We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(+) = {2}, Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[+](x1, x2) = [1] x1 + [1] x2 + [1]
[0] = [0]
[s](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[+(0(), y)] = [1] y + [1]
> [1] y + [0]
= [y]
[+(s(x), 0())] = [1] x + [1]
> [1] x + [0]
= [s(x)]
[+(s(x), s(y))] = [1] y + [1] x + [1]
? [1] y + [1] x + [2]
= [s(+(s(x), +(y, 0())))]
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(n^1)).
Strict Trs: { +(s(x), s(y)) -> s(+(s(x), +(y, 0()))) }
Weak Trs:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs: { +(s(x), s(y)) -> s(+(s(x), +(y, 0()))) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(+) = {2}, Uargs(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[+](x1, x2) = [1] x1 + [2] x2 + [0]
[0] = [0]
[s](x1) = [1] x1 + [4]
The order satisfies the following ordering constraints:
[+(0(), y)] = [2] y + [0]
>= [1] y + [0]
= [y]
[+(s(x), 0())] = [1] x + [4]
>= [1] x + [4]
= [s(x)]
[+(s(x), s(y))] = [2] y + [1] x + [12]
> [2] y + [1] x + [8]
= [s(+(s(x), +(y, 0())))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))