We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, fst(s(X), cons(Y, Z)) ->
cons(Y, n__fst(activate(X), activate(Z)))
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, add(s(X), Y) -> s(n__add(activate(X), Y))
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Arguments of following rules are not normal-forms:
{ fst(s(X), cons(Y, Z)) ->
cons(Y, n__fst(activate(X), activate(Z)))
, add(s(X), Y) -> s(n__add(activate(X), Y)) }
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:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
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(fst) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1},
Uargs(add) = {1, 2}, Uargs(len) = {1}, Uargs(n__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[fst](x1, x2) = [1] x1 + [1] x2 + [0]
[0] = [4]
[nil] = [0]
[s](x1) = [1] x1 + [0]
[cons](x1, x2) = [1] x2 + [0]
[n__fst](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]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[n__add](x1, x2) = [1] x1 + [1] x2 + [0]
[len](x1) = [1] x1 + [0]
[n__len](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[fst(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__fst(X1, X2)]
[fst(0(), Z)] = [1] Z + [4]
> [0]
= [nil()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [fst(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(X)]
[activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(activate(X1), activate(X2))]
[activate(n__len(X))] = [1] X + [0]
>= [1] X + [0]
= [len(activate(X))]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [n__from(X)]
[add(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [4]
> [1] X + [0]
= [X]
[len(X)] = [1] X + [0]
>= [1] X + [0]
= [n__len(X)]
[len(nil())] = [0]
? [4]
= [0()]
[len(cons(X, Z))] = [1] Z + [0]
>= [1] Z + [0]
= [s(n__len(activate(Z)))]
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:
{ fst(X1, X2) -> n__fst(X1, X2)
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, add(X1, X2) -> n__add(X1, X2)
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
Weak Trs:
{ fst(0(), Z) -> nil()
, 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(fst) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1},
Uargs(add) = {1, 2}, Uargs(len) = {1}, Uargs(n__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[fst](x1, x2) = [1] x1 + [1] x2 + [1]
[0] = [4]
[nil] = [0]
[s](x1) = [1] x1 + [0]
[cons](x1, x2) = [1] x2 + [0]
[n__fst](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]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[n__add](x1, x2) = [1] x1 + [1] x2 + [0]
[len](x1) = [1] x1 + [0]
[n__len](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[fst(X1, X2)] = [1] X1 + [1] X2 + [1]
> [1] X1 + [1] X2 + [0]
= [n__fst(X1, X2)]
[fst(0(), Z)] = [1] Z + [5]
> [0]
= [nil()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [1]
= [fst(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(X)]
[activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(activate(X1), activate(X2))]
[activate(n__len(X))] = [1] X + [0]
>= [1] X + [0]
= [len(activate(X))]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [n__from(X)]
[add(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [4]
> [1] X + [0]
= [X]
[len(X)] = [1] X + [0]
>= [1] X + [0]
= [n__len(X)]
[len(nil())] = [0]
? [4]
= [0()]
[len(cons(X, Z))] = [1] Z + [0]
>= [1] Z + [0]
= [s(n__len(activate(Z)))]
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__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, add(X1, X2) -> n__add(X1, X2)
