We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N)))
, terms(X) -> n__terms(X)
, sqr(X) -> n__sqr(X)
, sqr(0()) -> 0()
, sqr(s(X)) -> s(n__add(n__sqr(activate(X)), n__dbl(activate(X))))
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__terms(X)) -> terms(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__sqr(X)) -> sqr(activate(X))
, activate(n__dbl(X)) -> dbl(activate(X))
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, dbl(X) -> n__dbl(X)
, dbl(0()) -> 0()
, dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, add(s(X), Y) -> s(n__add(activate(X), Y))
, first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil()
, first(s(X), cons(Y, Z)) ->
cons(Y, n__first(activate(X), activate(Z))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Arguments of following rules are not normal-forms:
{ sqr(s(X)) -> s(n__add(n__sqr(activate(X)), n__dbl(activate(X))))
, dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
, add(s(X), Y) -> s(n__add(activate(X), Y))
, first(s(X), cons(Y, Z)) ->
cons(Y, n__first(activate(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:
{ terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N)))
, terms(X) -> n__terms(X)
, sqr(X) -> n__sqr(X)
, sqr(0()) -> 0()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__terms(X)) -> terms(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__sqr(X)) -> sqr(activate(X))
, activate(n__dbl(X)) -> dbl(activate(X))
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, dbl(X) -> n__dbl(X)
, dbl(0()) -> 0()
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, 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(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(sqr) = {1}, Uargs(dbl) = {1}, Uargs(add) = {1, 2},
Uargs(first) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[terms](x1) = [1] x1 + [4]
[cons](x1, x2) = [1] x1 + [1]
[recip](x1) = [1] x1 + [1]
[sqr](x1) = [1] x1 + [0]
[n__terms](x1) = [1] x1 + [0]
[n__s](x1) = [0]
[0] = [0]
[s](x1) = [0]
[n__add](x1, x2) = [1] x1 + [1] x2 + [0]
[n__sqr](x1) = [1] x1 + [0]
[activate](x1) = [1] x1 + [0]
[n__dbl](x1) = [1] x1 + [0]
[dbl](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[first](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[terms(N)] = [1] N + [4]
> [1] N + [2]
= [cons(recip(sqr(N)), n__terms(n__s(N)))]
[terms(X)] = [1] X + [4]
> [1] X + [0]
= [n__terms(X)]
[sqr(X)] = [1] X + [0]
>= [1] X + [0]
= [n__sqr(X)]
[sqr(0())] = [0]
>= [0]
= [0()]
[s(X)] = [0]
>= [0]
= [n__s(X)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__terms(X))] = [1] X + [0]
? [1] X + [4]
= [terms(activate(X))]
[activate(n__s(X))] = [0]
>= [0]
= [s(X)]
[activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(activate(X1), activate(X2))]
[activate(n__sqr(X))] = [1] X + [0]
>= [1] X + [0]
= [sqr(activate(X))]
[activate(n__dbl(X))] = [1] X + [0]
>= [1] X + [0]
= [dbl(activate(X))]
[activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [first(activate(X1), activate(X2))]
[dbl(X)] = [1] X + [0]
>= [1] X + [0]
= [n__dbl(X)]
[dbl(0())] = [0]
>= [0]
= [0()]
[add(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[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()]
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:
{ sqr(X) -> n__sqr(X)
, sqr(0()) -> 0()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__terms(X)) -> terms(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__sqr(X)) -> sqr(activate(X))
, activate(n__dbl(X)) -> dbl(activate(X))
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, dbl(X) -> n__dbl(X)
, dbl(0()) -> 0()
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil() }
Weak Trs:
{ terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N)))
, terms(X) -> n__terms(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(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(sqr) = {1}, Uargs(dbl) = {1}, Uargs(add) = {1, 2},
Uargs(first) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[terms](x1) = [1] x1 + [1]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [0]
