Tool Bounds
stdout:
TIMEOUT
We consider the following Problem:
  Strict Trs:
    {  a(b(b(b(a(x1))))) -> b(b(b(b(b(b(b(x1)))))))
     , a(b(b(a(a(x1))))) -> b(b(a(b(b(b(a(x1)))))))
     , a(b(a(b(a(x1))))) -> b(a(b(b(b(a(b(x1)))))))
     , a(b(a(a(a(x1))))) -> b(a(a(b(b(a(a(x1)))))))
     , a(a(b(b(a(x1))))) -> a(b(b(b(a(b(b(x1)))))))
     , a(a(b(a(a(x1))))) -> a(b(a(b(a(b(a(x1)))))))
     , a(a(a(b(a(x1))))) -> a(a(b(b(a(a(b(x1)))))))
     , a(a(a(a(a(x1))))) -> a(a(a(b(a(a(a(x1)))))))
     , a(b(b(a(x1)))) -> b(b(b(b(b(x1)))))
     , a(b(a(a(x1)))) -> b(a(b(b(a(x1)))))
     , a(a(b(a(x1)))) -> a(b(b(a(b(x1)))))
     , a(a(a(a(x1)))) -> a(a(b(a(a(x1)))))
     , a(b(a(x1))) -> b(b(b(x1)))
     , a(a(a(x1))) -> a(b(a(x1)))
     , a(a(x1)) -> b(x1)}
  StartTerms: all
  Strategy: none
Certificate: TIMEOUT
Proof:
  Computation stopped due to timeout after 60.0 seconds.
Arrrr..Tool CDI
stdout:
TIMEOUT
Statistics:
Number of monomials: 0
Last formula building started for bound 0
Last SAT solving started for bound 0Tool EDA
stdout:
TIMEOUT
We consider the following Problem:
  Strict Trs:
    {  a(b(b(b(a(x1))))) -> b(b(b(b(b(b(b(x1)))))))
     , a(b(b(a(a(x1))))) -> b(b(a(b(b(b(a(x1)))))))
     , a(b(a(b(a(x1))))) -> b(a(b(b(b(a(b(x1)))))))
     , a(b(a(a(a(x1))))) -> b(a(a(b(b(a(a(x1)))))))
     , a(a(b(b(a(x1))))) -> a(b(b(b(a(b(b(x1)))))))
     , a(a(b(a(a(x1))))) -> a(b(a(b(a(b(a(x1)))))))
     , a(a(a(b(a(x1))))) -> a(a(b(b(a(a(b(x1)))))))
     , a(a(a(a(a(x1))))) -> a(a(a(b(a(a(a(x1)))))))
     , a(b(b(a(x1)))) -> b(b(b(b(b(x1)))))
     , a(b(a(a(x1)))) -> b(a(b(b(a(x1)))))
     , a(a(b(a(x1)))) -> a(b(b(a(b(x1)))))
     , a(a(a(a(x1)))) -> a(a(b(a(a(x1)))))
     , a(b(a(x1))) -> b(b(b(x1)))
     , a(a(a(x1))) -> a(b(a(x1)))
     , a(a(x1)) -> b(x1)}
  StartTerms: all
  Strategy: none
Certificate: TIMEOUT
Proof:
  Computation stopped due to timeout after 60.0 seconds.
Arrrr..Tool IDA
stdout:
TIMEOUT
We consider the following Problem:
  Strict Trs:
    {  a(b(b(b(a(x1))))) -> b(b(b(b(b(b(b(x1)))))))
     , a(b(b(a(a(x1))))) -> b(b(a(b(b(b(a(x1)))))))
     , a(b(a(b(a(x1))))) -> b(a(b(b(b(a(b(x1)))))))
     , a(b(a(a(a(x1))))) -> b(a(a(b(b(a(a(x1)))))))
     , a(a(b(b(a(x1))))) -> a(b(b(b(a(b(b(x1)))))))
     , a(a(b(a(a(x1))))) -> a(b(a(b(a(b(a(x1)))))))
     , a(a(a(b(a(x1))))) -> a(a(b(b(a(a(b(x1)))))))
     , a(a(a(a(a(x1))))) -> a(a(a(b(a(a(a(x1)))))))
     , a(b(b(a(x1)))) -> b(b(b(b(b(x1)))))
     , a(b(a(a(x1)))) -> b(a(b(b(a(x1)))))
     , a(a(b(a(x1)))) -> a(b(b(a(b(x1)))))
     , a(a(a(a(x1)))) -> a(a(b(a(a(x1)))))
     , a(b(a(x1))) -> b(b(b(x1)))
     , a(a(a(x1))) -> a(b(a(x1)))
     , a(a(x1)) -> b(x1)}
  StartTerms: all
  Strategy: none
Certificate: TIMEOUT
Proof:
  Computation stopped due to timeout after 60.0 seconds.
