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:
{ f(a, empty()) -> g(a, empty())
, f(a, cons(x, k)) -> f(cons(x, a), k)
, g(empty(), d) -> d
, g(cons(x, k), d) -> g(k, cons(x, d)) }
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: { g(empty(), d) -> d }
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).
[f](x1, x2) = [2] x1 + [2] x2 + [0]
[empty] = [2]
[g](x1, x2) = [2] x1 + [2] x2 + [0]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
This order satisfies the following ordering constraints:
[f(a, empty())] = [2] a + [4]
>= [2] a + [4]
= [g(a, empty())]
[f(a, cons(x, k))] = [2] a + [2] x + [2] k + [0]
>= [2] a + [2] x + [2] k + [0]
= [f(cons(x, a), k)]
[g(empty(), d)] = [2] d + [4]
> [1] d + [0]
= [d]
[g(cons(x, k), d)] = [2] x + [2] k + [2] d + [0]
>= [2] x + [2] k + [2] d + [0]
= [g(k, cons(x, d))]
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:
{ f(a, empty()) -> g(a, empty())
, f(a, cons(x, k)) -> f(cons(x, a), k)
, g(cons(x, k), d) -> g(k, cons(x, d)) }
Weak Trs: { g(empty(), d) -> d }
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: { g(cons(x, k), d) -> g(k, cons(x, d)) }
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).
[f](x1, x2) = [2] x1 + [2] x2 + [0]
[empty] = [0]
[g](x1, x2) = [2] x1 + [1] x2 + [0]
[cons](x1, x2) = [1] x1 + [1] x2 + [2]
This order satisfies the following ordering constraints:
[f(a, empty())] = [2] a + [0]
>= [2] a + [0]
= [g(a, empty())]
[f(a, cons(x, k))] = [2] a + [2] x + [2] k + [4]
>= [2] a + [2] x + [2] k + [4]
= [f(cons(x, a), k)]
[g(empty(), d)] = [1] d + [0]
>= [1] d + [0]
= [d]
[g(cons(x, k), d)] = [2] x + [2] k + [1] d + [4]
> [1] x + [2] k + [1] d + [2]
= [g(k, cons(x, d))]
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:
{ f(a, empty()) -> g(a, empty())
, f(a, cons(x, k)) -> f(cons(x, a), k) }
Weak Trs:
{ g(empty(), d) -> d
, g(cons(x, k), d) -> g(k, cons(x, d)) }
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(a, empty()) -> g(a, empty()) }
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).
[f](x1, x2) = [1] x1 + [2] x2 + [0]
[empty] = [2]
[g](x1, x2) = [1] x1 + [1] x2 + [1]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
This order satisfies the following ordering constraints:
[f(a, empty())] = [1] a + [4]
> [1] a + [3]
= [g(a, empty())]
[f(a, cons(x, k))] = [1] a + [2] x + [2] k + [0]
>= [1] a + [1] x + [2] k + [0]
= [f(cons(x, a), k)]
[g(empty(), d)] = [1] d + [3]
> [1] d + [0]
= [d]
[g(cons(x, k), d)] = [1] x + [1] k + [1] d + [1]
>= [1] x + [1] k + [1] d + [1]
= [g(k, cons(x, d))]
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: { f(a, cons(x, k)) -> f(cons(x, a), k) }
Weak Trs:
{ f(a, empty()) -> g(a, empty())
, g(empty(), d) -> d
, g(cons(x, k), d) -> g(k, cons(x, d)) }
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(a, cons(x, k)) -> f(cons(x, a), k) }
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).
[f](x1, x2) = [1] x1 + [2] x2 + [2]
[empty] = [1]
[g](x1, x2) = [1] x1 + [1] x2 + [0]
[cons](x1, x2) = [1] x1 + [1] x2 + [1]
This order satisfies the following ordering constraints:
[f(a, empty())] = [1] a + [4]
> [1] a + [1]
= [g(a, empty())]
[f(a, cons(x, k))] = [1] a + [2] x + [2] k + [4]
> [1] a + [1] x + [2] k + [3]
= [f(cons(x, a), k)]
[g(empty(), d)] = [1] d + [1]
> [1] d + [0]
= [d]
[g(cons(x, k), d)] = [1] x + [1] k + [1] d + [1]
>= [1] x + [1] k + [1] d + [1]
= [g(k, cons(x, d))]
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:
{ f(a, empty()) -> g(a, empty())
, f(a, cons(x, k)) -> f(cons(x, a), k)
, g(empty(), d) -> d
, g(cons(x, k), d) -> g(k, cons(x, d)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))