We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(s(X)) -> s(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [0]
[0] = [0]
[s](x1) = [1] x1 + [0]
[a__geq](x1, x2) = [0]
[true] = [4]
[false] = [0]
[a__div](x1, x2) = [1] x1 + [0]
[a__if](x1, x2, x3) = [1] x1 + [0]
[div](x1, x2) = [0]
[minus](x1, x2) = [0]
[mark](x1) = [0]
[geq](x1, x2) = [0]
[if](x1, x2, x3) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [0]
>= [0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [0]
>= [0]
= [0()]
[a__minus(s(X), s(Y))] = [0]
>= [0]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [0]
>= [0]
= [geq(X1, X2)]
[a__geq(X, 0())] = [0]
? [4]
= [true()]
[a__geq(0(), s(Y))] = [0]
>= [0]
= [false()]
[a__geq(s(X), s(Y))] = [0]
>= [0]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [0]
>= [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [0]
>= [0]
= [0()]
[a__div(s(X), s(Y))] = [1] X + [0]
>= [0]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [0]
>= [1] X1 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [4]
> [0]
= [mark(X)]
[a__if(false(), X, Y)] = [0]
>= [0]
= [mark(Y)]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(s(X))] = [0]
>= [0]
= [s(mark(X))]
[mark(true())] = [0]
? [4]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(div(X1, X2))] = [0]
>= [0]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [0]
>= [0]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [0]
>= [0]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [0]
>= [0]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(s(X)) -> s(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs: { a__if(true(), X, Y) -> mark(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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [4]
[0] = [0]
[s](x1) = [1] x1 + [0]
[a__geq](x1, x2) = [0]
[true] = [4]
[false] = [0]
[a__div](x1, x2) = [1] x1 + [0]
[a__if](x1, x2, x3) = [1] x1 + [0]
[div](x1, x2) = [0]
[minus](x1, x2) = [0]
[mark](x1) = [0]
[geq](x1, x2) = [0]
[if](x1, x2, x3) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [4]
> [0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [4]
> [0]
= [0()]
[a__minus(s(X), s(Y))] = [4]
>= [4]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [0]
>= [0]
= [geq(X1, X2)]
[a__geq(X, 0())] = [0]
? [4]
= [true()]
[a__geq(0(), s(Y))] = [0]
>= [0]
= [false()]
[a__geq(s(X), s(Y))] = [0]
>= [0]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [0]
>= [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [0]
>= [0]
= [0()]
[a__div(s(X), s(Y))] = [1] X + [0]
>= [0]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [0]
>= [1] X1 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [4]
> [0]
= [mark(X)]
[a__if(false(), X, Y)] = [0]
>= [0]
= [mark(Y)]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(s(X))] = [0]
>= [0]
= [s(mark(X))]
[mark(true())] = [0]
? [4]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(div(X1, X2))] = [0]
>= [0]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [0]
? [4]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [0]
>= [0]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [0]
>= [0]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(s(X)) -> s(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__if(true(), X, Y) -> mark(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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [4]
[0] = [0]
[s](x1) = [1] x1 + [0]
[a__geq](x1, x2) = [4]
[true] = [4]
[false] = [0]
[a__div](x1, x2) = [1] x1 + [0]
[a__if](x1, x2, x3) = [1] x1 + [0]
[div](x1, x2) = [0]
[minus](x1, x2) = [0]
[mark](x1) = [0]
[geq](x1, x2) = [0]
[if](x1, x2, x3) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [4]
> [0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [4]
> [0]
= [0()]
[a__minus(s(X), s(Y))] = [4]
>= [4]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [4]
> [0]
= [geq(X1, X2)]
[a__geq(X, 0())] = [4]
>= [4]
= [true()]
[a__geq(0(), s(Y))] = [4]
> [0]
= [false()]
[a__geq(s(X), s(Y))] = [4]
>= [4]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [0]
>= [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [0]
>= [0]
= [0()]
[a__div(s(X), s(Y))] = [1] X + [0]
? [4]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [0]
>= [1] X1 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [4]
> [0]
= [mark(X)]
[a__if(false(), X, Y)] = [0]
>= [0]
= [mark(Y)]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(s(X))] = [0]
>= [0]
= [s(mark(X))]
[mark(true())] = [0]
? [4]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(div(X1, X2))] = [0]