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
Weak Trs:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, 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(fst) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1},
Uargs(add) = {1, 2}, Uargs(len) = {1}, Uargs(n__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[fst](x1, x2) = [1] x1 + [1] x2 + [0]
[0] = [4]
[nil] = [0]
[s](x1) = [1] x1 + [0]
[cons](x1, x2) = [1] x2 + [0]
[n__fst](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]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[n__add](x1, x2) = [1] x1 + [1] x2 + [0]
[len](x1) = [1] x1 + [1]
[n__len](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[fst(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__fst(X1, X2)]
[fst(0(), Z)] = [1] Z + [4]
> [0]
= [nil()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [fst(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(X)]
[activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(activate(X1), activate(X2))]
[activate(n__len(X))] = [1] X + [0]
? [1] X + [1]
= [len(activate(X))]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [n__from(X)]
[add(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [4]
> [1] X + [0]
= [X]
[len(X)] = [1] X + [1]
> [1] X + [0]
= [n__len(X)]
[len(nil())] = [1]
? [4]
= [0()]
[len(cons(X, Z))] = [1] Z + [1]
> [1] Z + [0]
= [s(n__len(activate(Z)))]
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__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, add(X1, X2) -> n__add(X1, X2)
, len(nil()) -> 0() }
Weak Trs:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, add(0(), X) -> X
, len(X) -> n__len(X)
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
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(fst) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1},
Uargs(add) = {1, 2}, Uargs(len) = {1}, Uargs(n__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[fst](x1, x2) = [1] x1 + [1] x2 + [4]
[0] = [4]
[nil] = [4]
[s](x1) = [1] x1 + [0]
[cons](x1, x2) = [1] x2 + [4]
[n__fst](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]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[n__add](x1, x2) = [1] x1 + [1] x2 + [0]
[len](x1) = [1] x1 + [4]
[n__len](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[fst(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [0]
= [n__fst(X1, X2)]
[fst(0(), Z)] = [1] Z + [8]
> [4]
= [nil()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [4]
= [fst(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(X)]
[activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(activate(X1), activate(X2))]
[activate(n__len(X))] = [1] X + [0]
? [1] X + [4]
= [len(activate(X))]
[from(X)] = [1] X + [0]
? [1] X + [4]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [n__from(X)]
[add(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [4]
> [1] X + [0]
= [X]
[len(X)] = [1] X + [4]
> [1] X + [0]
= [n__len(X)]
[len(nil())] = [8]
> [4]
= [0()]
[len(cons(X, Z))] = [1] Z + [8]
> [1] Z + [0]
= [s(n__len(activate(Z)))]
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__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, add(X1, X2) -> n__add(X1, X2) }
Weak Trs:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, add(0(), X) -> X
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
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(fst) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1},
