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:
{ 2nd(cons1(X, cons(Y, Z))) -> Y
, 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(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(2nd) = {1}, Uargs(cons1) = {2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[2nd](x1) = [1] x1 + [0]
[cons1](x1, x2) = [1] x1 + [1] x2 + [0]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[activate](x1) = [1] x1 + [0]
[from](x1) = [1] x1 + [5]
[n__from](x1) = [1] x1 + [0]
[s](x1) = [0]
The following symbols are considered usable
{2nd, activate, from}
The order satisfies the following ordering constraints:
[2nd(cons1(X, cons(Y, Z)))] = [1] X + [1] Y + [1] Z + [0]
>= [1] Y + [0]
= [Y]
[2nd(cons(X, X1))] = [1] X + [1] X1 + [0]
>= [1] X + [1] X1 + [0]
= [2nd(cons1(X, activate(X1)))]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__from(X))] = [1] X + [0]
? [1] X + [5]
= [from(X)]
[from(X)] = [1] X + [5]
> [1] X + [0]
= [cons(X, n__from(s(X)))]
[from(X)] = [1] X + [5]
> [1] X + [0]
= [n__from(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:
{ 2nd(cons1(X, cons(Y, Z))) -> Y
, 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
, activate(X) -> X
, activate(n__from(X)) -> from(X) }
Weak Trs:
{ from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(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(2nd) = {1}, Uargs(cons1) = {2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[2nd](x1) = [1] x1 + [0]
[cons1](x1, x2) = [1] x1 + [1] x2 + [0]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[activate](x1) = [1] x1 + [4]
[from](x1) = [1] x1 + [4]
[n__from](x1) = [1] x1 + [0]
[s](x1) = [0]
The following symbols are considered usable
{2nd, activate, from}
The order satisfies the following ordering constraints:
[2nd(cons1(X, cons(Y, Z)))] = [1] X + [1] Y + [1] Z + [0]
>= [1] Y + [0]
= [Y]
[2nd(cons(X, X1))] = [1] X + [1] X1 + [0]
? [1] X + [1] X1 + [4]
= [2nd(cons1(X, activate(X1)))]
[activate(X)] = [1] X + [4]
> [1] X + [0]
= [X]
[activate(n__from(X))] = [1] X + [4]
>= [1] X + [4]
= [from(X)]
[from(X)] = [1] X + [4]
> [1] X + [0]
= [cons(X, n__from(s(X)))]
[from(X)] = [1] X + [4]
> [1] X + [0]
= [n__from(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:
{ 2nd(cons1(X, cons(Y, Z))) -> Y
, 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
, activate(n__from(X)) -> from(X) }
Weak Trs:
{ activate(X) -> X
, from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(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(2nd) = {1}, Uargs(cons1) = {2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[2nd](x1) = [1] x1 + [0]
[cons1](x1, x2) = [1] x1 + [1] x2 + [0]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[activate](x1) = [1] x1 + [1]
[from](x1) = [1] x1 + [0]
[n__from](x1) = [1] x1 + [0]
[s](x1) = [0]
The following symbols are considered usable
{2nd, activate, from}
The order satisfies the following ordering constraints:
[2nd(cons1(X, cons(Y, Z)))] = [1] X + [1] Y + [1] Z + [0]
>= [1] Y + [0]
= [Y]
[2nd(cons(X, X1))] = [1] X + [1] X1 + [0]
? [1] X + [1] X1 + [1]
= [2nd(cons1(X, activate(X1)))]
[activate(X)] = [1] X + [1]
> [1] X + [0]
= [X]
[activate(n__from(X))] = [1] X + [1]
> [1] X + [0]
= [from(X)]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [cons(X, n__from(s(X)))]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [n__from(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:
{ 2nd(cons1(X, cons(Y, Z))) -> Y
, 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) }
Weak Trs:
{ activate(X) -> X
, activate(n__from(X)) -> from(X)
, from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(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(2nd) = {1}, Uargs(cons1) = {2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[2nd](x1) = [1] x1 + [0]
[cons1](x1, x2) = [1] x1 + [1] x2 + [4]
[cons](x1, x2) = [1] x1 + [1] x2 + [4]
[activate](x1) = [1] x1 + [4]
[from](x1) = [1] x1 + [4]
[n__from](x1) = [1] x1 + [0]
[s](x1) = [0]
The following symbols are considered usable
{2nd, activate, from}
The order satisfies the following ordering constraints:
[2nd(cons1(X, cons(Y, Z)))] = [1] X + [1] Y + [1] Z + [8]
> [1] Y + [0]
= [Y]
[2nd(cons(X, X1))] = [1] X + [1] X1 + [4]
? [1] X + [1] X1 + [8]
= [2nd(cons1(X, activate(X1)))]
[activate(X)] = [1] X + [4]
> [1] X + [0]
= [X]
[activate(n__from(X))] = [1] X + [4]
>= [1] X + [4]
= [from(X)]
[from(X)] = [1] X + [4]
>= [1] X + [4]
= [cons(X, n__from(s(X)))]
[from(X)] = [1] X + [4]
> [1] X + [0]
= [n__from(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: { 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) }
Weak Trs:
{ 2nd(cons1(X, cons(Y, Z))) -> Y
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(X) }
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: { 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1))) }
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(2nd) = {1}, Uargs(cons1) = {2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[2nd](x1) = [2 1] x1 + [0]
[1 0] [0]
[cons1](x1, x2) = [1 0] x2 + [0]
[0 0] [1]
[cons](x1, x2) = [1 1] x1 + [1 0] x2 + [0]
[0 0] [0 0] [4]
[activate](x1) = [1 0] x1 + [0]
[0 1] [4]
[from](x1) = [1 1] x1 + [0]
[0 0] [4]
[n__from](x1) = [1 1] x1 + [0]
[0 0] [0]
[s](x1) = [0]
[0]
The following symbols are considered usable
{2nd, activate, from}
The order satisfies the following ordering constraints:
[2nd(cons1(X, cons(Y, Z)))] = [2 2] Y + [2 0] Z + [1]
[1 1] [1 0] [0]
> [1 0] Y + [0]
[0 1] [0]
= [Y]
[2nd(cons(X, X1))] = [2 2] X + [2 0] X1 + [4]
[1 1] [1 0] [0]
> [2 0] X1 + [1]
[1 0] [0]
= [2nd(cons1(X, activate(X1)))]
[activate(X)] = [1 0] X + [0]
[0 1] [4]
>= [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__from(X))] = [1 1] X + [0]
[0 0] [4]
>= [1 1] X + [0]
[0 0] [4]
= [from(X)]
[from(X)] = [1 1] X + [0]
[0 0] [4]
>= [1 1] X + [0]
[0 0] [4]
= [cons(X, n__from(s(X)))]
[from(X)] = [1 1] X + [0]
[0 0] [4]
>= [1 1] X + [0]
[0 0] [0]
= [n__from(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:
{ 2nd(cons1(X, cons(Y, Z))) -> Y
, 2nd(cons(X, X1)) -> 2nd(cons1(X, activate(X1)))
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(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))