We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil()
, first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Arguments of following rules are not normal-forms:
{ first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) }
All above mentioned rules can be savely removed.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, from(X) -> cons(X, n__from(n__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 1' to
orient following rules strictly.
Trs: { first(0(), X) -> nil() }
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(first) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[first](x1, x2) = [1] x1 + [1] x2 + [0]
[0] = [1]
[nil] = [0]
[s](x1) = [1] x1 + [0]
[cons](x1, x2) = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [0]
[activate](x1) = [1] x1 + [0]
[from](x1) = [1] x1 + [0]
[n__from](x1) = [1] x1 + [0]
[n__s](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__first(X1, X2)]
[first(0(), X)] = [1] X + [1]
> [0]
= [nil()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [first(activate(X1), activate(X2))]
[activate(n__from(X))] = [1] X + [0]
>= [1] X + [0]
= [from(activate(X))]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[from(X)] = [1] X + [0]
>= [0]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1] X + [0]
>= [1] X + [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(n^1)).
Strict Trs:
{ first(X1, X2) -> n__first(X1, X2)
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X) }
Weak Trs: { first(0(), X) -> nil() }
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(first) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[first](x1, x2) = [1] x1 + [1] x2 + [1]
[0] = [0]
[nil] = [0]
[s](x1) = [1] x1 + [0]
[cons](x1, x2) = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [0]
[activate](x1) = [1] x1 + [0]
[from](x1) = [1] x1 + [0]
[n__from](x1) = [1] x1 + [0]
[n__s](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[first(X1, X2)] = [1] X1 + [1] X2 + [1]
> [1] X1 + [1] X2 + [0]
= [n__first(X1, X2)]
[first(0(), X)] = [1] X + [1]
> [0]
= [nil()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [1]
= [first(activate(X1), activate(X2))]
[activate(n__from(X))] = [1] X + [0]
>= [1] X + [0]
= [from(activate(X))]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[from(X)] = [1] X + [0]
>= [0]
= [cons(X, n__from(n__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:
{ s(X) -> n__s(X)
, activate(X) -> X
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X) }
Weak Trs:
{ first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil() }
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(first) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[first](x1, x2) = [1] x1 + [1] x2 + [0]
[0] = [0]
[nil] = [0]
[s](x1) = [1] x1 + [0]
[cons](x1, x2) = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [0]
[activate](x1) = [1] x1 + [0]
[from](x1) = [1] x1 + [1]
[n__from](x1) = [1] x1 + [0]
[n__s](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__first(X1, X2)]
[first(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [first(activate(X1), activate(X2))]
[activate(n__from(X))] = [1] X + [0]
? [1] X + [1]
= [from(activate(X))]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[from(X)] = [1] X + [1]
> [0]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1] X + [1]
> [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:
{ s(X) -> n__s(X)
, activate(X) -> X
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X)) }
Weak Trs:
{ first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil()
, from(X) -> cons(X, n__from(n__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(first) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[first](x1, x2) = [1] x1 + [1] x2 + [4]
[0] = [0]
[nil] = [0]
[s](x1) = [1] x1 + [4]
[cons](x1, x2) = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [4]
[activate](x1) = [1] x1 + [4]
[from](x1) = [1] x1 + [4]
[n__from](x1) = [1] x1 + [0]
[n__s](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[first(X1, X2)] = [1] X1 + [1] X2 + [4]
>= [1] X1 + [1] X2 + [4]
= [n__first(X1, X2)]
[first(0(), X)] = [1] X + [4]
> [0]
= [nil()]
[s(X)] = [1] X + [4]
> [1] X + [0]
= [n__s(X)]
[activate(X)] = [1] X + [4]
> [1] X + [0]
= [X]
[activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [8]
? [1] X1 + [1] X2 + [12]
= [first(activate(X1), activate(X2))]
[activate(n__from(X))] = [1] X + [4]
? [1] X + [8]
= [from(activate(X))]
[activate(n__s(X))] = [1] X + [4]
? [1] X + [8]
= [s(activate(X))]
[from(X)] = [1] X + [4]
> [0]
= [cons(X, n__from(n__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:
{ activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X)) }
Weak Trs:
{ first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil()
, s(X) -> n__s(X)
, activate(X) -> X
, from(X) -> cons(X, n__from(n__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 1' to
orient following rules strictly.
Trs:
{ activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(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(first) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[first](x1, x2) = [1] x1 + [1] x2 + [1]
[0] = [0]
[nil] = [0]
[s](x1) = [1] x1 + [0]
[cons](x1, x2) = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [1]
[activate](x1) = [4] x1 + [1]
[from](x1) = [1] x1 + [7]
[n__from](x1) = [1] x1 + [2]
[n__s](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[first(X1, X2)] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [n__first(X1, X2)]
[first(0(), X)] = [1] X + [1]
> [0]
= [nil()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[activate(X)] = [4] X + [1]
> [1] X + [0]
= [X]
[activate(n__first(X1, X2))] = [4] X1 + [4] X2 + [5]
> [4] X1 + [4] X2 + [3]
= [first(activate(X1), activate(X2))]
[activate(n__from(X))] = [4] X + [9]
> [4] X + [8]
= [from(activate(X))]
[activate(n__s(X))] = [4] X + [1]
>= [4] X + [1]
= [s(activate(X))]
[from(X)] = [1] X + [7]
> [0]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1] X + [7]
> [1] X + [2]
= [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(n^1)).
Strict Trs: { activate(n__s(X)) -> s(activate(X)) }
Weak Trs:
{ first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, from(X) -> cons(X, n__from(n__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 1' to
orient following rules strictly.
Trs: { activate(n__s(X)) -> s(activate(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(first) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[first](x1, x2) = [1] x1 + [1] x2 + [0]
[0] = [0]
[nil] = [0]
[s](x1) = [1] x1 + [3]
[cons](x1, x2) = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [0]
[activate](x1) = [2] x1 + [0]
[from](x1) = [1] x1 + [0]
[n__from](x1) = [1] x1 + [0]
[n__s](x1) = [1] x1 + [2]
The order satisfies the following ordering constraints:
[first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__first(X1, X2)]
[first(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
[s(X)] = [1] X + [3]
> [1] X + [2]
= [n__s(X)]
[activate(X)] = [2] X + [0]
>= [1] X + [0]
= [X]
[activate(n__first(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [first(activate(X1), activate(X2))]
[activate(n__from(X))] = [2] X + [0]
>= [2] X + [0]
= [from(activate(X))]
[activate(n__s(X))] = [2] X + [4]
> [2] X + [3]
= [s(activate(X))]
[from(X)] = [1] X + [0]
>= [0]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1] X + [0]
>= [1] X + [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:
{ first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(activate(X))
, from(X) -> cons(X, n__from(n__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))