Uargs(add) = {1, 2}, Uargs(len) = {1}, Uargs(n__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[fst](x1, x2) = [1] x1 + [1] x2 + [6]
[0] = [2]
[nil] = [0]
[s](x1) = [1] x1 + [0]
[cons](x1, x2) = [1] x2 + [4]
[n__fst](x1, x2) = [1] x1 + [1] x2 + [0]
[activate](x1) = [1] x1 + [1]
[from](x1) = [1] x1 + [0]
[n__from](x1) = [1] x1 + [0]
[n__s](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [2]
[n__add](x1, x2) = [1] x1 + [1] x2 + [0]
[len](x1) = [1] x1 + [4]
[n__len](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[fst(X1, X2)] = [1] X1 + [1] X2 + [6]
> [1] X1 + [1] X2 + [0]
= [n__fst(X1, X2)]
[fst(0(), Z)] = [1] Z + [8]
> [0]
= [nil()]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[activate(X)] = [1] X + [1]
> [1] X + [0]
= [X]
[activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [1]
? [1] X1 + [1] X2 + [8]
= [fst(activate(X1), activate(X2))]
[activate(n__from(X))] = [1] X + [1]
>= [1] X + [1]
= [from(activate(X))]
[activate(n__s(X))] = [1] X + [1]
> [1] X + [0]
= [s(X)]
[activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [1]
? [1] X1 + [1] X2 + [4]
= [add(activate(X1), activate(X2))]
[activate(n__len(X))] = [1] X + [1]
? [1] X + [5]
= [len(activate(X))]
[from(X)] = [1] X + [0]
? [1] X + [4]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [n__from(X)]
[add(X1, X2)] = [1] X1 + [1] X2 + [2]
> [1] X1 + [1] X2 + [0]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [4]
> [1] X + [0]
= [X]
[len(X)] = [1] X + [4]
> [1] X + [0]
= [n__len(X)]
[len(nil())] = [4]
> [2]
= [0()]
[len(cons(X, Z))] = [1] Z + [8]
> [1] Z + [1]
= [s(n__len(activate(Z)))]
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(n__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X) }
Weak Trs:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, activate(X) -> X
, activate(n__s(X)) -> s(X)
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
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(fst) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1},
Uargs(add) = {1, 2}, Uargs(len) = {1}, Uargs(n__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[fst](x1, x2) = [1] x1 + [1] x2 + [1]
[0] = [3]
[nil] = [0]
[s](x1) = [1] x1 + [0]
[cons](x1, x2) = [1] x2 + [4]
[n__fst](x1, x2) = [1] x1 + [1] x2 + [0]
[activate](x1) = [1] x1 + [0]
[from](x1) = [1] x1 + [0]
[n__from](x1) = [1] x1 + [4]
[n__s](x1) = [1] x1 + [4]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[n__add](x1, x2) = [1] x1 + [1] x2 + [0]
[len](x1) = [1] x1 + [4]
[n__len](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[fst(X1, X2)] = [1] X1 + [1] X2 + [1]
> [1] X1 + [1] X2 + [0]
= [n__fst(X1, X2)]
[fst(0(), Z)] = [1] Z + [4]
> [0]
= [nil()]
[s(X)] = [1] X + [0]
? [1] X + [4]
= [n__s(X)]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [1]
= [fst(activate(X1), activate(X2))]
[activate(n__from(X))] = [1] X + [4]
> [1] X + [0]
= [from(activate(X))]
[activate(n__s(X))] = [1] X + [4]
> [1] X + [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__len(X))] = [1] X + [0]
? [1] X + [4]
= [len(activate(X))]
[from(X)] = [1] X + [0]
? [1] X + [12]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1] X + [0]
? [1] X + [4]
= [n__from(X)]
[add(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__add(X1, X2)]
[add(0(), X)] = [1] X + [3]
> [1] X + [0]
= [X]
[len(X)] = [1] X + [4]
> [1] X + [0]
= [n__len(X)]
[len(nil())] = [4]
> [3]
= [0()]
[len(cons(X, Z))] = [1] Z + [8]
> [1] Z + [0]
= [s(n__len(activate(Z)))]
