2)) -> and(isList(V1), isList(V2))
     , isList(nil()) -> tt()
     , isList(V) -> isNeList(V)
     , and(tt(), X) -> X
     , __(nil(), X) -> X
     , __(X, nil()) -> X
     , __(__(X, Y), Z) -> __(X, __(Y, Z))}
  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:
   __(x1, x2) = [1 3] x1 + [1 0] x2 + [3]
                [0 1]      [0 1]      [1]
   nil() = [2]
           [0]
   and(x1, x2) = [1 0] x1 + [1 1] x2 + [0]
                 [0 0]      [0 1]      [0]
   tt() = [1]
          [0]
   isList(x1) = [1 0] x1 + [2]
                [0 0]      [0]
   isNeList(x1) = [1 0] x1 + [1]
                  [0 0]      [0]
   isQid(x1) = [1 0] x1 + [0]
               [0 0]      [0]
   isNePal(x1) = [1 2] x1 + [1]
                 [0 0]      [1]
   isPal(x1) = [1 2] x1 + [3]
               [0 0]      [1]
   a() = [2]
         [3]
   e() = [2]
         [3]
   i() = [2]
         [3]
   o() = [2]
         [3]
   u() = [2]
         [3]

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

Tool TRI

Execution Time	0.39046502ms
Answer	YES(?,O(n^2))
Input	Maude 06 PALINDROME nokinds

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  isQid(u()) -> tt()
     , isQid(o()) -> tt()
     , isQid(i()) -> tt()
     , isQid(e()) -> tt()
     , isQid(a()) -> tt()
     , isPal(nil()) -> tt()
     , isPal(V) -> isNePal(V)
     , isNePal(__(I, __(P, I))) -> and(isQid(I), isPal(P))
     , isNePal(V) -> isQid(V)
     , isNeList(__(V1, V2)) -> and(isNeList(V1), isList(V2))
     , isNeList(__(V1, V2)) -> and(isList(V1), isNeList(V2))
     , isNeList(V) -> isQid(V)
     , isList(__(V1, V2)) -> and(isList(V1), isList(V2))
     , isList(nil()) -> tt()
     , isList(V) -> isNeList(V)
     , and(tt(), X) -> X
     , __(nil(), X) -> X
     , __(X, nil()) -> X
     , __(__(X, Y), Z) -> __(X, __(Y, Z))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   __(x1, x2) = [1 3] x1 + [1 0] x2 + [1]
                [0 1]      [0 1]      [1]
   nil() = [0]
           [0]
   and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                 [0 0]      [0 1]      [0]
   tt() = [1]
          [0]
   isList(x1) = [1 2] x1 + [2]
                [0 1]      [3]
   isNeList(x1) = [1 2] x1 + [1]
                  [0 1]      [2]
   isQid(x1) = [1 0] x1 + [0]
               [0 0]      [0]
   isNePal(x1) = [1 2] x1 + [2]
                 [0 1]      [2]
   isPal(x1) = [1 2] x1 + [3]
               [0 1]      [2]
   a() = [2]
         [3]
   e() = [2]
         [3]
   i() = [2]
         [3]
   o() = [2]
         [3]
   u() = [2]
         [3]

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

Tool TRI2

Execution Time	0.338197ms
Answer	YES(?,O(n^2))
Input	Maude 06 PALINDROME nokinds

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  isQid(u()) -> tt()
     , isQid(o()) -> tt()
     , isQid(i()) -> tt()
     , isQid(e()) -> tt()
     , isQid(a()) -> tt()
     , isPal(nil()) -> tt()
     , isPal(V) -> isNePal(V)
     , isNePal(__(I, __(P, I))) -> and(isQid(I), isPal(P))
     , isNePal(V) -> isQid(V)
     , isNeList(__(V1, V2)) -> and(isNeList(V1), isList(V2))
     , isNeList(__(V1, V2)) -> and(isList(V1), isNeList(V2))
     , isNeList(V) -> isQid(V)
     , isList(__(V1, V2)) -> and(isList(V1), isList(V2))
     , isList(nil()) -> tt()
     , isList(V) -> isNeList(V)
     , and(tt(), X) -> X
     , __(nil(), X) -> X
     , __(X, nil()) -> X
     , __(__(X, Y), Z) -> __(X, __(Y, Z))}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   __(x1, x2) = [1 2] x1 + [1 0] x2 + [1]
                [0 1]      [0 1]      [1]
   nil() = [0]
           [1]
   and(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                 [0 0]      [0 1]      [0]
   tt() = [2]
          [0]
   isList(x1) = [1 2] x1 + [2]
                [0 0]      [0]
   isNeList(x1) = [1 2] x1 + [1]
                  [0 0]      [0]
   isQid(x1) = [1 0] x1 + [0]
               [0 0]      [0]
   isNePal(x1) = [1 0] x1 + [2]
                 [0 0]      [0]
   isPal(x1) = [1 0] x1 + [3]
               [0 0]      [0]
   a() = [3]
         [3]
   e() = [3]
         [3]
   i() = [3]
         [3]
   o() = [3]
         [3]
   u() = [3]
         [3]

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