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))
, lastbit(0()) -> 0()
, lastbit(s(0())) -> s(0())
, lastbit(s(s(x))) -> lastbit(x)
, conv(0()) -> cons(nil(), 0())
, conv(s(x)) -> cons(conv(half(s(x))), lastbit(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(conv) = {1}, Uargs(cons) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[half](x1) = [0]
[0] = [0]
[s](x1) = [1] x1 + [0]
[lastbit](x1) = [4]
[conv](x1) = [1] x1 + [0]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
The order satisfies the following ordering constraints:
[half(0())] = [0]
>= [0]
= [0()]
[half(s(0()))] = [0]
>= [0]
= [0()]
[half(s(s(x)))] = [0]
>= [0]
= [s(half(x))]
[lastbit(0())] = [4]
> [0]
= [0()]
[lastbit(s(0()))] = [4]
> [0]
= [s(0())]
[lastbit(s(s(x)))] = [4]
>= [4]
= [lastbit(x)]
[conv(0())] = [0]
>= [0]
= [cons(nil(), 0())]
[conv(s(x))] = [1] x + [0]
? [4]
= [cons(conv(half(s(x))), lastbit(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()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, lastbit(s(s(x))) -> lastbit(x)
, conv(0()) -> cons(nil(), 0())
, conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) }
Weak Trs:
{ lastbit(0()) -> 0()
, lastbit(s(0())) -> s(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: { conv(0()) -> cons(nil(), 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}, Uargs(conv) = {1}, Uargs(cons) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[half](x1) = [1] x1 + [0]
[0] = [0]
[s](x1) = [1] x1 + [0]
[lastbit](x1) = [0]
[conv](x1) = [3] x1 + [1]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
The order satisfies the following ordering constraints:
[half(0())] = [0]
>= [0]
= [0()]
[half(s(0()))] = [0]
>= [0]
= [0()]
[half(s(s(x)))] = [1] x + [0]
>= [1] x + [0]
= [s(half(x))]
[lastbit(0())] = [0]
>= [0]
= [0()]
[lastbit(s(0()))] = [0]
>= [0]
= [s(0())]
[lastbit(s(s(x)))] = [0]
>= [0]
= [lastbit(x)]
[conv(0())] = [1]
> [0]
= [cons(nil(), 0())]
[conv(s(x))] = [3] x + [1]
>= [3] x + [1]
= [cons(conv(half(s(x))), lastbit(s(x)))]
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:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, lastbit(s(s(x))) -> lastbit(x)
, conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) }
Weak Trs:
{ lastbit(0()) -> 0()
, lastbit(s(0())) -> s(0())
, conv(0()) -> cons(nil(), 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(s) = {1}, Uargs(conv) = {1}, Uargs(cons) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[half](x1) = [0]
[0] = [4]
[s](x1) = [1] x1 + [4]
[lastbit](x1) = [1] x1 + [1]
[conv](x1) = [1] x1 + [4]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
The order satisfies the following ordering constraints:
[half(0())] = [0]
? [4]
= [0()]
[half(s(0()))] = [0]
? [4]
= [0()]
[half(s(s(x)))] = [0]
? [4]
= [s(half(x))]
[lastbit(0())] = [5]
> [4]
= [0()]
[lastbit(s(0()))] = [9]
> [8]
= [s(0())]
[lastbit(s(s(x)))] = [1] x + [9]
> [1] x + [1]
= [lastbit(x)]
[conv(0())] = [8]
> [4]
= [cons(nil(), 0())]
[conv(s(x))] = [1] x + [8]
? [1] x + [9]
= [cons(conv(half(s(x))), lastbit(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()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) }
Weak Trs:
{ lastbit(0()) -> 0()
, lastbit(s(0())) -> s(0())
, lastbit(s(s(x))) -> lastbit(x)