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(n__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X) }
Weak Trs:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
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(fst) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1},
Uargs(add) = {1, 2}, Uargs(len) = {1}, Uargs(n__len) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[fst](x1, x2) = [1] x1 + [1] x2 + [6]
[0] = [0]
[nil] = [0]
[s](x1) = [1] x1 + [1]
[cons](x1, x2) = [1] x2 + [0]
[n__fst](x1, x2) = [1] x1 + [1] x2 + [0]
[activate](x1) = [1] x1 + [1]
[from](x1) = [1] x1 + [0]
[n__from](x1) = [1] x1 + [0]
[n__s](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[n__add](x1, x2) = [1] x1 + [1] x2 + [0]
[len](x1) = [1] x1 + [4]
[n__len](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[fst(X1, X2)] = [1] X1 + [1] X2 + [6]
> [1] X1 + [1] X2 + [0]
= [n__fst(X1, X2)]
[fst(0(), Z)] = [1] Z + [6]
> [0]
= [nil()]
[s(X)] = [1] X + [1]
> [1] X + [0]
= [n__s(X)]
[activate(X)] = [1] X + [1]
> [1] X + [0]
= [X]
[activate(n__fst(X1, X2))] = [1] X1 + [1] X2 + [1]
? [1] X1 + [1] X2 + [8]
= [fst(activate(X1), activate(X2))]
[activate(n__from(X))] = [1] X + [1]
>= [1] X + [1]
= [from(activate(X))]
[activate(n__s(X))] = [1] X + [1]
>= [1] X + [1]
= [s(X)]
[activate(n__add(X1, X2))] = [1] X1 + [1] X2 + [1]
? [1] X1 + [1] X2 + [2]
= [add(activate(X1), activate(X2))]
[activate(n__len(X))] = [1] X + [1]
? [1] X + [5]
= [len(activate(X))]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1] X + [0]
>= [1] X + [0]
= [n__from(X)]
[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]
[len(X)] = [1] X + [4]
> [1] X + [0]
= [n__len(X)]
[len(nil())] = [4]
> [0]
= [0()]
[len(cons(X, Z))] = [1] Z + [4]
> [1] Z + [2]
= [s(n__len(activate(Z)))]
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__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X) }
Weak Trs:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
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:
{ activate(n__add(X1, X2)) -> add(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(fst) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1},
Uargs(add) = {1, 2}, Uargs(len) = {1}, Uargs(n__len) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[fst](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [0]
[0] = [0]
[0]
[nil] = [0]
[0]
[s](x1) = [1 0] x1 + [0]
[0 0] [0]
[cons](x1, x2) = [0 0] x1 + [1 1] x2 + [0]
[0 1] [0 0] [0]
[n__fst](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [0]
[activate](x1) = [1 1] x1 + [0]
[0 1] [0]
[from](x1) = [1 0] x1 + [0]
[0 1] [0]
[n__from](x1) = [1 0] x1 + [0]
[0 1] [0]
[n__s](x1) = [1 0] x1 + [0]
[0 0] [0]
[add](x1, x2) = [1 0] x1 + [1 0] x2 + [4]
[0 1] [0 1] [1]
[n__add](x1, x2) = [1 0] x1 + [1 0] x2 + [4]
[0 1] [0 1] [1]
[len](x1) = [1 0] x1 + [1]
[0 1] [1]
[n__len](x1) = [1 0] x1 + [0]
[0 1] [1]
The order satisfies the following ordering constraints:
[fst(X1, X2)] = [1 0] X1 + [1 0] X2 + [0]
[0 1] [0 1] [0]
>= [1 0] X1 + [1 0] X2 + [0]
[0 1] [0 1] [0]
= [n__fst(X1, X2)]
[fst(0(), Z)] = [1 0] Z + [0]
[0 1] [0]
>= [0]
[0]
= [nil()]
[s(X)] = [1 0] X + [0]
[0 0] [0]
>= [1 0] X + [0]
[0 0] [0]
= [n__s(X)]
[activate(X)] = [1 1] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__fst(X1, X2))] = [1 1] X1 + [1 1] X2 + [0]
[0 1] [0 1] [0]
>= [1 1] X1 + [1 1] X2 + [0]
[0 1] [0 1] [0]
= [fst(activate(X1), activate(X2))]
[activate(n__from(X))] = [1 1] X + [0]
[0 1] [0]
>= [1 1] X + [0]
