We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(s(x), y) -> average(x, s(y))
, average(0(), s(s(0()))) -> s(0())
, average(0(), s(0())) -> 0()
, average(0(), 0()) -> 0() }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^1))
The input is overlay and right-linear. Switching to innermost
rewriting.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(s(x), y) -> average(x, s(y))
, average(0(), s(s(0()))) -> s(0())
, average(0(), s(0())) -> 0()
, average(0(), 0()) -> 0() }
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:
{ average(0(), s(s(0()))) -> s(0())
, average(0(), s(0())) -> 0()
, average(0(), 0()) -> 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(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[average](x1, x2) = [5] x1 + [4] x2 + [0]
[s](x1) = [1] x1 + [0]
[0] = [1]
The order satisfies the following ordering constraints:
[average(x, s(s(s(y))))] = [5] x + [4] y + [0]
>= [5] x + [4] y + [0]
= [s(average(s(x), y))]
[average(s(x), y)] = [5] x + [4] y + [0]
>= [5] x + [4] y + [0]
= [average(x, s(y))]
[average(0(), s(s(0())))] = [9]
> [1]
= [s(0())]
[average(0(), s(0()))] = [9]
> [1]
= [0()]
[average(0(), 0())] = [9]
> [1]
= [0()]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(s(x), y) -> average(x, s(y)) }
Weak Trs:
{ average(0(), s(s(0()))) -> s(0())
, average(0(), s(0())) -> 0()
, average(0(), 0()) -> 0() }
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: { average(x, s(s(s(y)))) -> s(average(s(x), y)) }
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(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[average](x1, x2) = [2] x1 + [2] x2 + [0]
[s](x1) = [1] x1 + [2]
[0] = [0]
The order satisfies the following ordering constraints:
[average(x, s(s(s(y))))] = [2] x + [2] y + [12]
> [2] x + [2] y + [6]
= [s(average(s(x), y))]
[average(s(x), y)] = [2] x + [2] y + [4]
>= [2] x + [2] y + [4]
= [average(x, s(y))]
[average(0(), s(s(0())))] = [8]
> [2]
= [s(0())]
[average(0(), s(0()))] = [4]
> [0]
= [0()]
[average(0(), 0())] = [0]
>= [0]
= [0()]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs: { average(s(x), y) -> average(x, s(y)) }
Weak Trs:
{ average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(0(), s(s(0()))) -> s(0())
, average(0(), s(0())) -> 0()
, average(0(), 0()) -> 0() }
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: { average(s(x), y) -> average(x, s(y)) }
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(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[average](x1, x2) = [4] x1 + [3] x2 + [4]
[s](x1) = [1] x1 + [1]
[0] = [0]
The order satisfies the following ordering constraints:
[average(x, s(s(s(y))))] = [4] x + [3] y + [13]
> [4] x + [3] y + [9]
= [s(average(s(x), y))]
[average(s(x), y)] = [4] x + [3] y + [8]
> [4] x + [3] y + [7]
= [average(x, s(y))]
[average(0(), s(s(0())))] = [10]
> [1]
= [s(0())]
[average(0(), s(0()))] = [7]
> [0]
= [0()]
[average(0(), 0())] = [4]
> [0]
= [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:
{ average(x, s(s(s(y)))) -> s(average(s(x), y))
, average(s(x), y) -> average(x, s(y))
, average(0(), s(s(0()))) -> s(0())
, average(0(), s(0())) -> 0()
, average(0(), 0()) -> 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))