We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, bits(0()) -> 0()
, bits(s(x)) -> s(bits(half(s(x)))) }
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(s) = {1}, Uargs(bits) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[half](x1) = [1] x1 + [0]
[0] = [0]
[s](x1) = [1] x1 + [4]
[bits](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[half(0())] = [0]
>= [0]
= [0()]
[half(s(0()))] = [4]
> [0]
= [0()]
[half(s(s(x)))] = [1] x + [8]
> [1] x + [4]
= [s(half(x))]
[bits(0())] = [0]
>= [0]
= [0()]
[bits(s(x))] = [1] x + [4]
? [1] x + [8]
= [s(bits(half(s(x))))]
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:
{ half(0()) -> 0()
, bits(0()) -> 0()
, bits(s(x)) -> s(bits(half(s(x)))) }
Weak Trs:
{ half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x)) }
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(s) = {1}, Uargs(bits) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[half](x1) = [1] x1 + [4]
[0] = [4]
[s](x1) = [1] x1 + [4]
[bits](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[half(0())] = [8]
> [4]
= [0()]
[half(s(0()))] = [12]
> [4]
= [0()]
[half(s(s(x)))] = [1] x + [12]
> [1] x + [8]
= [s(half(x))]
[bits(0())] = [4]
>= [4]
= [0()]
[bits(s(x))] = [1] x + [4]
? [1] x + [12]
= [s(bits(half(s(x))))]
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:
{ bits(0()) -> 0()
, bits(s(x)) -> s(bits(half(s(x)))) }
Weak Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x)) }
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(s) = {1}, Uargs(bits) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[half](x1) = [7]
[0] = [0]
[s](x1) = [1] x1 + [0]
[bits](x1) = [1] x1 + [1]
The order satisfies the following ordering constraints:
[half(0())] = [7]
> [0]
= [0()]
[half(s(0()))] = [7]
> [0]
= [0()]
[half(s(s(x)))] = [7]
>= [7]
= [s(half(x))]
[bits(0())] = [1]
> [0]
= [0()]
[bits(s(x))] = [1] x + [1]
? [8]
= [s(bits(half(s(x))))]
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: { bits(s(x)) -> s(bits(half(s(x)))) }
Weak Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, bits(0()) -> 0() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs: { bits(s(x)) -> s(bits(half(s(x)))) }
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}, Uargs(bits) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[half](x1) = [1 0] x1 + [0]
[1 0] [0]
[0] = [0]
[0]
[s](x1) = [1 0] x1 + [2]
[1 0] [4]
[bits](x1) = [1 3] x1 + [1]
[3 1] [3]
The order satisfies the following ordering constraints:
[half(0())] = [0]
[0]
>= [0]
[0]
= [0()]
[half(s(0()))] = [2]
[2]
> [0]
[0]
= [0()]
[half(s(s(x)))] = [1 0] x + [4]
[1 0] [4]
> [1 0] x + [2]
[1 0] [4]
= [s(half(x))]
[bits(0())] = [1]
[3]
> [0]
[0]
= [0()]
[bits(s(x))] = [4 0] x + [15]
[4 0] [13]
> [4 0] x + [11]
[4 0] [13]
= [s(bits(half(s(x))))]
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:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, bits(0()) -> 0()
, bits(s(x)) -> s(bits(half(s(x)))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))