Tool Bounds
stdout:
TIMEOUT
We consider the following Problem:
Strict Trs:
{ g(cons(x, k), d) -> g(k, cons(x, d))
, g(empty(), d) -> d
, f(a, cons(x, k)) -> f(cons(x, a), k)
, f(a, empty()) -> g(a, empty())}
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
cons(delta, X1, X0) = + 1*X0 + 0*X1 + 2 + 0*X0*delta + 1*X1*delta + 0*delta
f(delta, X1, X0) = + 1*X0 + 1*X1 + 0 + 2*X0*delta + 1*X1*delta + 0*delta
empty(delta) = + 2 + 0*delta
g(delta, X1, X0) = + 1*X0 + 1*X1 + 0 + 0*X0*delta + 1*X1*delta + 3*delta
cons_tau_1(delta) = delta/(0 + 1 * delta)
cons_tau_2(delta) = delta/(1 + 0 * delta)
f_tau_1(delta) = delta/(1 + 1 * delta)
f_tau_2(delta) = delta/(1 + 2 * delta)
g_tau_1(delta) = delta/(1 + 1 * delta)
g_tau_2(delta) = delta/(1 + 0 * delta)
Time: 0.220611 seconds
Statistics:
Number of monomials: 238
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:
{ g(cons(x, k), d) -> g(k, cons(x, d))
, g(empty(), d) -> d
, f(a, cons(x, k)) -> f(cons(x, a), k)
, f(a, empty()) -> g(a, empty())}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following EDA-non-satisfying matrix interpretation:
Interpretation Functions:
empty() = [0]
[0]
f(x1, x2) = [1 1] x1 + [1 2] x2 + [2]
[0 1] [0 1] [2]
g(x1, x2) = [1 1] x1 + [1 0] x2 + [1]
[0 1] [0 1] [2]
cons(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [3]
Hurray, we answered YES(?,O(n^2))Tool IDA
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ g(cons(x, k), d) -> g(k, cons(x, d))
, g(empty(), d) -> d
, f(a, cons(x, k)) -> f(cons(x, a), k)
, f(a, empty()) -> g(a, empty())}
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:
empty() = [0]
[0]
f(x1, x2) = [1 1] x1 + [1 2] x2 + [2]
[0 1] [0 1] [2]
g(x1, x2) = [1 1] x1 + [1 0] x2 + [1]
[0 1] [0 1] [1]
cons(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [3]
Hurray, we answered YES(?,O(n^2))Tool TRI
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ g(cons(x, k), d) -> g(k, cons(x, d))
, g(empty(), d) -> d
, f(a, cons(x, k)) -> f(cons(x, a), k)
, f(a, empty()) -> g(a, empty())}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following triangular matrix interpretation:
Interpretation Functions:
empty() = [0]
[0]
f(x1, x2) = [1 1] x1 + [1 2] x2 + [2]
[0 1] [0 1] [1]
g(x1, x2) = [1 1] x1 + [1 0] x2 + [1]
[0 1] [0 1] [1]
cons(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [3]
Hurray, we answered YES(?,O(n^2))Tool TRI2
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ g(cons(x, k), d) -> g(k, cons(x, d))
, g(empty(), d) -> d
, f(a, cons(x, k)) -> f(cons(x, a), k)
, f(a, empty()) -> g(a, empty())}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following triangular matrix interpretation:
Interpretation Functions:
empty() = [0]
[0]
f(x1, x2) = [1 1] x1 + [1 2] x2 + [2]
[0 1] [0 1] [1]
g(x1, x2) = [1 1] x1 + [1 0] x2 + [1]
[0 1] [0 1] [1]
cons(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [3]
Hurray, we answered YES(?,O(n^2))