We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ minus(X1, X2) -> n__minus(X1, X2)
, minus(n__0(), Y) -> 0()
, minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y))
, 0() -> n__0()
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__s(X)) -> s(activate(X))
, activate(n__div(X1, X2)) -> div(activate(X1), X2)
, activate(n__minus(X1, X2)) -> minus(X1, X2)
, geq(X, n__0()) -> true()
, geq(n__0(), n__s(Y)) -> false()
, geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y))
, div(X1, X2) -> n__div(X1, X2)
, div(0(), n__s(Y)) -> 0()
, div(s(X), n__s(Y)) ->
if(geq(X, activate(Y)),
n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))),
n__0())
, s(X) -> n__s(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
Arguments of following rules are not normal-forms:
{ div(0(), n__s(Y)) -> 0()
, div(s(X), n__s(Y)) ->
if(geq(X, activate(Y)),
n__s(n__div(n__minus(X, activate(Y)), n__s(activate(Y)))),
n__0()) }
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^2)).
Strict Trs:
{ minus(X1, X2) -> n__minus(X1, X2)
, minus(n__0(), Y) -> 0()
, minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y))
, 0() -> n__0()
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__s(X)) -> s(activate(X))
, activate(n__div(X1, X2)) -> div(activate(X1), X2)
, activate(n__minus(X1, X2)) -> minus(X1, X2)
, geq(X, n__0()) -> true()
, geq(n__0(), n__s(Y)) -> false()
, geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y))
, div(X1, X2) -> n__div(X1, X2)
, s(X) -> n__s(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(minus) = {1, 2}, Uargs(geq) = {1, 2}, Uargs(div) = {1},
Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[minus](x1, x2) = [1] x1 + [1] x2 + [0]
[n__0] = [0]
[0] = [0]
[n__s](x1) = [1] x1 + [0]
[activate](x1) = [1] x1 + [0]
[geq](x1, x2) = [1] x1 + [1] x2 + [0]
[true] = [4]
[false] = [0]
[div](x1, x2) = [1] x1 + [0]
[s](x1) = [1] x1 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1]
[n__div](x1, x2) = [1] x1 + [0]
[n__minus](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[minus(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [n__minus(X1, X2)]
[minus(n__0(), Y)] = [1] Y + [0]
>= [0]
= [0()]
[minus(n__s(X), n__s(Y))] = [1] Y + [1] X + [0]
>= [1] Y + [1] X + [0]
= [minus(activate(X), activate(Y))]
[0()] = [0]
>= [0]
= [n__0()]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__0())] = [0]
>= [0]
= [0()]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[activate(n__div(X1, X2))] = [1] X1 + [0]
>= [1] X1 + [0]
= [div(activate(X1), X2)]
[activate(n__minus(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [minus(X1, X2)]
[geq(X, n__0())] = [1] X + [0]
? [4]
= [true()]
[geq(n__0(), n__s(Y))] = [1] Y + [0]
>= [0]
= [false()]
[geq(n__s(X), n__s(Y))] = [1] Y + [1] X + [0]
>= [1] Y + [1] X + [0]
= [geq(activate(X), activate(Y))]
[div(X1, X2)] = [1] X1 + [0]
>= [1] X1 + [0]
= [n__div(X1, X2)]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[if(true(), X, Y)] = [1] Y + [1] X + [5]
> [1] X + [0]
= [activate(X)]
[if(false(), X, Y)] = [1] Y + [1] X + [1]
> [1] Y + [0]
= [activate(Y)]
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^2)).