>= [0]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [0]
? [4]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [0]
? [4]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [0]
>= [0]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(X, 0()) -> true()
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(s(X)) -> s(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(0(), s(Y)) -> false()
, a__if(true(), X, Y) -> mark(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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [4]
[0] = [0]
[s](x1) = [1] x1 + [0]
[a__geq](x1, x2) = [5]
[true] = [4]
[false] = [0]
[a__div](x1, x2) = [1] x1 + [0]
[a__if](x1, x2, x3) = [1] x1 + [0]
[div](x1, x2) = [0]
[minus](x1, x2) = [0]
[mark](x1) = [0]
[geq](x1, x2) = [0]
[if](x1, x2, x3) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [4]
> [0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [4]
> [0]
= [0()]
[a__minus(s(X), s(Y))] = [4]
>= [4]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [5]
> [0]
= [geq(X1, X2)]
[a__geq(X, 0())] = [5]
> [4]
= [true()]
[a__geq(0(), s(Y))] = [5]
> [0]
= [false()]
[a__geq(s(X), s(Y))] = [5]
>= [5]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [0]
>= [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [0]
>= [0]
= [0()]
[a__div(s(X), s(Y))] = [1] X + [0]
? [5]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [0]
>= [1] X1 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [4]
> [0]
= [mark(X)]
[a__if(false(), X, Y)] = [0]
>= [0]
= [mark(Y)]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(s(X))] = [0]
>= [0]
= [s(mark(X))]
[mark(true())] = [0]
? [4]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(div(X1, X2))] = [0]
>= [0]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [0]
? [4]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [0]
? [5]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [0]
>= [0]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(s(X)) -> s(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__if(true(), X, Y) -> mark(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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [4]
[0] = [0]
[s](x1) = [1] x1 + [0]
[a__geq](x1, x2) = [4]
[true] = [0]
[false] = [0]
[a__div](x1, x2) = [1] x1 + [1]
[a__if](x1, x2, x3) = [1] x1 + [4]
[div](x1, x2) = [0]
[minus](x1, x2) = [0]
[mark](x1) = [0]
[geq](x1, x2) = [0]
[if](x1, x2, x3) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [4]
> [0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [4]
> [0]
= [0()]
[a__minus(s(X), s(Y))] = [4]
>= [4]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [4]
> [0]
= [geq(X1, X2)]
[a__geq(X, 0())] = [4]
> [0]
= [true()]
[a__geq(0(), s(Y))] = [4]
> [0]
= [false()]
[a__geq(s(X), s(Y))] = [4]
>= [4]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [1]
> [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [1]
> [0]
= [0()]
[a__div(s(X), s(Y))] = [1] X + [1]
? [8]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [4]
> [1] X1 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [4]
> [0]
= [mark(X)]
[a__if(false(), X, Y)] = [4]
> [0]
= [mark(Y)]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(s(X))] = [0]
>= [0]
= [s(mark(X))]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(div(X1, X2))] = [0]
? [1]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [0]
? [4]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [0]
? [4]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [0]
? [4]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, mark(0()) -> 0()
, mark(s(X)) -> s(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y) }
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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [4]
[0] = [0]
[s](x1) = [1] x1 + [0]
[a__geq](x1, x2) = [1]
[true] = [1]
[false] = [1]
[a__div](x1, x2) = [1] x1 + [4]
[a__if](x1, x2, x3) = [1] x1 + [7]
[div](x1, x2) = [1] x1 + [0]
[minus](x1, x2) = [1]
[mark](x1) = [1]
[geq](x1, x2) = [0]
[if](x1, x2, x3) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [4]
> [1]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [4]
> [0]
= [0()]
[a__minus(s(X), s(Y))] = [4]
>= [4]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [1]
> [0]
= [geq(X1, X2)]
[a__geq(X, 0())] = [1]
>= [1]
= [true()]
[a__geq(0(), s(Y))] = [1]
>= [1]
= [false()]
[a__geq(s(X), s(Y))] = [1]
>= [1]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [4]
> [1] X1 + [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [4]