[sqr](x1) = [1] x1 + [1]
[n__terms](x1) = [1] x1 + [1]
[n__s](x1) = [0]
[0] = [0]
[s](x1) = [2]
[n__add](x1, x2) = [1] x1 + [1] x2 + [0]
[n__sqr](x1) = [1] x1 + [0]
[activate](x1) = [1] x1 + [0]
[n__dbl](x1) = [1] x1 + [0]
[dbl](x1) = [1] x1 + [1]
[add](x1, x2) = [1] x1 + [1] x2 + [1]
[first](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[terms(N)] = [1] N + [1]
>= [1] N + [1]
= [cons(recip(sqr(N)), n__terms(n__s(N)))]
[terms(X)] = [1] X + [1]
>= [1] X + [1]
= [n__terms(X)]
[sqr(X)] = [1] X + [1]
> [1] X + [0]
= [n__sqr(X)]
[sqr(0())] = [1]
> [0]
= [0()]
[s(X)] = [2]
> [0]
= [n__s(X)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__terms(X))] = [1] X + [1]
>= [1] X + [1]
= [terms(activate(X))]
[activate(n__s(X))] = [0]
? [2]
= [s(X)]
[activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [1]
= [add(activate(X1), activate(X2))]
[activate(n__sqr(X))] = [1] X + [0]
? [1] X + [1]
= [sqr(activate(X))]
[activate(n__dbl(X))] = [1] X + [0]
? [1] X + [1]
= [dbl(activate(X))]
[activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [first(activate(X1), activate(X2))]
[dbl(X)] = [1] X + [1]
> [1] X + [0]
= [n__dbl(X)]
[dbl(0())] = [1]
> [0]
= [0()]
[add(X1, X2)] = [1] X1 + [1] X2 + [1]
> [1] X1 + [1] X2 + [0]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [1]
> [1] X + [0]
= [X]
[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()]
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(X) -> X
, activate(n__terms(X)) -> terms(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__sqr(X)) -> sqr(activate(X))
, activate(n__dbl(X)) -> dbl(activate(X))
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil() }
Weak Trs:
{ terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N)))
, terms(X) -> n__terms(X)
, sqr(X) -> n__sqr(X)
, sqr(0()) -> 0()
, s(X) -> n__s(X)
, dbl(X) -> n__dbl(X)
, dbl(0()) -> 0()
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> 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(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(sqr) = {1}, Uargs(dbl) = {1}, Uargs(add) = {1, 2},
Uargs(first) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[terms](x1) = [1] x1 + [7]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [5]
[sqr](x1) = [1] x1 + [2]
[n__terms](x1) = [1] x1 + [0]
[n__s](x1) = [0]
[0] = [4]
[s](x1) = [1]
[n__add](x1, x2) = [1] x1 + [1] x2 + [0]
[n__sqr](x1) = [1] x1 + [0]
[activate](x1) = [1] x1 + [1]
[n__dbl](x1) = [1] x1 + [0]
[dbl](x1) = [1] x1 + [4]
[add](x1, x2) = [1] x1 + [1] x2 + [1]
[first](x1, x2) = [1] x1 + [1] x2 + [1]
[nil] = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[terms(N)] = [1] N + [7]
>= [1] N + [7]
= [cons(recip(sqr(N)), n__terms(n__s(N)))]
[terms(X)] = [1] X + [7]
> [1] X + [0]
= [n__terms(X)]
[sqr(X)] = [1] X + [2]
> [1] X + [0]
= [n__sqr(X)]
[sqr(0())] = [6]
> [4]
= [0()]
[s(X)] = [1]
> [0]
= [n__s(X)]
[activate(X)] = [1] X + [1]
> [1] X + [0]
= [X]
[activate(n__terms(X))] = [1] X + [1]
? [1] X + [8]
= [terms(activate(X))]
[activate(n__s(X))] = [1]
>= [1]
= [s(X)]
[activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [1]
? [1] X1 + [1] X2 + [3]
= [add(activate(X1), activate(X2))]
[activate(n__sqr(X))] = [1] X + [1]
? [1] X + [3]
= [sqr(activate(X))]
[activate(n__dbl(X))] = [1] X + [1]
? [1] X + [5]
= [dbl(activate(X))]
[activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [1]
? [1] X1 + [1] X2 + [3]
= [first(activate(X1), activate(X2))]
[dbl(X)] = [1] X + [4]
> [1] X + [0]
= [n__dbl(X)]
[dbl(0())] = [8]
> [4]
= [0()]
[add(X1, X2)] = [1] X1 + [1] X2 + [1]
> [1] X1 + [1] X2 + [0]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [5]
> [1] X + [0]
= [X]
[first(X1, X2)] = [1] X1 + [1] X2 + [1]
> [1] X1 + [1] X2 + [0]
= [n__first(X1, X2)]
[first(0(), X)] = [1] X + [5]
> [0]
= [nil()]
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__terms(X)) -> terms(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__sqr(X)) -> sqr(activate(X))
, activate(n__dbl(X)) -> dbl(activate(X))
