We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ f(x, c(y)) -> f(x, s(f(y, y)))
, f(s(x), s(y)) -> f(x, s(c(s(y)))) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
We add the following weak dependency pairs:
Strict DPs:
{ f^#(x, c(y)) -> c_1(f^#(x, s(f(y, y))))
, f^#(s(x), s(y)) -> c_2(f^#(x, s(c(s(y))))) }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ f^#(x, c(y)) -> c_1(f^#(x, s(f(y, y))))
, f^#(s(x), s(y)) -> c_2(f^#(x, s(c(s(y))))) }
Strict Trs:
{ f(x, c(y)) -> f(x, s(f(y, y)))
, f(s(x), s(y)) -> f(x, s(c(s(y)))) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
DPs: { f^#(s(x), s(y)) -> c_2(f^#(x, s(c(s(y))))) }
Trs:
{ f(x, c(y)) -> f(x, s(f(y, y)))
, f(s(x), s(y)) -> f(x, s(c(s(y)))) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(f) = {1, 2}, Uargs(c) = {1}, Uargs(s) = {1},
Uargs(f^#) = {2}, Uargs(c_1) = {1}, Uargs(c_2) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[f](x1, x2) = [2 0] x1 + [1 3] x2 + [3]
[0 0] [0 0] [0]
[c](x1) = [1 0] x1 + [0]
[1 1] [4]
[s](x1) = [1 0] x1 + [3]
[0 0] [1]
[f^#](x1, x2) = [3 0] x1 + [1 3] x2 + [0]
[0 0] [0 0] [1]
[c_1](x1) = [1 3] x1 + [0]
[0 0] [0]
[c_2](x1) = [1 5] x1 + [0]
[0 0] [0]
The order satisfies the following ordering constraints:
[f(x, c(y))] = [2 0] x + [4 3] y + [15]
[0 0] [0 0] [0]
> [2 0] x + [3 3] y + [12]
[0 0] [0 0] [0]
= [f(x, s(f(y, y)))]
[f(s(x), s(y))] = [2 0] x + [1 0] y + [15]
[0 0] [0 0] [0]
> [2 0] x + [1 0] y + [12]
[0 0] [0 0] [0]
= [f(x, s(c(s(y))))]
[f^#(x, c(y))] = [3 0] x + [4 3] y + [12]
[0 0] [0 0] [1]
>= [3 0] x + [3 3] y + [12]
[0 0] [0 0] [0]
= [c_1(f^#(x, s(f(y, y))))]
[f^#(s(x), s(y))] = [3 0] x + [1 0] y + [15]
[0 0] [0 0] [1]
> [3 0] x + [1 0] y + [14]
[0 0] [0 0] [0]
= [c_2(f^#(x, s(c(s(y)))))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs: { f^#(x, c(y)) -> c_1(f^#(x, s(f(y, y)))) }
Weak DPs: { f^#(s(x), s(y)) -> c_2(f^#(x, s(c(s(y))))) }
Weak Trs:
{ f(x, c(y)) -> f(x, s(f(y, y)))
, f(s(x), s(y)) -> f(x, s(c(s(y)))) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following constant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(f) = {1, 2}, Uargs(c) = {1}, Uargs(s) = {1},
Uargs(f^#) = {2}, Uargs(c_1) = {1}, Uargs(c_2) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[f](x1, x2) = [1 0] x1 + [1 3] x2 + [0]
[0 0] [0 1] [0]
[c](x1) = [1 0] x1 + [0]
[1 1] [2]
[s](x1) = [1 0] x1 + [0]
[0 0] [1]
[f^#](x1, x2) = [2 7] x2 + [0]
[0 0] [0]
[c_1](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_2](x1) = [1 0] x1 + [0]
[0 1] [0]
The order satisfies the following ordering constraints:
[f(x, c(y))] = [1 0] x + [4 3] y + [6]
[0 0] [1 1] [2]
> [1 0] x + [2 3] y + [3]
[0 0] [0 0] [1]
= [f(x, s(f(y, y)))]
[f(s(x), s(y))] = [1 0] x + [1 0] y + [3]
[0 0] [0 0] [1]
>= [1 0] x + [1 0] y + [3]
[0 0] [0 0] [1]
= [f(x, s(c(s(y))))]
[f^#(x, c(y))] = [9 7] y + [14]
[0 0] [0]
> [4 6] y + [7]
[0 0] [0]
= [c_1(f^#(x, s(f(y, y))))]
[f^#(s(x), s(y))] = [2 0] y + [7]
[0 0] [0]
>= [2 0] y + [7]
[0 0] [0]
= [c_2(f^#(x, s(c(s(y)))))]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak DPs:
{ f^#(x, c(y)) -> c_1(f^#(x, s(f(y, y))))
, f^#(s(x), s(y)) -> c_2(f^#(x, s(c(s(y))))) }
Weak Trs:
{ f(x, c(y)) -> f(x, s(f(y, y)))
, f(s(x), s(y)) -> f(x, s(c(s(y)))) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^2))