[0 1] [0]
= [from(activate(X))]
[activate(n__s(X))] = [1 0] X + [0]
[0 0] [0]
>= [1 0] X + [0]
[0 0] [0]
= [s(X)]
[activate(n__add(X1, X2))] = [1 1] X1 + [1 1] X2 + [5]
[0 1] [0 1] [1]
> [1 1] X1 + [1 1] X2 + [4]
[0 1] [0 1] [1]
= [add(activate(X1), activate(X2))]
[activate(n__len(X))] = [1 1] X + [1]
[0 1] [1]
>= [1 1] X + [1]
[0 1] [1]
= [len(activate(X))]
[from(X)] = [1 0] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1 0] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [n__from(X)]
[add(X1, X2)] = [1 0] X1 + [1 0] X2 + [4]
[0 1] [0 1] [1]
>= [1 0] X1 + [1 0] X2 + [4]
[0 1] [0 1] [1]
= [n__add(X1, X2)]
[add(0(), X)] = [1 0] X + [4]
[0 1] [1]
> [1 0] X + [0]
[0 1] [0]
= [X]
[len(X)] = [1 0] X + [1]
[0 1] [1]
> [1 0] X + [0]
[0 1] [1]
= [n__len(X)]
[len(nil())] = [1]
[1]
> [0]
[0]
= [0()]
[len(cons(X, Z))] = [1 1] Z + [0 0] X + [1]
[0 0] [0 1] [1]
> [1 1] Z + [0]
[0 0] [0]
= [s(n__len(activate(Z)))]
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__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X) }
Weak Trs:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
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:
{ activate(n__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__len(X)) -> len(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(fst) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1},
Uargs(add) = {1, 2}, Uargs(len) = {1}, Uargs(n__len) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[fst](x1, x2) = [1 0] x1 + [1 0] x2 + [5]
[0 1] [0 1] [2]
[0] = [4]
[0]
[nil] = [6]
[0]
[s](x1) = [1 0] x1 + [1]
[0 0] [0]
[cons](x1, x2) = [1 6] x2 + [0]
[0 0] [0]
[n__fst](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [2]
[activate](x1) = [1 5] x1 + [1]
[1 1] [0]
[from](x1) = [1 0] x1 + [1]
[0 1] [1]
[n__from](x1) = [1 0] x1 + [1]
[0 1] [0]
[n__s](x1) = [1 0] x1 + [0]
[0 0] [0]
[add](x1, x2) = [1 0] x1 + [1 0] x2 + [7]
[0 1] [0 1] [1]
[n__add](x1, x2) = [1 0] x1 + [1 0] x2 + [7]
[0 1] [0 1] [1]
[len](x1) = [1 0] x1 + [2]
[0 1] [1]
[n__len](x1) = [1 0] x1 + [0]
[0 1] [1]
The order satisfies the following ordering constraints:
[fst(X1, X2)] = [1 0] X1 + [1 0] X2 + [5]
[0 1] [0 1] [2]
> [1 0] X1 + [1 0] X2 + [0]
[0 1] [0 1] [2]
= [n__fst(X1, X2)]
[fst(0(), Z)] = [1 0] Z + [9]
[0 1] [2]
> [6]
[0]
= [nil()]
[s(X)] = [1 0] X + [1]
[0 0] [0]
> [1 0] X + [0]
[0 0] [0]
= [n__s(X)]
[activate(X)] = [1 5] X + [1]
[1 1] [0]
> [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__fst(X1, X2))] = [1 5] X1 + [1 5] X2 + [11]
[1 1] [1 1] [2]
> [1 5] X1 + [1 5] X2 + [7]
[1 1] [1 1] [2]
= [fst(activate(X1), activate(X2))]
[activate(n__from(X))] = [1 5] X + [2]
[1 1] [1]
>= [1 5] X + [2]
[1 1] [1]
= [from(activate(X))]
[activate(n__s(X))] = [1 0] X + [1]
[1 0] [0]
>= [1 0] X + [1]
[0 0] [0]
= [s(X)]
[activate(n__add(X1, X2))] = [1 5] X1 + [1 5] X2 + [13]
[1 1] [1 1] [8]
> [1 5] X1 + [1 5] X2 + [9]
[1 1] [1 1] [1]
= [add(activate(X1), activate(X2))]
[activate(n__len(X))] = [1 5] X + [6]
[1 1] [1]
> [1 5] X + [3]
[1 1] [1]
= [len(activate(X))]
[from(X)] = [1 0] X + [1]
[0 1] [1]
>= [1 0] X + [1]
[0 0] [0]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1 0] X + [1]
[0 1] [1]
>= [1 0] X + [1]
[0 1] [0]
= [n__from(X)]
[add(X1, X2)] = [1 0] X1 + [1 0] X2 + [7]
[0 1] [0 1] [1]
>= [1 0] X1 + [1 0] X2 + [7]
[0 1] [0 1] [1]
= [n__add(X1, X2)]
[add(0(), X)] = [1 0] X + [11]
[0 1] [1]
> [1 0] X + [0]
[0 1] [0]
= [X]
[len(X)] = [1 0] X + [2]
[0 1] [1]
> [1 0] X + [0]
[0 1] [1]
= [n__len(X)]