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) }
Weak Trs:
{ terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N)))
, terms(X) -> n__terms(X)
, sqr(X) -> n__sqr(X)
, sqr(0()) -> 0()
, s(X) -> n__s(X)
, activate(X) -> X
, dbl(X) -> n__dbl(X)
, dbl(0()) -> 0()
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, 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(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(sqr) = {1}, Uargs(dbl) = {1}, Uargs(add) = {1, 2},
Uargs(first) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[terms](x1) = [1] x1 + [7]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [1]
[sqr](x1) = [1] x1 + [4]
[n__terms](x1) = [1] x1 + [0]
[n__s](x1) = [0]
[0] = [0]
[s](x1) = [0]
[n__add](x1, x2) = [1] x1 + [1] x2 + [6]
[n__sqr](x1) = [1] x1 + [2]
[activate](x1) = [1] x1 + [1]
[n__dbl](x1) = [1] x1 + [4]
[dbl](x1) = [1] x1 + [4]
[add](x1, x2) = [1] x1 + [1] x2 + [6]
[first](x1, x2) = [1] x1 + [1] x2 + [4]
[nil] = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [4]
The order satisfies the following ordering constraints:
[terms(N)] = [1] N + [7]
> [1] N + [5]
= [cons(recip(sqr(N)), n__terms(n__s(N)))]
[terms(X)] = [1] X + [7]
> [1] X + [0]
= [n__terms(X)]
[sqr(X)] = [1] X + [4]
> [1] X + [2]
= [n__sqr(X)]
[sqr(0())] = [4]
> [0]
= [0()]
[s(X)] = [0]
>= [0]
= [n__s(X)]
[activate(X)] = [1] X + [1]
> [1] X + [0]
= [X]
[activate(n__terms(X))] = [1] X + [1]
? [1] X + [8]
= [terms(activate(X))]
[activate(n__s(X))] = [1]
> [0]
= [s(X)]
[activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [7]
? [1] X1 + [1] X2 + [8]
= [add(activate(X1), activate(X2))]
[activate(n__sqr(X))] = [1] X + [3]
? [1] X + [5]
= [sqr(activate(X))]
[activate(n__dbl(X))] = [1] X + [5]
>= [1] X + [5]
= [dbl(activate(X))]
[activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [5]
? [1] X1 + [1] X2 + [6]
= [first(activate(X1), activate(X2))]
[dbl(X)] = [1] X + [4]
>= [1] X + [4]
= [n__dbl(X)]
[dbl(0())] = [4]
> [0]
= [0()]
[add(X1, X2)] = [1] X1 + [1] X2 + [6]
>= [1] X1 + [1] X2 + [6]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [6]
> [1] X + [0]
= [X]
[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()]
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__terms(X)) -> terms(activate(X))
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__sqr(X)) -> sqr(activate(X))
, activate(n__dbl(X)) -> dbl(activate(X))
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) }
Weak Trs:
{ terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N)))
, terms(X) -> n__terms(X)
, sqr(X) -> n__sqr(X)
, sqr(0()) -> 0()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__s(X)) -> s(X)
, dbl(X) -> n__dbl(X)
, dbl(0()) -> 0()
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil() }
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__terms(X)) -> terms(activate(X))
, activate(n__sqr(X)) -> sqr(activate(X))
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) }
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(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(sqr) = {1}, Uargs(dbl) = {1}, Uargs(add) = {1, 2},
Uargs(first) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[terms](x1) = [1] x1 + [5]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [0]
[sqr](x1) = [1] x1 + [5]
[n__terms](x1) = [1] x1 + [5]
[n__s](x1) = [0]
[0] = [0]
[s](x1) = [1]
[n__add](x1, x2) = [1] x1 + [1] x2 + [3]
[n__sqr](x1) = [1] x1 + [5]
[activate](x1) = [2] x1 + [3]
[n__dbl](x1) = [1] x1 + [0]
[dbl](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [3]
[first](x1, x2) = [1] x1 + [1] x2 + [4]
[nil] = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [4]
The order satisfies the following ordering constraints:
[terms(N)] = [1] N + [5]
>= [1] N + [5]
= [cons(recip(sqr(N)), n__terms(n__s(N)))]
[terms(X)] = [1] X + [5]
>= [1] X + [5]
= [n__terms(X)]
[sqr(X)] = [1] X + [5]
>= [1] X + [5]
= [n__sqr(X)]
[sqr(0())] = [5]
> [0]
= [0()]
[s(X)] = [1]
> [0]
= [n__s(X)]
[activate(X)] = [2] X + [3]
> [1] X + [0]
= [X]
[activate(n__terms(X))] = [2] X + [13]
