Problem Maude 06 PALINDROME nokinds-noand

Tool Bounds

Execution Time	60.0841ms
Answer	TIMEOUT
Input	Maude 06 PALINDROME nokinds-noand

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) -> U81(isNePal(V))
     , isNePal(__(I, __(P, I))) -> U71(isQid(I), P)
     , isNePal(V) -> U61(isQid(V))
     , isNeList(__(V1, V2)) -> U51(isNeList(V1), V2)
     , isNeList(__(V1, V2)) -> U41(isList(V1), V2)
     , isNeList(V) -> U31(isQid(V))
     , isList(__(V1, V2)) -> U21(isList(V1), V2)
     , isList(nil()) -> tt()
     , isList(V) -> U11(isNeList(V))
     , U81(tt()) -> tt()
     , U72(tt()) -> tt()
     , U71(tt(), P) -> U72(isPal(P))
     , U61(tt()) -> tt()
     , U52(tt()) -> tt()
     , U51(tt(), V2) -> U52(isList(V2))
     , U42(tt()) -> tt()
     , U41(tt(), V2) -> U42(isNeList(V2))
     , U31(tt()) -> tt()
     , U22(tt()) -> tt()
     , U21(tt(), V2) -> U22(isList(V2))
     , U11(tt()) -> tt()
     , __(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	1.883477ms
Answer	MAYBE
Input	Maude 06 PALINDROME nokinds-noand

stdout:

MAYBE

Statistics:
Number of monomials: 1249
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	1.2104912ms
Answer	YES(?,O(n^2))
Input	Maude 06 PALINDROME nokinds-noand

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) -> U81(isNePal(V))
     , isNePal(__(I, __(P, I))) -> U71(isQid(I), P)
     , isNePal(V) -> U61(isQid(V))
     , isNeList(__(V1, V2)) -> U51(isNeList(V1), V2)
     , isNeList(__(V1, V2)) -> U41(isList(V1), V2)
     , isNeList(V) -> U31(isQid(V))
     , isList(__(V1, V2)) -> U21(isList(V1), V2)
     , isList(nil()) -> tt()
     , isList(V) -> U11(isNeList(V))
     , U81(tt()) -> tt()
     , U72(tt()) -> tt()
     , U71(tt(), P) -> U72(isPal(P))
     , U61(tt()) -> tt()
     , U52(tt()) -> tt()
     , U51(tt(), V2) -> U52(isList(V2))
     , U42(tt()) -> tt()
     , U41(tt(), V2) -> U42(isNeList(V2))
     , U31(tt()) -> tt()
     , U22(tt()) -> tt()
     , U21(tt(), V2) -> U22(isList(V2))
     , U11(tt()) -> tt()
     , __(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 3] x1 + [1 0] x2 + [1]
                [0 1]      [0 1]      [2]
   nil() = [2]
           [2]
   U11(x1) = [1 0] x1 + [1]
             [0 1]      [0]
   tt() = [2]
          [2]
   U21(x1, x2) = [1 3] x1 + [1 1] x2 + [0]
                 [0 0]      [0 0]      [2]
   U22(x1) = [1 0] x1 + [1]
             [0 0]      [2]
   isList(x1) = [1 1] x1 + [3]
                [0 1]      [0]
   U31(x1) = [1 1] x1 + [0]
             [0 1]      [0]
   U41(x1, x2) = [1 1] x1 + [1 1] x2 + [0]
                 [0 0]      [0 1]      [0]
   U42(x1) = [1 0] x1 + [2]
             [0 1]      [0]
   isNeList(x1) = [1 1] x1 + [1]
                  [0 1]      [0]
   U51(x1, x2) = [1 3] x1 + [1 1] x2 + [0]
                 [0 0]      [0 1]      [0]
   U52(x1) = [1 0] x1 + [2]
             [0 1]      [0]
   U61(x1) = [1 0] x1 + [1]
             [0 1]      [0]
   U71(x1, x2) = [1 2] x1 + [1 3] x2 + [2]
                 [0 0]      [0 0]      [2]
   U72(x1) = [1 0] x1 + [2]
             [0 0]      [2]
   isPal(x1) = [1 3] x1 + [3]
               [0 0]      [2]
   U81(x1) = [1 3] x1 + [0]
             [0 0]      [2]
   isQid(x1) = [1 0] x1 + [0]
               [0 1]      [0]
   isNePal(x1) = [1 0] x1 + [2]
                 [0 1]      [0]
   a() = [3]
         [2]
   e() = [3]
         [2]
   i() = [3]
         [2]
   o() = [3]
         [2]
   u() = [3]
         [2]