Strict Trs:
{ minus(X1, X2) -> n__minus(X1, X2)
, minus(n__0(), Y) -> 0()
, minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y))
, 0() -> n__0()
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__s(X)) -> s(activate(X))
, activate(n__div(X1, X2)) -> div(activate(X1), X2)
, activate(n__minus(X1, X2)) -> minus(X1, X2)
, geq(X, n__0()) -> true()
, geq(n__0(), n__s(Y)) -> false()
, geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y))
, div(X1, X2) -> n__div(X1, X2)
, s(X) -> n__s(X) }
Weak Trs:
{ if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(minus) = {1, 2}, Uargs(geq) = {1, 2}, Uargs(div) = {1},
Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[minus](x1, x2) = [1] x1 + [1] x2 + [4]
[n__0] = [0]
[0] = [0]
[n__s](x1) = [1] x1 + [0]
[activate](x1) = [1] x1 + [0]
[geq](x1, x2) = [1] x1 + [1] x2 + [0]
[true] = [0]
[false] = [0]
[div](x1, x2) = [1] x1 + [0]
[s](x1) = [1] x1 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [5]
[n__div](x1, x2) = [1] x1 + [0]
[n__minus](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[minus(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [0]
= [n__minus(X1, X2)]
[minus(n__0(), Y)] = [1] Y + [4]
> [0]
= [0()]
[minus(n__s(X), n__s(Y))] = [1] Y + [1] X + [4]
>= [1] Y + [1] X + [4]
= [minus(activate(X), activate(Y))]
[0()] = [0]
>= [0]
= [n__0()]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__0())] = [0]
>= [0]
= [0()]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[activate(n__div(X1, X2))] = [1] X1 + [0]
>= [1] X1 + [0]
= [div(activate(X1), X2)]
[activate(n__minus(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [4]
= [minus(X1, X2)]
[geq(X, n__0())] = [1] X + [0]
>= [0]
= [true()]
[geq(n__0(), n__s(Y))] = [1] Y + [0]
>= [0]
= [false()]
[geq(n__s(X), n__s(Y))] = [1] Y + [1] X + [0]
>= [1] Y + [1] X + [0]
= [geq(activate(X), activate(Y))]
[div(X1, X2)] = [1] X1 + [0]
>= [1] X1 + [0]
= [n__div(X1, X2)]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[if(true(), X, Y)] = [1] Y + [1] X + [5]
> [1] X + [0]
= [activate(X)]
[if(false(), X, Y)] = [1] Y + [1] X + [5]
> [1] Y + [0]
= [activate(Y)]
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^2)).
Strict Trs:
{ minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y))
, 0() -> n__0()
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__s(X)) -> s(activate(X))
, activate(n__div(X1, X2)) -> div(activate(X1), X2)
, activate(n__minus(X1, X2)) -> minus(X1, X2)
, geq(X, n__0()) -> true()
, geq(n__0(), n__s(Y)) -> false()
, geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y))
, div(X1, X2) -> n__div(X1, X2)
, s(X) -> n__s(X) }
Weak Trs:
{ minus(X1, X2) -> n__minus(X1, X2)
, minus(n__0(), Y) -> 0()
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(minus) = {1, 2}, Uargs(geq) = {1, 2}, Uargs(div) = {1},
Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[minus](x1, x2) = [1] x1 + [1] x2 + [4]
[n__0] = [0]
[0] = [2]
[n__s](x1) = [1] x1 + [0]
[activate](x1) = [1] x1 + [0]
[geq](x1, x2) = [1] x1 + [1] x2 + [0]
[true] = [0]
[false] = [0]
[div](x1, x2) = [1] x1 + [0]
[s](x1) = [1] x1 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [5]
[n__div](x1, x2) = [1] x1 + [0]
[n__minus](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[minus(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [0]
= [n__minus(X1, X2)]
[minus(n__0(), Y)] = [1] Y + [4]
> [2]
= [0()]
[minus(n__s(X), n__s(Y))] = [1] Y + [1] X + [4]
>= [1] Y + [1] X + [4]
= [minus(activate(X), activate(Y))]
[0()] = [2]
> [0]
= [n__0()]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__0())] = [0]
? [2]
= [0()]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[activate(n__div(X1, X2))] = [1] X1 + [0]
>= [1] X1 + [0]
= [div(activate(X1), X2)]
[activate(n__minus(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [4]
= [minus(X1, X2)]
[geq(X, n__0())] = [1] X + [0]
>= [0]
= [true()]
[geq(n__0(), n__s(Y))] = [1] Y + [0]
>= [0]
= [false()]
[geq(n__s(X), n__s(Y))] = [1] Y + [1] X + [0]
>= [1] Y + [1] X + [0]
= [geq(activate(X), activate(Y))]
[div(X1, X2)] = [1] X1 + [0]
>= [1] X1 + [0]
= [n__div(X1, X2)]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[if(true(), X, Y)] = [1] Y + [1] X + [5]
> [1] X + [0]
= [activate(X)]
[if(false(), X, Y)] = [1] Y + [1] X + [5]
> [1] Y + [0]
= [activate(Y)]
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^2)).
