We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))
, p(s(X)) -> X
, div(0(), s(Y)) -> 0()
, div(s(X), s(Y)) -> s(div(minus(X, Y), s(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:
{ minus(s(X), s(Y)) -> p(minus(X, Y))
, p(s(X)) -> X
, div(0(), s(Y)) -> 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(p) = {1}, Uargs(div) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[minus](x1, x2) = [1] x1 + [0]
[0] = [0]
[s](x1) = [1] x1 + [2]
[p](x1) = [1] x1 + [0]
[div](x1, x2) = [1] x1 + [4] x2 + [0]
The order satisfies the following ordering constraints:
[minus(X, 0())] = [1] X + [0]
>= [1] X + [0]
= [X]
[minus(s(X), s(Y))] = [1] X + [2]
> [1] X + [0]
= [p(minus(X, Y))]
[p(s(X))] = [1] X + [2]
> [1] X + [0]
= [X]
[div(0(), s(Y))] = [4] Y + [8]
> [0]
= [0()]
[div(s(X), s(Y))] = [1] X + [4] Y + [10]
>= [1] X + [4] Y + [10]
= [s(div(minus(X, Y), s(Y)))]
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:
{ minus(X, 0()) -> X
, div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y))) }
Weak Trs:
{ minus(s(X), s(Y)) -> p(minus(X, Y))
, p(s(X)) -> X
, div(0(), s(Y)) -> 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: { div(s(X), s(Y)) -> s(div(minus(X, Y), 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}, Uargs(p) = {1}, Uargs(div) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[minus](x1, x2) = [1] x1 + [0]
[0] = [0]
[s](x1) = [1] x1 + [1]
[p](x1) = [1] x1 + [0]
[div](x1, x2) = [3] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[minus(X, 0())] = [1] X + [0]
>= [1] X + [0]
= [X]
[minus(s(X), s(Y))] = [1] X + [1]
> [1] X + [0]
= [p(minus(X, Y))]
[p(s(X))] = [1] X + [1]
> [1] X + [0]
= [X]
[div(0(), s(Y))] = [1] Y + [1]
> [0]
= [0()]
[div(s(X), s(Y))] = [3] X + [1] Y + [4]
> [3] X + [1] Y + [2]
= [s(div(minus(X, Y), s(Y)))]
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: { minus(X, 0()) -> X }
Weak Trs:
{ minus(s(X), s(Y)) -> p(minus(X, Y))
, p(s(X)) -> X
, div(0(), s(Y)) -> 0()
, div(s(X), s(Y)) -> s(div(minus(X, Y), s(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: { minus(X, 0()) -> 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(p) = {1}, Uargs(div) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[minus](x1, x2) = [1] x1 + [1]
[0] = [0]
[s](x1) = [1] x1 + [4]
[p](x1) = [1] x1 + [0]
[div](x1, x2) = [2] x1 + [0]
The order satisfies the following ordering constraints:
[minus(X, 0())] = [1] X + [1]
> [1] X + [0]
= [X]
[minus(s(X), s(Y))] = [1] X + [5]
> [1] X + [1]
= [p(minus(X, Y))]
[p(s(X))] = [1] X + [4]
> [1] X + [0]
= [X]
[div(0(), s(Y))] = [0]
>= [0]
= [0()]
[div(s(X), s(Y))] = [2] X + [8]
> [2] X + [6]
= [s(div(minus(X, Y), s(Y)))]
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:
{ minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))
, p(s(X)) -> X
, div(0(), s(Y)) -> 0()
, div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))