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

Tool IDA

Execution Time	2.3238628ms
Answer	YES(?,O(n^2))
Input	Maude 06 PALINDROME nokinds-noand

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) -> U81(isNePal(V))
     , isNePal(__(I, __(P, I))) -> U71(isQid(I), P)
     , isNePal(V) -> U61(isQid(V))
     , isNeList(__(V1, V2)) -> U51(isNeList(V1), V2)
     , isNeList(__(V1, V2)) -> U41(isList(V1), V2)
     , isNeList(V) -> U31(isQid(V))
     , isList(__(V1, V2)) -> U21(isList(V1), V2)
     , isList(nil()) -> tt()
     , isList(V) -> U11(isNeList(V))
     , U81(tt()) -> tt()
     , U72(tt()) -> tt()
     , U71(tt(), P) -> U72(isPal(P))
     , U61(tt()) -> tt()
     , U52(tt()) -> tt()
     , U51(tt(), V2) -> U52(isList(V2))
     , U42(tt()) -> tt()
     , U41(tt(), V2) -> U42(isNeList(V2))
     , U31(tt()) -> tt()
     , U22(tt()) -> tt()
     , U21(tt(), V2) -> U22(isList(V2))
     , U11(tt()) -> tt()
     , __(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 2] x1 + [1 0] x2 + [3]
                [0 1]      [0 1]      [1]
   nil() = [0]
           [3]
   U11(x1) = [1 0] x1 + [1]
             [0 1]      [0]
   tt() = [3]
          [1]
   U21(x1, x2) = [1 2] x1 + [1 3] x2 + [0]
                 [0 1]      [0 0]      [0]
   U22(x1) = [1 0] x1 + [1]
             [0 0]      [1]
   isList(x1) = [1 3] x1 + [3]
                [0 1]      [0]
   U31(x1) = [1 1] x1 + [0]
             [0 1]      [0]
   U41(x1, x2) = [1 2] x1 + [1 3] x2 + [0]
                 [0 0]      [0 1]      [0]
   U42(x1) = [1 0] x1 + [2]
             [0 1]      [0]
   isNeList(x1) = [1 3] x1 + [1]
                  [0 1]      [0]
   U51(x1, x2) = [1 2] x1 + [1 3] x2 + [1]
                 [0 0]      [0 1]      [0]
   U52(x1) = [1 0] x1 + [2]
             [0 1]      [0]
   U61(x1) = [1 0] x1 + [1]
             [0 1]      [0]
   U71(x1, x2) = [1 0] x1 + [1 3] x2 + [2]
                 [0 0]      [0 1]      [0]
   U72(x1) = [1 0] x1 + [1]
             [0 1]      [0]
   isPal(x1) = [1 3] x1 + [3]
               [0 1]      [0]
   U81(x1) = [1 1] x1 + [0]
             [0 1]      [0]
   isQid(x1) = [1 2] x1 + [0]
               [0 1]      [0]
   isNePal(x1) = [1 2] x1 + [2]
                 [0 1]      [0]
   a() = [2]
         [1]
   e() = [2]
         [1]
   i() = [2]
         [1]
   o() = [0]
         [2]
   u() = [2]
         [1]

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

Tool TRI

Execution Time	0.60312295ms
Answer	YES(?,O(n^2))
Input	Maude 06 PALINDROME nokinds-noand