> [2] X + [8]
= [terms(activate(X))]
[activate(n__s(X))] = [3]
> [1]
= [s(X)]
[activate(n__add(X1, X2))] = [2] X1 + [2] X2 + [9]
>= [2] X1 + [2] X2 + [9]
= [add(activate(X1), activate(X2))]
[activate(n__sqr(X))] = [2] X + [13]
> [2] X + [8]
= [sqr(activate(X))]
[activate(n__dbl(X))] = [2] X + [3]
>= [2] X + [3]
= [dbl(activate(X))]
[activate(n__first(X1, X2))] = [2] X1 + [2] X2 + [11]
> [2] X1 + [2] X2 + [10]
= [first(activate(X1), activate(X2))]
[dbl(X)] = [1] X + [0]
>= [1] X + [0]
= [n__dbl(X)]
[dbl(0())] = [0]
>= [0]
= [0()]
[add(X1, X2)] = [1] X1 + [1] X2 + [3]
>= [1] X1 + [1] X2 + [3]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [3]
> [1] X + [0]
= [X]
[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()]
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__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__dbl(X)) -> dbl(activate(X)) }
Weak Trs:
{ terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N)))
, terms(X) -> n__terms(X)
, sqr(X) -> n__sqr(X)
, sqr(0()) -> 0()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__terms(X)) -> terms(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__sqr(X)) -> sqr(activate(X))
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, dbl(X) -> n__dbl(X)
, dbl(0()) -> 0()
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil() }
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__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__dbl(X)) -> dbl(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(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(sqr) = {1}, Uargs(dbl) = {1}, Uargs(add) = {1, 2},
Uargs(first) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[terms](x1) = [1] x1 + [0]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [0]
[sqr](x1) = [1] x1 + [0]
[n__terms](x1) = [1] x1 + [0]
[n__s](x1) = [0]
[0] = [0]
[s](x1) = [1]
[n__add](x1, x2) = [1] x1 + [1] x2 + [3]
[n__sqr](x1) = [1] x1 + [0]
[activate](x1) = [3] x1 + [1]
[n__dbl](x1) = [1] x1 + [2]
[dbl](x1) = [1] x1 + [5]
[add](x1, x2) = [1] x1 + [1] x2 + [4]
[first](x1, x2) = [1] x1 + [1] x2 + [6]
[nil] = [0]
[n__first](x1, x2) = [1] x1 + [1] x2 + [4]
The order satisfies the following ordering constraints:
[terms(N)] = [1] N + [0]
>= [1] N + [0]
= [cons(recip(sqr(N)), n__terms(n__s(N)))]
[terms(X)] = [1] X + [0]
>= [1] X + [0]
= [n__terms(X)]
[sqr(X)] = [1] X + [0]
>= [1] X + [0]
= [n__sqr(X)]
[sqr(0())] = [0]
>= [0]
= [0()]
[s(X)] = [1]
> [0]
= [n__s(X)]
[activate(X)] = [3] X + [1]
> [1] X + [0]
= [X]
[activate(n__terms(X))] = [3] X + [1]
>= [3] X + [1]
= [terms(activate(X))]
[activate(n__s(X))] = [1]
>= [1]
= [s(X)]
[activate(n__add(X1, X2))] = [3] X1 + [3] X2 + [10]
> [3] X1 + [3] X2 + [6]
= [add(activate(X1), activate(X2))]
[activate(n__sqr(X))] = [3] X + [1]
>= [3] X + [1]
= [sqr(activate(X))]
[activate(n__dbl(X))] = [3] X + [7]
> [3] X + [6]
= [dbl(activate(X))]
[activate(n__first(X1, X2))] = [3] X1 + [3] X2 + [13]
> [3] X1 + [3] X2 + [8]
= [first(activate(X1), activate(X2))]
[dbl(X)] = [1] X + [5]
> [1] X + [2]
= [n__dbl(X)]
[dbl(0())] = [5]
> [0]
= [0()]
[add(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [3]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [4]
> [1] X + [0]
= [X]
[first(X1, X2)] = [1] X1 + [1] X2 + [6]
> [1] X1 + [1] X2 + [4]
= [n__first(X1, X2)]
[first(0(), X)] = [1] X + [6]
> [0]
= [nil()]
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:
{ terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N)))
, terms(X) -> n__terms(X)
, sqr(X) -> n__sqr(X)
, sqr(0()) -> 0()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__terms(X)) -> terms(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__sqr(X)) -> sqr(activate(X))
, activate(n__dbl(X)) -> dbl(activate(X))
, activate(n__first(X1, X2)) -> first(activate(X1), activate(X2))
, dbl(X) -> n__dbl(X)
, dbl(0()) -> 0()
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, first(X1, X2) -> n__first(X1, X2)
, first(0(), X) -> nil() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))