Problem Transformed CSR 04 PEANO nosorts FR

Tool Bounds

Execution Time	60.03121ms
Answer	TIMEOUT
Input	Transformed CSR 04 PEANO nosorts FR

stdout:

TIMEOUT

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , plus(N, s(M)) -> s(plus(N, M))
     , plus(N, 0()) -> N
     , and(tt(), X) -> activate(X)}
  StartTerms: all
  Strategy: none

Certificate: TIMEOUT

Proof:
  Computation stopped due to timeout after 60.0 seconds.

Arrrr..

Tool CDI

Execution Time	0.10820103ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 PEANO nosorts FR

stdout:

YES(?,O(n^2))
QUADRATIC upper bound on the derivational complexity

This TRS is terminating using the deltarestricted interpretation
s(delta, X0) =  + 1*X0 + 2 + 0*X0*delta + 0*delta
0(delta) =  + 0 + 1*delta
plus(delta, X1, X0) =  + 1*X0 + 1*X1 + 0 + 1*X0*delta + 0*X1*delta + 0*delta
tt(delta) =  + 2 + 0*delta
and(delta, X1, X0) =  + 1*X0 + 0*X1 + 0 + 0*X0*delta + 3*X1*delta + 0*delta
activate(delta, X0) =  + 1*X0 + 0 + 0*X0*delta + 3*delta
s_tau_1(delta) = delta/(1 + 0 * delta)
plus_tau_1(delta) = delta/(1 + 0 * delta)
plus_tau_2(delta) = delta/(1 + 1 * delta)
and_tau_1(delta) = delta/(0 + 3 * delta)
and_tau_2(delta) = delta/(1 + 0 * delta)
activate_tau_1(delta) = delta/(1 + 0 * delta)

Time: 0.069886 seconds
Statistics:
Number of monomials: 129
Last formula building started for bound 3
Last SAT solving started for bound 3

Tool EDA

Execution Time	0.27042198ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 PEANO nosorts FR

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , plus(N, s(M)) -> s(plus(N, M))
     , plus(N, 0()) -> N
     , and(tt(), X) -> activate(X)}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following EDA-non-satisfying matrix interpretation:
  Interpretation Functions:
   tt() = [3]
          [3]
   and(x1, x2) = [1 0] x1 + [1 0] x2 + [3]
                 [0 0]      [0 1]      [3]
   activate(x1) = [1 0] x1 + [1]
                  [0 1]      [3]
   0() = [1]
         [0]
   plus(x1, x2) = [1 0] x1 + [1 2] x2 + [0]
                  [0 1]      [0 1]      [2]
   s(x1) = [1 0] x1 + [0]
           [0 1]      [2]

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

Tool IDA

Execution Time	0.50924397ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 PEANO nosorts FR

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , plus(N, s(M)) -> s(plus(N, M))
     , plus(N, 0()) -> N
     , and(tt(), X) -> activate(X)}
  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:
   tt() = [3]
          [3]
   and(x1, x2) = [1 0] x1 + [1 0] x2 + [3]
                 [0 0]      [0 1]      [3]
   activate(x1) = [1 0] x1 + [1]
                  [0 1]      [3]
   0() = [3]
         [3]
   plus(x1, x2) = [1 0] x1 + [1 2] x2 + [0]
                  [0 1]      [0 1]      [2]
   s(x1) = [1 0] x1 + [0]
           [0 1]      [2]

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

Tool TRI

Execution Time	0.12893105ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 PEANO nosorts FR

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , plus(N, s(M)) -> s(plus(N, M))
     , plus(N, 0()) -> N
     , and(tt(), X) -> activate(X)}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   tt() = [1]
          [3]
   and(x1, x2) = [1 2] x1 + [1 3] x2 + [3]
                 [0 1]      [0 1]      [3]
   activate(x1) = [1 3] x1 + [1]
                  [0 1]      [1]
   0() = [1]
         [0]
   plus(x1, x2) = [1 3] x1 + [1 2] x2 + [0]
                  [0 1]      [0 1]      [0]
   s(x1) = [1 0] x1 + [0]
           [0 1]      [2]

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

Tool TRI2

Execution Time	7.807493e-2ms
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 PEANO nosorts FR

stdout:

YES(?,O(n^2))

We consider the following Problem:

  Strict Trs:
    {  activate(X) -> X
     , plus(N, s(M)) -> s(plus(N, M))
     , plus(N, 0()) -> N
     , and(tt(), X) -> activate(X)}
  StartTerms: all
  Strategy: none

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

Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   tt() = [1]
          [3]
   and(x1, x2) = [1 2] x1 + [1 3] x2 + [3]
                 [0 1]      [0 1]      [3]
   activate(x1) = [1 3] x1 + [1]
                  [0 1]      [1]
   0() = [1]
         [0]
   plus(x1, x2) = [1 3] x1 + [1 2] x2 + [0]
                  [0 1]      [0 1]      [0]
   s(x1) = [1 0] x1 + [0]
           [0 1]      [2]

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