[len(nil())] = [8]
[1]
> [4]
[0]
= [0()]
[len(cons(X, Z))] = [1 6] Z + [2]
[0 0] [1]
>= [1 5] Z + [2]
[0 0] [0]
= [s(n__len(activate(Z)))]
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:
{ from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X) }
Weak Trs:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
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: { from(X) -> n__from(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(fst) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1},
Uargs(add) = {1, 2}, Uargs(len) = {1}, Uargs(n__len) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[fst](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [0]
[0] = [0]
[4]
[nil] = [0]
[4]
[s](x1) = [1 0] x1 + [0]
[0 0] [0]
[cons](x1, x2) = [1 1] x2 + [0]
[0 0] [0]
[n__fst](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [0]
[activate](x1) = [1 1] x1 + [0]
[0 1] [0]
[from](x1) = [1 0] x1 + [1]
[0 1] [1]
[n__from](x1) = [1 0] x1 + [0]
[0 1] [1]
[n__s](x1) = [1 0] x1 + [0]
[0 0] [0]
[add](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [1]
[n__add](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [1]
[len](x1) = [1 0] x1 + [0]
[0 1] [0]
[n__len](x1) = [1 0] x1 + [0]
[0 1] [0]
The order satisfies the following ordering constraints:
[fst(X1, X2)] = [1 0] X1 + [1 0] X2 + [0]
[0 1] [0 1] [0]
>= [1 0] X1 + [1 0] X2 + [0]
[0 1] [0 1] [0]
= [n__fst(X1, X2)]
[fst(0(), Z)] = [1 0] Z + [0]
[0 1] [4]
>= [0]
[4]
= [nil()]
[s(X)] = [1 0] X + [0]
[0 0] [0]
>= [1 0] X + [0]
[0 0] [0]
= [n__s(X)]
[activate(X)] = [1 1] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__fst(X1, X2))] = [1 1] X1 + [1 1] X2 + [0]
[0 1] [0 1] [0]
>= [1 1] X1 + [1 1] X2 + [0]
[0 1] [0 1] [0]
= [fst(activate(X1), activate(X2))]
[activate(n__from(X))] = [1 1] X + [1]
[0 1] [1]
>= [1 1] X + [1]
[0 1] [1]
= [from(activate(X))]
[activate(n__s(X))] = [1 0] X + [0]
[0 0] [0]
>= [1 0] X + [0]
[0 0] [0]
= [s(X)]
[activate(n__add(X1, X2))] = [1 1] X1 + [1 1] X2 + [1]
[0 1] [0 1] [1]
> [1 1] X1 + [1 1] X2 + [0]
[0 1] [0 1] [1]
= [add(activate(X1), activate(X2))]
[activate(n__len(X))] = [1 1] X + [0]
[0 1] [0]
>= [1 1] X + [0]
[0 1] [0]
= [len(activate(X))]
[from(X)] = [1 0] X + [1]
[0 1] [1]
>= [1 0] X + [1]
[0 0] [0]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1 0] X + [1]
[0 1] [1]
> [1 0] X + [0]
[0 1] [1]
= [n__from(X)]
[add(X1, X2)] = [1 0] X1 + [1 0] X2 + [0]
[0 1] [0 1] [1]
>= [1 0] X1 + [1 0] X2 + [0]
[0 1] [0 1] [1]
= [n__add(X1, X2)]
[add(0(), X)] = [1 0] X + [0]
[0 1] [5]
>= [1 0] X + [0]
[0 1] [0]
= [X]
[len(X)] = [1 0] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [n__len(X)]
[len(nil())] = [0]
[4]
>= [0]
[4]
= [0()]
[len(cons(X, Z))] = [1 1] Z + [0]
[0 0] [0]
>= [1 1] Z + [0]
[0 0] [0]
= [s(n__len(activate(Z)))]
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: { from(X) -> cons(X, n__from(n__s(X))) }
Weak Trs:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> n__from(X)
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 3' to
orient following rules strictly.
Trs: { from(X) -> cons(X, n__from(n__s(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(fst) = {1, 2}, Uargs(s) = {1}, Uargs(from) = {1},
Uargs(add) = {1, 2}, Uargs(len) = {1}, Uargs(n__len) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[1 0 0] [1 0 0] [6]
[fst](x1, x2) = [0 1 0] x1 + [0 1 0] x2 + [1]
[0 0 1] [0 0 1] [1]
[3]
[0] = [0]
[0]
[4]
[nil] = [0]
[0]
[1 0 0] [4]
[s](x1) = [0 0 0] x1 + [0]
[0 0 0] [2]