Arrrr..Tool TRI
stdout:
YES(?,O(n^2))
We consider the following Problem:
  Strict Trs:
    {  a(b(b(b(a(x1))))) -> b(b(b(b(b(b(b(x1)))))))
     , a(b(b(a(a(x1))))) -> b(b(a(b(b(b(a(x1)))))))
     , a(b(a(b(a(x1))))) -> b(a(b(b(b(a(b(x1)))))))
     , a(b(a(a(a(x1))))) -> b(a(a(b(b(a(a(x1)))))))
     , a(a(b(b(a(x1))))) -> a(b(b(b(a(b(b(x1)))))))
     , a(a(b(a(a(x1))))) -> a(b(a(b(a(b(a(x1)))))))
     , a(a(a(b(a(x1))))) -> a(a(b(b(a(a(b(x1)))))))
     , a(a(a(a(a(x1))))) -> a(a(a(b(a(a(a(x1)))))))
     , a(b(b(a(x1)))) -> b(b(b(b(b(x1)))))
     , a(b(a(a(x1)))) -> b(a(b(b(a(x1)))))
     , a(a(b(a(x1)))) -> a(b(b(a(b(x1)))))
     , a(a(a(a(x1)))) -> a(a(b(a(a(x1)))))
     , a(b(a(x1))) -> b(b(b(x1)))
     , a(a(a(x1))) -> a(b(a(x1)))
     , a(a(x1)) -> b(x1)}
  StartTerms: all
  Strategy: none
Certificate: YES(?,O(n^2))
Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   a(x1) = [1 1] x1 + [1]
           [0 1]      [1]
   b(x1) = [1 0] x1 + [0]
           [0 0]      [0]
Hurray, we answered YES(?,O(n^2))Tool TRI2
stdout:
YES(?,O(n^2))
We consider the following Problem:
  Strict Trs:
    {  a(b(b(b(a(x1))))) -> b(b(b(b(b(b(b(x1)))))))
     , a(b(b(a(a(x1))))) -> b(b(a(b(b(b(a(x1)))))))
     , a(b(a(b(a(x1))))) -> b(a(b(b(b(a(b(x1)))))))
     , a(b(a(a(a(x1))))) -> b(a(a(b(b(a(a(x1)))))))
     , a(a(b(b(a(x1))))) -> a(b(b(b(a(b(b(x1)))))))
     , a(a(b(a(a(x1))))) -> a(b(a(b(a(b(a(x1)))))))
     , a(a(a(b(a(x1))))) -> a(a(b(b(a(a(b(x1)))))))
     , a(a(a(a(a(x1))))) -> a(a(a(b(a(a(a(x1)))))))
     , a(b(b(a(x1)))) -> b(b(b(b(b(x1)))))
     , a(b(a(a(x1)))) -> b(a(b(b(a(x1)))))
     , a(a(b(a(x1)))) -> a(b(b(a(b(x1)))))
     , a(a(a(a(x1)))) -> a(a(b(a(a(x1)))))
     , a(b(a(x1))) -> b(b(b(x1)))
     , a(a(a(x1))) -> a(b(a(x1)))
     , a(a(x1)) -> b(x1)}
  StartTerms: all
  Strategy: none
Certificate: YES(?,O(n^2))
Proof:
  We have the following triangular matrix interpretation:
  Interpretation Functions:
   a(x1) = [1 1] x1 + [2]
           [0 1]      [2]
   b(x1) = [1 0] x1 + [0]
           [0 0]      [0]
Hurray, we answered YES(?,O(n^2))