Strict Trs:
{ minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y))
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__s(X)) -> s(activate(X))
, activate(n__div(X1, X2)) -> div(activate(X1), X2)
, activate(n__minus(X1, X2)) -> minus(X1, X2)
, geq(X, n__0()) -> true()
, geq(n__0(), n__s(Y)) -> false()
, geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y))
, div(X1, X2) -> n__div(X1, X2)
, s(X) -> n__s(X) }
Weak Trs:
{ minus(X1, X2) -> n__minus(X1, X2)
, minus(n__0(), Y) -> 0()
, 0() -> n__0()
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(minus) = {1, 2}, Uargs(geq) = {1, 2}, Uargs(div) = {1},
Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[minus](x1, x2) = [1] x1 + [1] x2 + [4]
[n__0] = [0]
[0] = [4]
[n__s](x1) = [1] x1 + [0]
[activate](x1) = [1] x1 + [0]
[geq](x1, x2) = [1] x1 + [1] x2 + [0]
[true] = [0]
[false] = [0]
[div](x1, x2) = [1] x1 + [0]
[s](x1) = [1] x1 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [5]
[n__div](x1, x2) = [1] x1 + [1]
[n__minus](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[minus(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [0]
= [n__minus(X1, X2)]
[minus(n__0(), Y)] = [1] Y + [4]
>= [4]
= [0()]
[minus(n__s(X), n__s(Y))] = [1] Y + [1] X + [4]
>= [1] Y + [1] X + [4]
= [minus(activate(X), activate(Y))]
[0()] = [4]
> [0]
= [n__0()]
[activate(X)] = [1] X + [0]
>= [1] X + [0]
= [X]
[activate(n__0())] = [0]
? [4]
= [0()]
[activate(n__s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(activate(X))]
[activate(n__div(X1, X2))] = [1] X1 + [1]
> [1] X1 + [0]
= [div(activate(X1), X2)]
[activate(n__minus(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [4]
= [minus(X1, X2)]
[geq(X, n__0())] = [1] X + [0]
>= [0]
= [true()]
[geq(n__0(), n__s(Y))] = [1] Y + [0]
>= [0]
= [false()]
[geq(n__s(X), n__s(Y))] = [1] Y + [1] X + [0]
>= [1] Y + [1] X + [0]
= [geq(activate(X), activate(Y))]
[div(X1, X2)] = [1] X1 + [0]
? [1] X1 + [1]
= [n__div(X1, X2)]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[if(true(), X, Y)] = [1] Y + [1] X + [5]
> [1] X + [0]
= [activate(X)]
[if(false(), X, Y)] = [1] Y + [1] X + [5]
> [1] Y + [0]
= [activate(Y)]
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^2)).
Strict Trs:
{ minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y))
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__s(X)) -> s(activate(X))
, activate(n__minus(X1, X2)) -> minus(X1, X2)
, geq(X, n__0()) -> true()
, geq(n__0(), n__s(Y)) -> false()
, geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y))
, div(X1, X2) -> n__div(X1, X2)
, s(X) -> n__s(X) }
Weak Trs:
{ minus(X1, X2) -> n__minus(X1, X2)
, minus(n__0(), Y) -> 0()
, 0() -> n__0()
, activate(n__div(X1, X2)) -> div(activate(X1), X2)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(minus) = {1, 2}, Uargs(geq) = {1, 2}, Uargs(div) = {1},
Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[minus](x1, x2) = [1] x1 + [1] x2 + [4]
[n__0] = [2]
[0] = [4]
[n__s](x1) = [1] x1 + [2]
[activate](x1) = [1] x1 + [2]
[geq](x1, x2) = [1] x1 + [1] x2 + [0]
[true] = [0]
[false] = [0]
[div](x1, x2) = [1] x1 + [6]
[s](x1) = [1] x1 + [6]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [7]
[n__div](x1, x2) = [1] x1 + [7]
[n__minus](x1, x2) = [1] x1 + [1] x2 + [2]
The order satisfies the following ordering constraints:
