Tool Bounds
stdout:
TIMEOUT
We consider the following Problem:
Strict Trs:
{ q3(bl(x1)) -> bl(q4(x1))
, q3(y(x1)) -> y(q3(x1))
, q0(y(x1)) -> y(q3(x1))
, q2(x(x1)) -> x(q0(x1))
, y(q2(y(x1))) -> q2(y(y(x1)))
, y(q2(a(x1))) -> q2(y(a(x1)))
, y(q1(b(x1))) -> q2(y(y(x1)))
, a(q2(y(x1))) -> q2(a(y(x1)))
, a(q2(a(x1))) -> q2(a(a(x1)))
, a(q1(b(x1))) -> q2(a(y(x1)))
, q1(y(x1)) -> y(q1(x1))
, q1(a(x1)) -> a(q1(x1))
, q0(a(x1)) -> x(q1(x1))}
StartTerms: all
Strategy: none
Certificate: TIMEOUT
Proof:
Computation stopped due to timeout after 60.0 seconds.
Arrrr..Tool CDI
stdout:
YES(?,O(n^2))
QUADRATIC upper bound on the derivational complexity
This TRS is terminating using the deltarestricted interpretation
q4(delta, X0) = + 1*X0 + 0 + 2*X0*delta + 0*delta
bl(delta, X0) = + 1*X0 + 0 + 0*X0*delta + 0*delta
q3(delta, X0) = + 1*X0 + 0 + 2*X0*delta + 2*delta
b(delta, X0) = + 1*X0 + 3 + 2*X0*delta + 0*delta
q2(delta, X0) = + 1*X0 + 2 + 2*X0*delta + 2*delta
y(delta, X0) = + 1*X0 + 1 + 2*X0*delta + 0*delta
a(delta, X0) = + 1*X0 + 1 + 2*X0*delta + 0*delta
q0(delta, X0) = + 1*X0 + 0 + 2*X0*delta + 1*delta
q1(delta, X0) = + 1*X0 + 0 + 2*X0*delta + 1*delta
x(delta, X0) = + 1*X0 + 1 + 2*X0*delta + 1*delta
q4_tau_1(delta) = delta/(1 + 2 * delta)
bl_tau_1(delta) = delta/(1 + 0 * delta)
q3_tau_1(delta) = delta/(1 + 2 * delta)
b_tau_1(delta) = delta/(1 + 2 * delta)
q2_tau_1(delta) = delta/(1 + 2 * delta)
y_tau_1(delta) = delta/(1 + 2 * delta)
a_tau_1(delta) = delta/(1 + 2 * delta)
q0_tau_1(delta) = delta/(1 + 2 * delta)
q1_tau_1(delta) = delta/(1 + 2 * delta)
x_tau_1(delta) = delta/(1 + 2 * delta)
Time: 34.362599 seconds
Statistics:
Number of monomials: 2260
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:
{ q3(bl(x1)) -> bl(q4(x1))
, q3(y(x1)) -> y(q3(x1))
, q0(y(x1)) -> y(q3(x1))
, q2(x(x1)) -> x(q0(x1))
, y(q2(y(x1))) -> q2(y(y(x1)))
, y(q2(a(x1))) -> q2(y(a(x1)))
, y(q1(b(x1))) -> q2(y(y(x1)))
, a(q2(y(x1))) -> q2(a(y(x1)))
, a(q2(a(x1))) -> q2(a(a(x1)))
, a(q1(b(x1))) -> q2(a(y(x1)))
, q1(y(x1)) -> y(q1(x1))
, q1(a(x1)) -> a(q1(x1))
, q0(a(x1)) -> x(q1(x1))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following EDA-non-satisfying matrix interpretation:
Interpretation Functions:
a(x1) = [1 3] x1 + [1]
[0 1] [2]
q0(x1) = [1 1] x1 + [1]
[0 1] [0]
q1(x1) = [1 1] x1 + [0]
[0 1] [0]
x(x1) = [1 0] x1 + [3]
[0 1] [2]
y(x1) = [1 3] x1 + [2]
[0 1] [1]
b(x1) = [1 3] x1 + [1]
[0 1] [2]
q2(x1) = [1 1] x1 + [0]
[0 1] [1]
q3(x1) = [1 1] x1 + [1]
[0 1] [0]
bl(x1) = [1 0] x1 + [0]
[0 0] [0]
q4(x1) = [1 0] x1 + [0]
[0 0] [3]
Hurray, we answered YES(?,O(n^2))Tool IDA