, conv(0()) -> cons(nil(), 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(s) = {1}, Uargs(conv) = {1}, Uargs(cons) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[half](x1) = [1] x1 + [1]
[0] = [0]
[s](x1) = [1] x1 + [0]
[lastbit](x1) = [6]
[conv](x1) = [1] x1 + [1]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
The order satisfies the following ordering constraints:
[half(0())] = [1]
> [0]
= [0()]
[half(s(0()))] = [1]
> [0]
= [0()]
[half(s(s(x)))] = [1] x + [1]
>= [1] x + [1]
= [s(half(x))]
[lastbit(0())] = [6]
> [0]
= [0()]
[lastbit(s(0()))] = [6]
> [0]
= [s(0())]
[lastbit(s(s(x)))] = [6]
>= [6]
= [lastbit(x)]
[conv(0())] = [1]
> [0]
= [cons(nil(), 0())]
[conv(s(x))] = [1] x + [1]
? [1] x + [8]
= [cons(conv(half(s(x))), lastbit(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(s(s(x))) -> s(half(x))
, conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) }
Weak Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, lastbit(0()) -> 0()
, lastbit(s(0())) -> s(0())
, lastbit(s(s(x))) -> lastbit(x)
, conv(0()) -> cons(nil(), 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(s) = {1}, Uargs(conv) = {1}, Uargs(cons) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[half](x1) = [1] x1 + [1]
[0] = [0]
[s](x1) = [1] x1 + [4]
[lastbit](x1) = [6]
[conv](x1) = [1] x1 + [1]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
The order satisfies the following ordering constraints:
[half(0())] = [1]
> [0]
= [0()]
[half(s(0()))] = [5]
> [0]
= [0()]
[half(s(s(x)))] = [1] x + [9]
> [1] x + [5]
= [s(half(x))]
[lastbit(0())] = [6]
> [0]
= [0()]
[lastbit(s(0()))] = [6]
> [4]
= [s(0())]
[lastbit(s(s(x)))] = [6]
>= [6]
= [lastbit(x)]
[conv(0())] = [1]
> [0]
= [cons(nil(), 0())]
[conv(s(x))] = [1] x + [5]
? [1] x + [12]
= [cons(conv(half(s(x))), lastbit(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: { conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x))) }
Weak Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, lastbit(0()) -> 0()
, lastbit(s(0())) -> s(0())
, lastbit(s(s(x))) -> lastbit(x)
, conv(0()) -> cons(nil(), 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: { conv(s(x)) -> cons(conv(half(s(x))), lastbit(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(conv) = {1}, Uargs(cons) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[half](x1) = [1 0] x1 + [0]
[1 0] [1]
[0] = [0]
[1]
[s](x1) = [1 0] x1 + [2]
[1 0] [5]
[lastbit](x1) = [0 0] x1 + [2]
[1 0] [7]
[conv](x1) = [1 2] x1 + [0]
[1 0] [3]
[cons](x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [0]
[nil] = [1]
[0]
The order satisfies the following ordering constraints:
[half(0())] = [0]
[1]
>= [0]
[1]
= [0()]
[half(s(0()))] = [2]
[3]
> [0]
[1]
= [0()]
[half(s(s(x)))] = [1 0] x + [4]
[1 0] [5]
> [1 0] x + [2]
[1 0] [5]
= [s(half(x))]
[lastbit(0())] = [2]
[7]
> [0]
[1]
= [0()]
[lastbit(s(0()))] = [2]
[9]
>= [2]
[5]
= [s(0())]
[lastbit(s(s(x)))] = [0 0] x + [2]
[1 0] [11]
>= [0 0] x + [2]
[1 0] [7]
= [lastbit(x)]
[conv(0())] = [2]
[3]
>= [2]
[0]
= [cons(nil(), 0())]
[conv(s(x))] = [3 0] x + [12]
[1 0] [5]
> [3 0] x + [11]
[0 0] [0]
= [cons(conv(half(s(x))), lastbit(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))
, lastbit(0()) -> 0()
, lastbit(s(0())) -> s(0())
, lastbit(s(s(x))) -> lastbit(x)
, conv(0()) -> cons(nil(), 0())
, conv(s(x)) -> cons(conv(half(s(x))), lastbit(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))