[0 0 0] [1 7 0] [0]
[cons](x1, x2) = [0 1 0] x1 + [0 0 0] x2 + [4]
[0 0 1] [0 0 1] [1]
[1 0 0] [1 0 0] [5]
[n__fst](x1, x2) = [0 1 0] x1 + [0 1 0] x2 + [0]
[0 0 1] [0 0 1] [1]
[1 7 2] [1]
[activate](x1) = [0 7 2] x1 + [1]
[0 0 2] [0]
[1 0 0] [6]
[from](x1) = [0 1 0] x1 + [4]
[0 0 1] [4]
[1 0 0] [3]
[n__from](x1) = [0 1 0] x1 + [0]
[0 0 1] [2]
[1 0 0] [2]
[n__s](x1) = [0 0 0] x1 + [0]
[0 0 0] [1]
[1 0 0] [1 0 0] [6]
[add](x1, x2) = [0 1 2] x1 + [0 1 0] x2 + [2]
[0 0 0] [0 0 1] [6]
[1 0 0] [1 0 0] [0]
[n__add](x1, x2) = [0 1 2] x1 + [0 1 0] x2 + [0]
[0 0 0] [0 0 1] [4]
[1 0 3] [7]
[len](x1) = [0 1 5] x1 + [0]
[0 0 0] [3]
[1 0 0] [4]
[n__len](x1) = [0 1 2] x1 + [0]
[0 0 0] [3]
The order satisfies the following ordering constraints:
[fst(X1, X2)] = [1 0 0] [1 0 0] [6]
[0 1 0] X1 + [0 1 0] X2 + [1]
[0 0 1] [0 0 1] [1]
> [1 0 0] [1 0 0] [5]
[0 1 0] X1 + [0 1 0] X2 + [0]
[0 0 1] [0 0 1] [1]
= [n__fst(X1, X2)]
[fst(0(), Z)] = [1 0 0] [9]
[0 1 0] Z + [1]
[0 0 1] [1]
> [4]
[0]
[0]
= [nil()]
[s(X)] = [1 0 0] [4]
[0 0 0] X + [0]
[0 0 0] [2]
> [1 0 0] [2]
[0 0 0] X + [0]
[0 0 0] [1]
= [n__s(X)]
[activate(X)] = [1 7 2] [1]
[0 7 2] X + [1]
[0 0 2] [0]
> [1 0 0] [0]
[0 1 0] X + [0]
[0 0 1] [0]
= [X]
[activate(n__fst(X1, X2))] = [1 7 2] [1 7 2] [8]
[0 7 2] X1 + [0 7 2] X2 + [3]
[0 0 2] [0 0 2] [2]
>= [1 7 2] [1 7 2] [8]
[0 7 2] X1 + [0 7 2] X2 + [3]
[0 0 2] [0 0 2] [1]
= [fst(activate(X1), activate(X2))]
[activate(n__from(X))] = [1 7 2] [8]
[0 7 2] X + [5]
[0 0 2] [4]
> [1 7 2] [7]
[0 7 2] X + [5]
[0 0 2] [4]
= [from(activate(X))]
[activate(n__s(X))] = [1 0 0] [5]
[0 0 0] X + [3]
[0 0 0] [2]
> [1 0 0] [4]
[0 0 0] X + [0]
[0 0 0] [2]
= [s(X)]
[activate(n__add(X1, X2))] = [1 7 14] [1 7 2] [9]
[0 7 14] X1 + [0 7 2] X2 + [9]
[0 0 0] [0 0 2] [8]
> [1 7 2] [1 7 2] [8]
[0 7 6] X1 + [0 7 2] X2 + [4]
[0 0 0] [0 0 2] [6]
= [add(activate(X1), activate(X2))]
[activate(n__len(X))] = [1 7 14] [11]
[0 7 14] X + [7]
[0 0 0] [6]
> [1 7 8] [8]
[0 7 12] X + [1]
[0 0 0] [3]
= [len(activate(X))]
[from(X)] = [1 0 0] [6]
[0 1 0] X + [4]
[0 0 1] [4]
> [1 0 0] [5]
[0 1 0] X + [4]
[0 0 1] [4]
= [cons(X, n__from(n__s(X)))]
[from(X)] = [1 0 0] [6]
[0 1 0] X + [4]
[0 0 1] [4]
> [1 0 0] [3]
[0 1 0] X + [0]
[0 0 1] [2]
= [n__from(X)]
[add(X1, X2)] = [1 0 0] [1 0 0] [6]
[0 1 2] X1 + [0 1 0] X2 + [2]
[0 0 0] [0 0 1] [6]
> [1 0 0] [1 0 0] [0]
[0 1 2] X1 + [0 1 0] X2 + [0]
[0 0 0] [0 0 1] [4]
= [n__add(X1, X2)]
[add(0(), X)] = [1 0 0] [9]
[0 1 0] X + [2]
[0 0 1] [6]
> [1 0 0] [0]
[0 1 0] X + [0]
[0 0 1] [0]
= [X]
[len(X)] = [1 0 3] [7]
[0 1 5] X + [0]
[0 0 0] [3]
> [1 0 0] [4]
[0 1 2] X + [0]
[0 0 0] [3]
= [n__len(X)]
[len(nil())] = [11]
[0]
[3]
> [3]
[0]
[0]
= [0()]
[len(cons(X, Z))] = [1 7 3] [0 0 3] [10]
[0 0 5] Z + [0 1 5] X + [9]
[0 0 0] [0 0 0] [3]
> [1 7 2] [9]
[0 0 0] Z + [0]
[0 0 0] [2]
= [s(n__len(activate(Z)))]
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:
{ fst(X1, X2) -> n__fst(X1, X2)
, fst(0(), Z) -> nil()
, s(X) -> n__s(X)
, activate(X) -> X
, activate(n__fst(X1, X2)) -> fst(activate(X1), activate(X2))
, activate(n__from(X)) -> from(activate(X))
, activate(n__s(X)) -> s(X)
, activate(n__add(X1, X2)) -> add(activate(X1), activate(X2))
, activate(n__len(X)) -> len(activate(X))
, from(X) -> cons(X, n__from(n__s(X)))
, from(X) -> n__from(X)
, add(X1, X2) -> n__add(X1, X2)
, add(0(), X) -> X
, len(X) -> n__len(X)
, len(nil()) -> 0()
, len(cons(X, Z)) -> s(n__len(activate(Z))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))