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ q3(bl(x1)) -> bl(q4(x1))
, q3(y(x1)) -> y(q3(x1))
, q0(y(x1)) -> y(q3(x1))
, q2(x(x1)) -> x(q0(x1))
, y(q2(y(x1))) -> q2(y(y(x1)))
, y(q2(a(x1))) -> q2(y(a(x1)))
, y(q1(b(x1))) -> q2(y(y(x1)))
, a(q2(y(x1))) -> q2(a(y(x1)))
, a(q2(a(x1))) -> q2(a(a(x1)))
, a(q1(b(x1))) -> q2(a(y(x1)))
, q1(y(x1)) -> y(q1(x1))
, q1(a(x1)) -> a(q1(x1))
, q0(a(x1)) -> x(q1(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] x1 + [3]
[0 1] [1]
q0(x1) = [1 1] x1 + [0]
[0 1] [0]
q1(x1) = [1 3] x1 + [1]
[0 1] [0]
x(x1) = [1 0] x1 + [0]
[0 1] [1]
y(x1) = [1 2] x1 + [0]
[0 1] [1]
b(x1) = [1 0] x1 + [1]
[0 1] [2]
q2(x1) = [1 1] x1 + [3]
[0 1] [1]
q3(x1) = [1 1] x1 + [0]
[0 1] [0]
bl(x1) = [1 0] x1 + [0]
[0 0] [1]
q4(x1) = [1 0] x1 + [0]
[0 0] [3]
Hurray, we answered YES(?,O(n^2))Tool TRI
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ q3(bl(x1)) -> bl(q4(x1))
, q3(y(x1)) -> y(q3(x1))
, q0(y(x1)) -> y(q3(x1))
, q2(x(x1)) -> x(q0(x1))
, y(q2(y(x1))) -> q2(y(y(x1)))
, y(q2(a(x1))) -> q2(y(a(x1)))
, y(q1(b(x1))) -> q2(y(y(x1)))
, a(q2(y(x1))) -> q2(a(y(x1)))
, a(q2(a(x1))) -> q2(a(a(x1)))
, a(q1(b(x1))) -> q2(a(y(x1)))
, q1(y(x1)) -> y(q1(x1))
, q1(a(x1)) -> a(q1(x1))
, q0(a(x1)) -> x(q1(x1))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following triangular matrix interpretation:
Interpretation Functions:
a(x1) = [1 2] x1 + [0]
[0 1] [2]
q0(x1) = [1 1] x1 + [0]
[0 1] [2]
q1(x1) = [1 3] x1 + [1]
[0 1] [2]
x(x1) = [1 0] x1 + [0]
[0 1] [2]
y(x1) = [1 1] x1 + [0]
[0 1] [1]
b(x1) = [1 0] x1 + [0]
[0 1] [2]
q2(x1) = [1 1] x1 + [0]
[0 1] [3]
q3(x1) = [1 1] x1 + [0]
[0 1] [0]
bl(x1) = [1 3] x1 + [0]
[0 1] [2]
q4(x1) = [1 1] x1 + [0]
[0 1] [0]
Hurray, we answered YES(?,O(n^2))Tool TRI2
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ q3(bl(x1)) -> bl(q4(x1))
, q3(y(x1)) -> y(q3(x1))
, q0(y(x1)) -> y(q3(x1))
, q2(x(x1)) -> x(q0(x1))
, y(q2(y(x1))) -> q2(y(y(x1)))
, y(q2(a(x1))) -> q2(y(a(x1)))
, y(q1(b(x1))) -> q2(y(y(x1)))
, a(q2(y(x1))) -> q2(a(y(x1)))
, a(q2(a(x1))) -> q2(a(a(x1)))
, a(q1(b(x1))) -> q2(a(y(x1)))
, q1(y(x1)) -> y(q1(x1))
, q1(a(x1)) -> a(q1(x1))
, q0(a(x1)) -> x(q1(x1))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following triangular matrix interpretation:
Interpretation Functions:
a(x1) = [1 2] x1 + [3]
[0 1] [1]
q0(x1) = [1 1] x1 + [3]
[0 1] [2]
q1(x1) = [1 3] x1 + [1]
[0 1] [0]
x(x1) = [1 0] x1 + [0]
[0 1] [2]
y(x1) = [1 2] x1 + [0]
[0 1] [1]
b(x1) = [1 3] x1 + [0]
[0 1] [3]
q2(x1) = [1 2] x1 + [0]
[0 1] [2]
q3(x1) = [1 1] x1 + [0]
[0 1] [0]
bl(x1) = [1 3] x1 + [0]
[0 1] [1]
q4(x1) = [1 1] x1 + [0]
[0 1] [0]
Hurray, we answered YES(?,O(n^2))