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) -> U81(isNePal(V))
     , isNePal(__(I, __(P, I))) -> U71(isQid(I), P)
     , isNePal(V) -> U61(isQid(V))
     , isNeList(__(V1, V2)) -> U51(isNeList(V1), V2)
     , isNeList(__(V1, V2)) -> U41(isList(V1), V2)
     , isNeList(V) -> U31(isQid(V))
     , isList(__(V1, V2)) -> U21(isList(V1), V2)
     , isList(nil()) -> tt()
     , isList(V) -> U11(isNeList(V))
     , U81(tt()) -> tt()
     , U72(tt()) -> tt()
     , U71(tt(), P) -> U72(isPal(P))
     , U61(tt()) -> tt()
     , U52(tt()) -> tt()
     , U51(tt(), V2) -> U52(isList(V2))
     , U42(tt()) -> tt()
     , U41(tt(), V2) -> U42(isNeList(V2))
     , U31(tt()) -> tt()
     , U22(tt()) -> tt()
     , U21(tt(), V2) -> U22(isList(V2))
     , U11(tt()) -> tt()
     , __(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 + [0]
                [0 1]      [0 1]      [3]
   nil() = [1]
           [2]
   U11(x1) = [1 1] x1 + [0]
             [0 0]      [1]
   tt() = [2]
          [1]
   U21(x1, x2) = [1 1] x1 + [1 3] x2 + [3]
                 [0 0]      [0 0]      [1]
   U22(x1) = [1 0] x1 + [2]
             [0 0]      [1]
   isList(x1) = [1 3] x1 + [3]
                [0 0]      [1]
   U31(x1) = [1 1] x1 + [0]
             [0 0]      [1]
   U41(x1, x2) = [1 0] x1 + [1 3] x2 + [3]
                 [0 1]      [0 0]      [0]
   U42(x1) = [1 3] x1 + [0]
             [0 0]      [1]
   isNeList(x1) = [1 3] x1 + [1]
                  [0 0]      [1]
   U51(x1, x2) = [1 3] x1 + [1 3] x2 + [2]
                 [0 1]      [0 0]      [0]
   U52(x1) = [1 2] x1 + [0]
             [0 0]      [1]
   U61(x1) = [1 1] x1 + [0]
             [0 1]      [0]
   U71(x1, x2) = [1 0] x1 + [1 3] x2 + [1]
                 [0 0]      [0 1]      [0]
   U72(x1) = [1 1] x1 + [0]
             [0 1]      [0]
   isPal(x1) = [1 2] x1 + [2]
               [0 1]      [0]
   U81(x1) = [1 1] x1 + [0]
             [0 1]      [0]
   isQid(x1) = [1 0] x1 + [0]
               [0 1]      [0]
   isNePal(x1) = [1 1] x1 + [1]
                 [0 1]      [0]
   a() = [3]
         [1]
   e() = [3]
         [1]
   i() = [3]
         [2]
   o() = [3]
         [1]
   u() = [3]
         [1]

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

Tool TRI2

Execution Time	0.50017214ms
Answer	YES(?,O(n^2))
Input	Maude 06 PALINDROME nokinds-noand

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) -> U81(isNePal(V))
     , isNePal(__(I, __(P, I))) -> U71(isQid(I), P)
     , isNePal(V) -> U61(isQid(V))
     , isNeList(__(V1, V2)) -> U51(isNeList(V1), V2)
     , isNeList(__(V1, V2)) -> U41(isList(V1), V2)
     , isNeList(V) -> U31(isQid(V))
     , isList(__(V1, V2)) -> U21(isList(V1), V2)
     , isList(nil()) -> tt()
     , isList(V) -> U11(isNeList(V))
     , U81(tt()) -> tt()
     , U72(tt()) -> tt()
     , U71(tt(), P) -> U72(isPal(P))
     , U61(tt()) -> tt()
     , U52(tt()) -> tt()
     , U51(tt(), V2) -> U52(isList(V2))
     , U42(tt()) -> tt()
     , U41(tt(), V2) -> U42(isNeList(V2))
     , U31(tt()) -> tt()
     , U22(tt()) -> tt()
     , U21(tt(), V2) -> U22(isList(V2))
     , U11(tt()) -> tt()
     , __(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 + [2]
                [0 1]      [0 1]      [3]
   nil() = [0]
           [2]
   U11(x1) = [1 0] x1 + [1]
             [0 1]      [0]
   tt() = [2]
          [2]
   U21(x1, x2) = [1 3] x1 + [1 1] x2 + [0]
                 [0 0]      [0 1]      [2]
   U22(x1) = [1 0] x1 + [2]
             [0 0]      [2]
   isList(x1) = [1 1] x1 + [3]
                [0 1]      [0]
   U31(x1) = [1 1] x1 + [0]
             [0 1]      [0]
   U41(x1, x2) = [1 1] x1 + [1 1] x2 + [0]
                 [0 0]      [0 1]      [2]
   U42(x1) = [1 0] x1 + [1]
             [0 1]      [2]
   isNeList(x1) = [1 1] x1 + [1]
                  [0 1]      [0]
   U51(x1, x2) = [1 1] x1 + [1 1] x2 + [2]
                 [0 1]      [0 1]      [0]
   U52(x1) = [1 0] x1 + [2]
             [0 0]      [2]
   U61(x1) = [1 0] x1 + [1]
             [0 1]      [0]
   U71(x1, x2) = [1 2] x1 + [1 3] x2 + [2]
                 [0 1]      [0 1]      [2]
   U72(x1) = [1 2] x1 + [0]
             [0 0]      [2]
   isPal(x1) = [1 1] x1 + [3]
               [0 1]      [0]
   U81(x1) = [1 1] x1 + [0]
             [0 1]      [0]
   isQid(x1) = [1 0] x1 + [0]
               [0 1]      [0]
   isNePal(x1) = [1 0] x1 + [2]
                 [0 1]      [0]
   a() = [3]
         [2]
   e() = [3]
         [2]
   i() = [3]
         [2]
   o() = [3]
         [2]
   u() = [3]
         [2]

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