[minus(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [2]
= [n__minus(X1, X2)]
[minus(n__0(), Y)] = [1] Y + [6]
> [4]
= [0()]
[minus(n__s(X), n__s(Y))] = [1] Y + [1] X + [8]
>= [1] Y + [1] X + [8]
= [minus(activate(X), activate(Y))]
[0()] = [4]
> [2]
= [n__0()]
[activate(X)] = [1] X + [2]
> [1] X + [0]
= [X]
[activate(n__0())] = [4]
>= [4]
= [0()]
[activate(n__s(X))] = [1] X + [4]
? [1] X + [8]
= [s(activate(X))]
[activate(n__div(X1, X2))] = [1] X1 + [9]
> [1] X1 + [8]
= [div(activate(X1), X2)]
[activate(n__minus(X1, X2))] = [1] X1 + [1] X2 + [4]
>= [1] X1 + [1] X2 + [4]
= [minus(X1, X2)]
[geq(X, n__0())] = [1] X + [2]
> [0]
= [true()]
[geq(n__0(), n__s(Y))] = [1] Y + [4]
> [0]
= [false()]
[geq(n__s(X), n__s(Y))] = [1] Y + [1] X + [4]
>= [1] Y + [1] X + [4]
= [geq(activate(X), activate(Y))]
[div(X1, X2)] = [1] X1 + [6]
? [1] X1 + [7]
= [n__div(X1, X2)]
[s(X)] = [1] X + [6]
> [1] X + [2]
= [n__s(X)]
[if(true(), X, Y)] = [1] Y + [1] X + [7]
> [1] X + [2]
= [activate(X)]
[if(false(), X, Y)] = [1] Y + [1] X + [7]
> [1] Y + [2]
= [activate(Y)]
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^2)).
Strict Trs:
{ minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y))
, activate(n__0()) -> 0()
, activate(n__s(X)) -> s(activate(X))
, activate(n__minus(X1, X2)) -> minus(X1, X2)
, geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y))
, div(X1, X2) -> n__div(X1, X2) }
Weak Trs:
{ minus(X1, X2) -> n__minus(X1, X2)
, minus(n__0(), Y) -> 0()
, 0() -> n__0()
, activate(X) -> X
, activate(n__div(X1, X2)) -> div(activate(X1), X2)
, geq(X, n__0()) -> true()
, geq(n__0(), n__s(Y)) -> false()
, s(X) -> n__s(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(minus) = {1, 2}, Uargs(geq) = {1, 2}, Uargs(div) = {1},
Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[minus](x1, x2) = [1] x1 + [1] x2 + [4]
[n__0] = [6]
[0] = [6]
[n__s](x1) = [1] x1 + [0]
[activate](x1) = [1] x1 + [3]
[geq](x1, x2) = [1] x1 + [1] x2 + [6]
[true] = [0]
[false] = [2]
[div](x1, x2) = [1] x1 + [0]
[s](x1) = [1] x1 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [7]
[n__div](x1, x2) = [1] x1 + [0]
[n__minus](x1, x2) = [1] x1 + [1] x2 + [2]
The order satisfies the following ordering constraints:
[minus(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [2]
= [n__minus(X1, X2)]
[minus(n__0(), Y)] = [1] Y + [10]
> [6]
= [0()]
[minus(n__s(X), n__s(Y))] = [1] Y + [1] X + [4]
? [1] Y + [1] X + [10]
= [minus(activate(X), activate(Y))]
[0()] = [6]
>= [6]
= [n__0()]
[activate(X)] = [1] X + [3]
> [1] X + [0]
= [X]
[activate(n__0())] = [9]
> [6]
= [0()]
[activate(n__s(X))] = [1] X + [3]
>= [1] X + [3]
= [s(activate(X))]
[activate(n__div(X1, X2))] = [1] X1 + [3]
>= [1] X1 + [3]
= [div(activate(X1), X2)]
[activate(n__minus(X1, X2))] = [1] X1 + [1] X2 + [5]
> [1] X1 + [1] X2 + [4]
= [minus(X1, X2)]
[geq(X, n__0())] = [1] X + [12]
> [0]
= [true()]
[geq(n__0(), n__s(Y))] = [1] Y + [12]
> [2]
= [false()]
[geq(n__s(X), n__s(Y))] = [1] Y + [1] X + [6]
? [1] Y + [1] X + [12]
= [geq(activate(X), activate(Y))]
[div(X1, X2)] = [1] X1 + [0]
>= [1] X1 + [0]
= [n__div(X1, X2)]
[s(X)] = [1] X + [0]
>= [1] X + [0]
= [n__s(X)]
[if(true(), X, Y)] = [1] Y + [1] X + [7]
> [1] X + [3]
= [activate(X)]
[if(false(), X, Y)] = [1] Y + [1] X + [9]
> [1] Y + [3]
= [activate(Y)]
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^2)).
