Tool Bounds
stdout:
TIMEOUT
We consider the following Problem:
Strict Trs:
{ d(d(x1)) -> b(a(c(x1)))
, c(d(x1)) -> a(a(x1))
, c(c(x1)) -> b(x1)
, b(b(x1)) -> a(c(c(x1)))
, a(b(x1)) -> c(c(c(x1)))
, a(a(x1)) -> d(c(x1))}
StartTerms: all
Strategy: none
Certificate: TIMEOUT
Proof:
Computation stopped due to timeout after 60.0 seconds.
Arrrr..Tool CDI
stdout:
MAYBE
Statistics:
Number of monomials: 656
Last formula building started for bound 3
Last SAT solving started for bound 3Tool EDA
stdout:
TIMEOUT
We consider the following Problem:
Strict Trs:
{ d(d(x1)) -> b(a(c(x1)))
, c(d(x1)) -> a(a(x1))
, c(c(x1)) -> b(x1)
, b(b(x1)) -> a(c(c(x1)))
, a(b(x1)) -> c(c(c(x1)))
, a(a(x1)) -> d(c(x1))}
StartTerms: all
Strategy: none
Certificate: TIMEOUT
Proof:
Computation stopped due to timeout after 60.0 seconds.
Arrrr..Tool IDA
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ d(d(x1)) -> b(a(c(x1)))
, c(d(x1)) -> a(a(x1))
, c(c(x1)) -> b(x1)
, b(b(x1)) -> a(c(c(x1)))
, a(b(x1)) -> c(c(c(x1)))
, a(a(x1)) -> d(c(x1))}
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) = [1 2 2 1] x1 + [0]
[0 0 0 1] [1]
[0 0 0 1] [0]
[0 0 0 1] [2]
c(x1) = [1 0 2 1] x1 + [0]
[0 0 0 1] [0]
[0 1 0 0] [0]
[0 0 0 1] [1]
d(x1) = [1 2 2 3] x1 + [0]
[0 0 0 1] [2]
[0 0 0 1] [1]
[0 0 0 1] [3]
b(x1) = [1 2 2 2] x1 + [0]
[0 0 0 1] [1]
[0 0 0 1] [0]
[0 0 0 1] [2]
Hurray, we answered YES(?,O(n^2))Tool TRI
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ d(d(x1)) -> b(a(c(x1)))
, c(d(x1)) -> a(a(x1))
, c(c(x1)) -> b(x1)
, b(b(x1)) -> a(c(c(x1)))
, a(b(x1)) -> c(c(c(x1)))
, a(a(x1)) -> d(c(x1))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following triangular matrix interpretation:
Interpretation Functions:
a(x1) = [1 1 2 1] x1 + [0]
[0 0 0 2] [0]
[0 0 0 1] [1]
[0 0 0 1] [2]
c(x1) = [1 1 0 1] x1 + [0]
[0 0 2 0] [0]
[0 0 0 1] [0]
[0 0 0 1] [1]
d(x1) = [1 1 2 3] x1 + [0]
[0 0 0 2] [2]
[0 0 0 1] [2]
[0 0 0 1] [3]
b(x1) = [1 1 2 2] x1 + [0]
[0 0 0 2] [0]
[0 0 0 1] [1]
[0 0 0 1] [2]
Hurray, we answered YES(?,O(n^2))Tool TRI2
stdout:
MAYBE
We consider the following Problem:
Strict Trs:
{ d(d(x1)) -> b(a(c(x1)))
, c(d(x1)) -> a(a(x1))
, c(c(x1)) -> b(x1)
, b(b(x1)) -> a(c(c(x1)))
, a(b(x1)) -> c(c(c(x1)))
, a(a(x1)) -> d(c(x1))}
StartTerms: all
Strategy: none
Certificate: MAYBE
Proof:
The input cannot be shown compatible
Arrrr..