> [0]
= [0()]
[a__div(s(X), s(Y))] = [1] X + [4]
? [8]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [7]
> [1] X1 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [8]
> [1]
= [mark(X)]
[a__if(false(), X, Y)] = [8]
> [1]
= [mark(Y)]
[mark(0())] = [1]
> [0]
= [0()]
[mark(s(X))] = [1]
>= [1]
= [s(mark(X))]
[mark(true())] = [1]
>= [1]
= [true()]
[mark(false())] = [1]
>= [1]
= [false()]
[mark(div(X1, X2))] = [1]
? [5]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [1]
? [4]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [1]
>= [1]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [1]
? [8]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, mark(s(X)) -> s(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0() }
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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [4]
[0] = [1]
[s](x1) = [1] x1 + [0]
[a__geq](x1, x2) = [4]
[true] = [0]
[false] = [0]
[a__div](x1, x2) = [1] x1 + [0]
[a__if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [4]
[div](x1, x2) = [1] x1 + [0]
[minus](x1, x2) = [0]
[mark](x1) = [1] x1 + [1]
[geq](x1, x2) = [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [4]
> [0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [4]
> [1]
= [0()]
[a__minus(s(X), s(Y))] = [4]
>= [4]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [4]
> [0]
= [geq(X1, X2)]
[a__geq(X, 0())] = [4]
> [0]
= [true()]
[a__geq(0(), s(Y))] = [4]
> [0]
= [false()]
[a__geq(s(X), s(Y))] = [4]
>= [4]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [0]
>= [1] X1 + [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [1]
>= [1]
= [0()]
[a__div(s(X), s(Y))] = [1] X + [0]
? [9]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [1] X2 + [1] X3 + [4]
> [1] X1 + [1] X2 + [1] X3 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [1] Y + [1] X + [4]
> [1] X + [1]
= [mark(X)]
[a__if(false(), X, Y)] = [1] Y + [1] X + [4]
> [1] Y + [1]
= [mark(Y)]
[mark(0())] = [2]
> [1]
= [0()]
[mark(s(X))] = [1] X + [1]
>= [1] X + [1]
= [s(mark(X))]
[mark(true())] = [1]
> [0]
= [true()]
[mark(false())] = [1]
> [0]
= [false()]
[mark(div(X1, X2))] = [1] X1 + [1]
>= [1] X1 + [1]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [1]
? [4]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [1]
? [4]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [1] X1 + [1] X2 + [1] X3 + [1]
? [1] X1 + [1] X2 + [1] X3 + [5]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, mark(s(X)) -> s(mark(X))
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(false()) -> false() }
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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [0]
[0] = [0]
[s](x1) = [1] x1 + [0]
[a__geq](x1, x2) = [1] x1 + [0]
[true] = [0]
[false] = [0]
[a__div](x1, x2) = [1] x1 + [4]
[a__if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1]
[div](x1, x2) = [1] x1 + [0]
[minus](x1, x2) = [0]
[mark](x1) = [1] x1 + [0]
[geq](x1, x2) = [1] x1 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [0]
>= [0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [0]
>= [0]
= [0()]
[a__minus(s(X), s(Y))] = [0]
>= [0]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [1] X1 + [0]
>= [1] X1 + [0]
= [geq(X1, X2)]
[a__geq(X, 0())] = [1] X + [0]
>= [0]
= [true()]
[a__geq(0(), s(Y))] = [0]
>= [0]
= [false()]
[a__geq(s(X), s(Y))] = [1] X + [0]
>= [1] X + [0]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [4]
> [1] X1 + [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [4]
> [0]
= [0()]
[a__div(s(X), s(Y))] = [1] X + [4]
> [1] X + [1]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [1] X2 + [1] X3 + [1]
> [1] X1 + [1] X2 + [1] X3 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [1] Y + [1] X + [1]
> [1] X + [0]
= [mark(X)]
[a__if(false(), X, Y)] = [1] Y + [1] X + [1]
> [1] Y + [0]
= [mark(Y)]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(s(X))] = [1] X + [0]
>= [1] X + [0]
= [s(mark(X))]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(div(X1, X2))] = [1] X1 + [0]
? [1] X1 + [4]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [0]
>= [0]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [1] X1 + [0]
>= [1] X1 + [0]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [1] X1 + [1] X2 + [1] X3 + [0]
? [1] X1 + [1] X2 + [1] X3 + [1]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, mark(s(X)) -> s(mark(X))
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(false()) -> false() }