Strict Trs:
{ minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y))
, activate(n__s(X)) -> s(activate(X))
, geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y))
, div(X1, X2) -> n__div(X1, X2) }
Weak Trs:
{ minus(X1, X2) -> n__minus(X1, X2)
, minus(n__0(), Y) -> 0()
, 0() -> n__0()
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__div(X1, X2)) -> div(activate(X1), X2)
, activate(n__minus(X1, X2)) -> minus(X1, X2)
, geq(X, n__0()) -> true()
, geq(n__0(), n__s(Y)) -> false()
, s(X) -> n__s(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(minus) = {1, 2}, Uargs(geq) = {1, 2}, Uargs(div) = {1},
Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[minus](x1, x2) = [1] x1 + [1] x2 + [4]
[n__0] = [3]
[0] = [4]
[n__s](x1) = [1] x1 + [2]
[activate](x1) = [1] x1 + [1]
[geq](x1, x2) = [1] x1 + [1] x2 + [6]
[true] = [0]
[false] = [0]
[div](x1, x2) = [1] x1 + [0]
[s](x1) = [1] x1 + [2]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [4]
[n__div](x1, x2) = [1] x1 + [0]
[n__minus](x1, x2) = [1] x1 + [1] x2 + [4]
The order satisfies the following ordering constraints:
[minus(X1, X2)] = [1] X1 + [1] X2 + [4]
>= [1] X1 + [1] X2 + [4]
= [n__minus(X1, X2)]
[minus(n__0(), Y)] = [1] Y + [7]
> [4]
= [0()]
[minus(n__s(X), n__s(Y))] = [1] Y + [1] X + [8]
> [1] Y + [1] X + [6]
= [minus(activate(X), activate(Y))]
[0()] = [4]
> [3]
= [n__0()]
[activate(X)] = [1] X + [1]
> [1] X + [0]
= [X]
[activate(n__0())] = [4]
>= [4]
= [0()]
[activate(n__s(X))] = [1] X + [3]
>= [1] X + [3]
= [s(activate(X))]
[activate(n__div(X1, X2))] = [1] X1 + [1]
>= [1] X1 + [1]
= [div(activate(X1), X2)]
[activate(n__minus(X1, X2))] = [1] X1 + [1] X2 + [5]
> [1] X1 + [1] X2 + [4]
= [minus(X1, X2)]
[geq(X, n__0())] = [1] X + [9]
> [0]
= [true()]
[geq(n__0(), n__s(Y))] = [1] Y + [11]
> [0]
= [false()]
[geq(n__s(X), n__s(Y))] = [1] Y + [1] X + [10]
> [1] Y + [1] X + [8]
= [geq(activate(X), activate(Y))]
[div(X1, X2)] = [1] X1 + [0]
>= [1] X1 + [0]
= [n__div(X1, X2)]
[s(X)] = [1] X + [2]
>= [1] X + [2]
= [n__s(X)]
[if(true(), X, Y)] = [1] Y + [1] X + [4]
> [1] X + [1]
= [activate(X)]
[if(false(), X, Y)] = [1] Y + [1] X + [4]
> [1] Y + [1]
= [activate(Y)]
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^2)).
