YES(O(1),O(n^1))
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z) }
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:
{ rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(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:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[norm](x1) = [2] x1 + [0]
[nil] = [0]
[0] = [0]
[g](x1, x2) = [1] x1 + [1] x2 + [0]
[s](x1) = [1] x1 + [0]
[f](x1, x2) = [3] x1 + [2] x2 + [0]
[rem](x1, x2) = [2] x1 + [1] x2 + [1]
This order satisfies the following ordering constraints:
[norm(nil())] = [0]
>= [0]
= [0()]
[norm(g(x, y))] = [2] x + [2] y + [0]
>= [2] x + [0]
= [s(norm(x))]
[f(x, nil())] = [3] x + [0]
>= [1] x + [0]
= [g(nil(), x)]
[f(x, g(y, z))] = [3] x + [2] y + [2] z + [0]
>= [3] x + [2] y + [1] z + [0]
= [g(f(x, y), z)]
[rem(nil(), y)] = [1] y + [1]
> [0]
= [nil()]
[rem(g(x, y), 0())] = [2] x + [2] y + [1]
> [1] x + [1] y + [0]
= [g(x, y)]
[rem(g(x, y), s(z))] = [2] x + [2] y + [1] z + [1]
>= [2] x + [1] z + [1]
= [rem(x, z)]
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:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(g(x, y), s(z)) -> rem(x, z) }
Weak Trs:
{ rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y) }
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: { norm(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:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[norm](x1) = [1] x1 + [0]
[nil] = [1]
[0] = [0]
[g](x1, x2) = [1] x1 + [1] x2 + [0]
[s](x1) = [1] x1 + [0]
[f](x1, x2) = [3] x1 + [1] x2 + [0]
[rem](x1, x2) = [1] x1 + [2] x2 + [0]
This order satisfies the following ordering constraints:
[norm(nil())] = [1]
> [0]
= [0()]
[norm(g(x, y))] = [1] x + [1] y + [0]
>= [1] x + [0]
= [s(norm(x))]
[f(x, nil())] = [3] x + [1]
>= [1] x + [1]
= [g(nil(), x)]
[f(x, g(y, z))] = [3] x + [1] y + [1] z + [0]
>= [3] x + [1] y + [1] z + [0]
= [g(f(x, y), z)]
[rem(nil(), y)] = [2] y + [1]
>= [1]
= [nil()]
[rem(g(x, y), 0())] = [1] x + [1] y + [0]
>= [1] x + [1] y + [0]
= [g(x, y)]
[rem(g(x, y), s(z))] = [1] x + [1] y + [2] z + [0]
>= [1] x + [2] z + [0]
= [rem(x, z)]
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:
{ norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(g(x, y), s(z)) -> rem(x, z) }
Weak Trs:
{ norm(nil()) -> 0()
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y) }
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: { f(x, nil()) -> g(nil(), 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:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[norm](x1) = [1] x1 + [0]
[nil] = [0]
[0] = [0]
[g](x1, x2) = [1] x1 + [1] x2 + [0]
[s](x1) = [1] x1 + [0]
[f](x1, x2) = [3] x1 + [1] x2 + [2]
[rem](x1, x2) = [2] x1 + [2] x2 + [0]
This order satisfies the following ordering constraints:
[norm(nil())] = [0]
>= [0]
= [0()]
[norm(g(x, y))] = [1] x + [1] y + [0]
>= [1] x + [0]
= [s(norm(x))]
[f(x, nil())] = [3] x + [2]
> [1] x + [0]
= [g(nil(), x)]
[f(x, g(y, z))] = [3] x + [1] y + [1] z + [2]
>= [3] x + [1] y + [1] z + [2]
= [g(f(x, y), z)]
[rem(nil(), y)] = [2] y + [0]
>= [0]
= [nil()]
[rem(g(x, y), 0())] = [2] x + [2] y + [0]
>= [1] x + [1] y + [0]
= [g(x, y)]
[rem(g(x, y), s(z))] = [2] x + [2] y + [2] z + [0]
>= [2] x + [2] z + [0]
= [rem(x, z)]
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:
{ norm(g(x, y)) -> s(norm(x))
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(g(x, y), s(z)) -> rem(x, z) }
Weak Trs:
{ norm(nil()) -> 0()
, f(x, nil()) -> g(nil(), x)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y) }
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:
{ norm(g(x, y)) -> s(norm(x))
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(g(x, y), s(z)) -> rem(x, z) }
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:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[norm](x1) = [1] x1 + [0]
[nil] = [0]
[0] = [0]
[g](x1, x2) = [1] x1 + [1] x2 + [1]
[s](x1) = [1] x1 + [0]
[f](x1, x2) = [3] x1 + [3] x2 + [1]
[rem](x1, x2) = [1] x1 + [1] x2 + [0]
This order satisfies the following ordering constraints:
[norm(nil())] = [0]
>= [0]
= [0()]
[norm(g(x, y))] = [1] x + [1] y + [1]
> [1] x + [0]
= [s(norm(x))]
[f(x, nil())] = [3] x + [1]
>= [1] x + [1]
= [g(nil(), x)]
[f(x, g(y, z))] = [3] x + [3] y + [3] z + [4]
> [3] x + [3] y + [1] z + [2]
= [g(f(x, y), z)]
[rem(nil(), y)] = [1] y + [0]
>= [0]
= [nil()]
[rem(g(x, y), 0())] = [1] x + [1] y + [1]
>= [1] x + [1] y + [1]
= [g(x, y)]
[rem(g(x, y), s(z))] = [1] x + [1] y + [1] z + [1]
> [1] x + [1] z + [0]
= [rem(x, z)]
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:
{ norm(nil()) -> 0()
, norm(g(x, y)) -> s(norm(x))
, f(x, nil()) -> g(nil(), x)
, f(x, g(y, z)) -> g(f(x, y), z)
, rem(nil(), y) -> nil()
, rem(g(x, y), 0()) -> g(x, y)
, rem(g(x, y), s(z)) -> rem(x, z) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))