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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [0]
[0] = [0]
[s](x1) = [1] x1 + [3]
[a__geq](x1, x2) = [1]
[true] = [1]
[false] = [1]
[a__div](x1, x2) = [1] x1 + [0]
[a__if](x1, x2, x3) = [1] x1 + [0]
[div](x1, x2) = [1] x1 + [0]
[minus](x1, x2) = [0]
[mark](x1) = [1]
[geq](x1, x2) = [0]
[if](x1, x2, x3) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [0]
>= [0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [0]
>= [0]
= [0()]
[a__minus(s(X), s(Y))] = [0]
>= [0]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [1]
> [0]
= [geq(X1, X2)]
[a__geq(X, 0())] = [1]
>= [1]
= [true()]
[a__geq(0(), s(Y))] = [1]
>= [1]
= [false()]
[a__geq(s(X), s(Y))] = [1]
>= [1]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [0]
>= [1] X1 + [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [0]
>= [0]
= [0()]
[a__div(s(X), s(Y))] = [1] X + [3]
> [1]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [0]
>= [1] X1 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [1]
>= [1]
= [mark(X)]
[a__if(false(), X, Y)] = [1]
>= [1]
= [mark(Y)]
[mark(0())] = [1]
> [0]
= [0()]
[mark(s(X))] = [1]
? [4]
= [s(mark(X))]
[mark(true())] = [1]
>= [1]
= [true()]
[mark(false())] = [1]
>= [1]
= [false()]
[mark(div(X1, X2))] = [1]
>= [1]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [1]
> [0]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [1]
>= [1]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [1]
>= [1]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, mark(s(X)) -> s(mark(X))
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(minus(X1, X2)) -> a__minus(X1, X2) }
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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [0]
[0] = [0]
[s](x1) = [1] x1 + [4]
[a__geq](x1, x2) = [0]
[true] = [0]
[false] = [0]
[a__div](x1, x2) = [1] x1 + [4]
[a__if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [4]
[div](x1, x2) = [1] x1 + [0]
[minus](x1, x2) = [0]
[mark](x1) = [1] x1 + [4]
[geq](x1, x2) = [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [4]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [0]
>= [0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [0]
>= [0]
= [0()]
[a__minus(s(X), s(Y))] = [0]
>= [0]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [0]
>= [0]
= [geq(X1, X2)]
[a__geq(X, 0())] = [0]
>= [0]
= [true()]
[a__geq(0(), s(Y))] = [0]
>= [0]
= [false()]
[a__geq(s(X), s(Y))] = [0]
>= [0]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [4]
> [1] X1 + [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [4]
> [0]
= [0()]
[a__div(s(X), s(Y))] = [1] X + [8]
>= [8]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [1] X2 + [1] X3 + [4]
>= [1] X1 + [1] X2 + [1] X3 + [4]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [1] Y + [1] X + [4]
>= [1] X + [4]
= [mark(X)]
[a__if(false(), X, Y)] = [1] Y + [1] X + [4]
>= [1] Y + [4]
= [mark(Y)]
[mark(0())] = [4]
> [0]
= [0()]
[mark(s(X))] = [1] X + [8]
>= [1] X + [8]
= [s(mark(X))]
[mark(true())] = [4]
> [0]
= [true()]
[mark(false())] = [4]
> [0]
= [false()]
[mark(div(X1, X2))] = [1] X1 + [4]
? [1] X1 + [8]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [4]
> [0]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [4]
> [0]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [1] X1 + [1] X2 + [1] X3 + [8]
>= [1] X1 + [1] X2 + [1] X3 + [8]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, mark(s(X)) -> s(mark(X))
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2) }
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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [0]
[0] = [0]
[s](x1) = [1] x1 + [1]
[a__geq](x1, x2) = [1] x1 + [1]
[true] = [0]
[false] = [1]
[a__div](x1, x2) = [1] x1 + [1] x2 + [1]
[a__if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
[div](x1, x2) = [1] x1 + [1] x2 + [0]
[minus](x1, x2) = [0]
[mark](x1) = [1] x1 + [0]
[geq](x1, x2) = [1] x1 + [1]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [0]
>= [0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [0]
>= [0]
= [0()]
[a__minus(s(X), s(Y))] = [0]
>= [0]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [1] X1 + [1]
>= [1] X1 + [1]
= [geq(X1, X2)]
[a__geq(X, 0())] = [1] X + [1]
> [0]
= [true()]
[a__geq(0(), s(Y))] = [1]
>= [1]
= [false()]
[a__geq(s(X), s(Y))] = [1] X + [2]
> [1] X + [1]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [1] X2 + [1]
> [1] X1 + [1] X2 + [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [1] Y + [2]
> [0]
= [0()]
[a__div(s(X), s(Y))] = [1] Y + [1] X + [3]