Strict Trs:
{ activate(n__s(X)) -> s(activate(X))
, div(X1, X2) -> n__div(X1, X2) }
Weak Trs:
{ minus(X1, X2) -> n__minus(X1, X2)
, minus(n__0(), Y) -> 0()
, minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y))
, 0() -> n__0()
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__div(X1, X2)) -> div(activate(X1), X2)
, activate(n__minus(X1, X2)) -> minus(X1, X2)
, geq(X, n__0()) -> true()
, geq(n__0(), n__s(Y)) -> false()
, geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y))
, s(X) -> n__s(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs: { div(X1, X2) -> n__div(X1, X2) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(minus) = {1, 2}, Uargs(geq) = {1, 2}, Uargs(div) = {1},
Uargs(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[minus](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [4]
[n__0] = [0]
[0]
[0] = [0]
[0]
[n__s](x1) = [1 1] x1 + [6]
[0 1] [0]
[activate](x1) = [1 1] x1 + [5]
[0 1] [0]
[geq](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [4]
[true] = [0]
[4]
[false] = [0]
[4]
[div](x1, x2) = [1 0] x1 + [4]
[0 1] [4]
[s](x1) = [1 1] x1 + [6]
[0 1] [0]
[if](x1, x2, x3) = [0 2] x1 + [7 7] x2 + [7 7] x3 + [1]
[0 0] [7 7] [7 7] [5]
[n__div](x1, x2) = [1 0] x1 + [3]
[0 1] [4]
[n__minus](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [4]
The order satisfies the following ordering constraints:
[minus(X1, X2)] = [1 0] X1 + [1 0] X2 + [0]
[0 0] [0 0] [4]
>= [1 0] X1 + [1 0] X2 + [0]
[0 0] [0 0] [4]
= [n__minus(X1, X2)]
[minus(n__0(), Y)] = [1 0] Y + [0]
[0 0] [4]
>= [0]
[0]
= [0()]
[minus(n__s(X), n__s(Y))] = [1 1] Y + [1 1] X + [12]
[0 0] [0 0] [4]
> [1 1] Y + [1 1] X + [10]
[0 0] [0 0] [4]
= [minus(activate(X), activate(Y))]
[0()] = [0]
[0]
>= [0]
[0]
= [n__0()]
[activate(X)] = [1 1] X + [5]
[0 1] [0]
> [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__0())] = [5]
[0]
> [0]
[0]
= [0()]
[activate(n__s(X))] = [1 2] X + [11]
[0 1] [0]
>= [1 2] X + [11]
[0 1] [0]
= [s(activate(X))]
[activate(n__div(X1, X2))] = [1 1] X1 + [12]
[0 1] [4]
> [1 1] X1 + [9]
[0 1] [4]
= [div(activate(X1), X2)]
[activate(n__minus(X1, X2))] = [1 0] X1 + [1 0] X2 + [9]
[0 0] [0 0] [4]
> [1 0] X1 + [1 0] X2 + [0]
[0 0] [0 0] [4]
= [minus(X1, X2)]
[geq(X, n__0())] = [1 0] X + [0]
[0 0] [4]
>= [0]
[4]
= [true()]
[geq(n__0(), n__s(Y))] = [1 1] Y + [6]
[0 0] [4]
> [0]
[4]
= [false()]
[geq(n__s(X), n__s(Y))] = [1 1] Y + [1 1] X + [12]
[0 0] [0 0] [4]
> [1 1] Y + [1 1] X + [10]
[0 0] [0 0] [4]
= [geq(activate(X), activate(Y))]
[div(X1, X2)] = [1 0] X1 + [4]
[0 1] [4]
> [1 0] X1 + [3]
[0 1] [4]
= [n__div(X1, X2)]
[s(X)] = [1 1] X + [6]
[0 1] [0]
>= [1 1] X + [6]
[0 1] [0]
= [n__s(X)]
[if(true(), X, Y)] = [7 7] Y + [7 7] X + [9]
[7 7] [7 7] [5]
> [1 1] X + [5]
[0 1] [0]
= [activate(X)]
[if(false(), X, Y)] = [7 7] Y + [7 7] X + [9]
[7 7] [7 7] [5]
> [1 1] Y + [5]
[0 1] [0]
= [activate(Y)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs: { activate(n__s(X)) -> s(activate(X)) }
Weak Trs:
{ minus(X1, X2) -> n__minus(X1, X2)
, minus(n__0(), Y) -> 0()
, minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y))
, 0() -> n__0()
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__div(X1, X2)) -> div(activate(X1), X2)
