Tool Bounds
stdout:
MAYBE
We consider the following Problem:
Strict Trs:
{ a(a(a(a(x1)))) -> b(c(x1))
, c(b(x1)) -> a(a(a(x1)))
, b(c(x1)) -> c(b(x1))}
StartTerms: all
Strategy: none
Certificate: MAYBE
Proof:
None of the processors succeeded.
Arrrr..Tool CDI
stdout:
MAYBE
Statistics:
Number of monomials: 534
Last formula building started for bound 3
Last SAT solving started for bound 3Tool EDA
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ a(a(a(a(x1)))) -> b(c(x1))
, c(b(x1)) -> a(a(a(x1)))
, b(c(x1)) -> c(b(x1))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following EDA-non-satisfying matrix interpretation:
Interpretation Functions:
c(x1) = [1 3] x1 + [3]
[0 1] [2]
b(x1) = [1 3] x1 + [1]
[0 1] [1]
a(x1) = [1 2] x1 + [0]
[0 1] [1]
Hurray, we answered YES(?,O(n^2))Tool IDA
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ a(a(a(a(x1)))) -> b(c(x1))
, c(b(x1)) -> a(a(a(x1)))
, b(c(x1)) -> c(b(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:
c(x1) = [1 3] x1 + [2]
[0 1] [2]
b(x1) = [1 3] x1 + [2]
[0 1] [1]
a(x1) = [1 2] x1 + [0]
[0 1] [1]
Hurray, we answered YES(?,O(n^2))Tool TRI
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ a(a(a(a(x1)))) -> b(c(x1))
, c(b(x1)) -> a(a(a(x1)))
, b(c(x1)) -> c(b(x1))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following triangular matrix interpretation:
Interpretation Functions:
c(x1) = [1 3] x1 + [2]
[0 1] [2]
b(x1) = [1 3] x1 + [2]
[0 1] [1]
a(x1) = [1 2] x1 + [0]
[0 1] [1]
Hurray, we answered YES(?,O(n^2))Tool TRI2
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ a(a(a(a(x1)))) -> b(c(x1))
, c(b(x1)) -> a(a(a(x1)))
, b(c(x1)) -> c(b(x1))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following triangular matrix interpretation:
Interpretation Functions:
c(x1) = [1 1] x1 + [0]
[0 1] [1]
b(x1) = [1 3] x1 + [2]
[0 1] [2]
a(x1) = [1 1] x1 + [0]
[0 1] [1]
Hurray, we answered YES(?,O(n^2))