Tool Bounds
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{f(f(a(), f(a(), a())), x) -> f(x, f(f(a(), a()), a()))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^1))
Proof:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ a_0() -> 1
, a_1() -> 5
, a_1() -> 6
, a_1() -> 7
, f_0(1, 1) -> 2
, f_0(1, 2) -> 2
, f_0(2, 1) -> 2
, f_0(2, 2) -> 2
, f_1(1, 3) -> 2
, f_1(2, 3) -> 2
, f_1(3, 3) -> 2
, f_1(4, 5) -> 3
, f_1(6, 7) -> 4}
Hurray, we answered YES(?,O(n^1))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:
YES(?,O(n^3))
We consider the following Problem:
Strict Trs:
{f(f(a(), f(a(), a())), x) -> f(x, f(f(a(), a()), a()))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^3))
Proof:
We have the following EDA-non-satisfying matrix interpretation:
Interpretation Functions:
a() = [0]
[0]
[0]
f(x1, x2) = [1 3 1] x1 + [1 3 2] x2 + [0]
[0 1 2] [0 1 3] [0]
[0 0 0] [0 0 0] [1]
Hurray, we answered YES(?,O(n^3))Tool IDA
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{f(f(a(), f(a(), a())), x) -> f(x, f(f(a(), a()), a()))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^1))
Proof:
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
a() = [0]
[2]
[0]
f(x1, x2) = [1 0 0] x1 + [1 0 1] x2 + [0]
[0 0 0] [0 0 0] [0]
[0 1 0] [0 1 0] [0]
Hurray, we answered YES(?,O(n^1))Tool TRI
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{f(f(a(), f(a(), a())), x) -> f(x, f(f(a(), a()), a()))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following triangular matrix interpretation:
Interpretation Functions:
a() = [0]
[0]
[0]
f(x1, x2) = [1 1 0] x1 + [1 2 0] x2 + [0]
[0 1 0] [0 1 1] [0]
[0 0 0] [0 0 0] [1]
Hurray, we answered YES(?,O(n^2))Tool TRI2
stdout:
MAYBE
We consider the following Problem:
Strict Trs:
{f(f(a(), f(a(), a())), x) -> f(x, f(f(a(), a()), a()))}
StartTerms: all
Strategy: none
Certificate: MAYBE
Proof:
The input cannot be shown compatible
Arrrr..