, activate(n__minus(X1, X2)) -> minus(X1, X2)
, geq(X, n__0()) -> true()
, geq(n__0(), n__s(Y)) -> false()
, geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y))
, div(X1, X2) -> n__div(X1, X2)
, s(X) -> n__s(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 2' 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^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(minus) = {1, 2}, Uargs(geq) = {1, 2}, Uargs(div) = {1},
Uargs(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[minus](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[n__0] = [0]
[0]
[0] = [0]
[0]
[n__s](x1) = [1 1] x1 + [0]
[0 1] [1]
[activate](x1) = [1 1] x1 + [0]
[0 1] [0]
[geq](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[true] = [0]
[0]
[false] = [0]
[0]
[div](x1, x2) = [1 0] x1 + [0]
[0 1] [0]
[s](x1) = [1 1] x1 + [0]
[0 1] [1]
[if](x1, x2, x3) = [7 7] x2 + [7 7] x3 + [0]
[7 7] [7 7] [0]
[n__div](x1, x2) = [1 0] x1 + [0]
[0 1] [0]
[n__minus](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
The order satisfies the following ordering constraints:
[minus(X1, X2)] = [1 0] X1 + [1 0] X2 + [0]
[0 0] [0 0] [0]
>= [1 0] X1 + [1 0] X2 + [0]
[0 0] [0 0] [0]
= [n__minus(X1, X2)]
[minus(n__0(), Y)] = [1 0] Y + [0]
[0 0] [0]
>= [0]
[0]
= [0()]
[minus(n__s(X), n__s(Y))] = [1 1] Y + [1 1] X + [0]
[0 0] [0 0] [0]
>= [1 1] Y + [1 1] X + [0]
[0 0] [0 0] [0]
= [minus(activate(X), activate(Y))]
[0()] = [0]
[0]
>= [0]
[0]
= [n__0()]
[activate(X)] = [1 1] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__0())] = [0]
[0]
>= [0]
[0]
= [0()]
[activate(n__s(X))] = [1 2] X + [1]
[0 1] [1]
> [1 2] X + [0]
[0 1] [1]
= [s(activate(X))]
[activate(n__div(X1, X2))] = [1 1] X1 + [0]
[0 1] [0]
>= [1 1] X1 + [0]
[0 1] [0]
= [div(activate(X1), X2)]
[activate(n__minus(X1, X2))] = [1 0] X1 + [1 0] X2 + [0]
[0 0] [0 0] [0]
>= [1 0] X1 + [1 0] X2 + [0]
[0 0] [0 0] [0]
= [minus(X1, X2)]
[geq(X, n__0())] = [1 0] X + [0]
[0 0] [0]
>= [0]
[0]
= [true()]
[geq(n__0(), n__s(Y))] = [1 1] Y + [0]
[0 0] [0]
>= [0]
[0]
= [false()]
[geq(n__s(X), n__s(Y))] = [1 1] Y + [1 1] X + [0]
[0 0] [0 0] [0]
>= [1 1] Y + [1 1] X + [0]
[0 0] [0 0] [0]
= [geq(activate(X), activate(Y))]
[div(X1, X2)] = [1 0] X1 + [0]
[0 1] [0]
>= [1 0] X1 + [0]
[0 1] [0]
= [n__div(X1, X2)]
[s(X)] = [1 1] X + [0]
[0 1] [1]
>= [1 1] X + [0]
[0 1] [1]
= [n__s(X)]
[if(true(), X, Y)] = [7 7] Y + [7 7] X + [0]
[7 7] [7 7] [0]
>= [1 1] X + [0]
[0 1] [0]
= [activate(X)]
[if(false(), X, Y)] = [7 7] Y + [7 7] X + [0]
[7 7] [7 7] [0]
>= [1 1] Y + [0]
[0 1] [0]
= [activate(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(X1, X2) -> n__minus(X1, X2)
, minus(n__0(), Y) -> 0()
, minus(n__s(X), n__s(Y)) -> minus(activate(X), activate(Y))
, 0() -> n__0()
, activate(X) -> X
, activate(n__0()) -> 0()
, activate(n__s(X)) -> s(activate(X))
, activate(n__div(X1, X2)) -> div(activate(X1), X2)
, activate(n__minus(X1, X2)) -> minus(X1, X2)
, geq(X, n__0()) -> true()
, geq(n__0(), n__s(Y)) -> false()
, geq(n__s(X), n__s(Y)) -> geq(activate(X), activate(Y))
, div(X1, X2) -> n__div(X1, X2)
, s(X) -> n__s(X)
, if(true(), X, Y) -> activate(X)
, if(false(), X, Y) -> activate(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^2))