>= [1] Y + [1] X + [3]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [1] X2 + [1] X3 + [0]
>= [1] X1 + [1] X2 + [1] X3 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [1] Y + [1] X + [0]
>= [1] X + [0]
= [mark(X)]
[a__if(false(), X, Y)] = [1] Y + [1] X + [1]
> [1] Y + [0]
= [mark(Y)]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(s(X))] = [1] X + [1]
>= [1] X + [1]
= [s(mark(X))]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(false())] = [1]
>= [1]
= [false()]
[mark(div(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [1]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [0]
>= [0]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [1] X1 + [1]
>= [1] X1 + [1]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [1] X1 + [1] X2 + [1] X3 + [0]
>= [1] X1 + [1] X2 + [1] X3 + [0]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(s(X), s(Y)) -> a__minus(X, Y)
, mark(s(X)) -> s(mark(X))
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2) }
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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [1] x1 + [0]
[0] = [0]
[s](x1) = [1] x1 + [4]
[a__geq](x1, x2) = [0]
[true] = [0]
[false] = [0]
[a__div](x1, x2) = [1] x1 + [4]
[a__if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [4]
[div](x1, x2) = [1] x1 + [0]
[minus](x1, x2) = [1] x1 + [0]
[mark](x1) = [1] x1 + [4]
[geq](x1, x2) = [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [1] X1 + [0]
>= [1] X1 + [0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [0]
>= [0]
= [0()]
[a__minus(s(X), s(Y))] = [1] X + [4]
> [1] X + [0]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [0]
>= [0]
= [geq(X1, X2)]
[a__geq(X, 0())] = [0]
>= [0]
= [true()]
[a__geq(0(), s(Y))] = [0]
>= [0]
= [false()]
[a__geq(s(X), s(Y))] = [0]
>= [0]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1] X1 + [4]
> [1] X1 + [0]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [4]
> [0]
= [0()]
[a__div(s(X), s(Y))] = [1] X + [8]
>= [1] X + [8]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1] X1 + [1] X2 + [1] X3 + [4]
> [1] X1 + [1] X2 + [1] X3 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [1] Y + [1] X + [4]
>= [1] X + [4]
= [mark(X)]
[a__if(false(), X, Y)] = [1] Y + [1] X + [4]
>= [1] Y + [4]
= [mark(Y)]
[mark(0())] = [4]
> [0]
= [0()]
[mark(s(X))] = [1] X + [8]
>= [1] X + [8]
= [s(mark(X))]
[mark(true())] = [4]
> [0]
= [true()]
[mark(false())] = [4]
> [0]
= [false()]
[mark(div(X1, X2))] = [1] X1 + [4]
? [1] X1 + [8]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [1] X1 + [4]
> [1] X1 + [0]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [4]
> [0]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [1] X1 + [1] X2 + [1] X3 + [4]
? [1] X1 + [1] X2 + [1] X3 + [8]
= [a__if(mark(X1), X2, X3)]
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:
{ mark(s(X)) -> s(mark(X))
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2) }
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: { mark(s(X)) -> s(mark(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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [0]
[0]
[0] = [0]
[0]
[s](x1) = [1 1] x1 + [0]
[0 0] [1]
[a__geq](x1, x2) = [1 0] x1 + [0]
[0 0] [1]
[true] = [0]
[1]
[false] = [0]
[1]
[a__div](x1, x2) = [1 7] x1 + [0 3] x2 + [5]
[0 0] [0 0] [1]
[a__if](x1, x2, x3) = [1 4] x1 + [3 2] x2 + [3 2] x3 + [0]
[0 0] [0 1] [0 1] [0]
[div](x1, x2) = [1 3] x1 + [0 1] x2 + [1]
[0 0] [0 0] [1]
[minus](x1, x2) = [0]
[0]
[mark](x1) = [3 2] x1 + [4]
[0 1] [0]
[geq](x1, x2) = [1 0] x1 + [0]
[0 0] [1]
[if](x1, x2, x3) = [1 4] x1 + [1 0] x2 + [1 0] x3 + [0]
[0 0] [0 1] [0 1] [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [0]
[0]
>= [0]
[0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [0]
[0]
>= [0]
[0]
= [0()]
[a__minus(s(X), s(Y))] = [0]
[0]
>= [0]
[0]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [1 0] X1 + [0]
[0 0] [1]
>= [1 0] X1 + [0]
[0 0] [1]
= [geq(X1, X2)]
[a__geq(X, 0())] = [1 0] X + [0]
[0 0] [1]
>= [0]
[1]
= [true()]
[a__geq(0(), s(Y))] = [0]
[1]
>= [0]
[1]
= [false()]
[a__geq(s(X), s(Y))] = [1 1] X + [0]
[0 0] [1]
>= [1 0] X + [0]
[0 0] [1]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1 7] X1 + [0 3] X2 + [5]
[0 0] [0 0] [1]
> [1 3] X1 + [0 1] X2 + [1]
[0 0] [0 0] [1]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [8]
[1]
> [0]
[0]
= [0()]
[a__div(s(X), s(Y))] = [1 1] X + [15]
[0 0] [1]
>= [1 0] X + [15]
[0 0] [1]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1 4] X1 + [3 2] X2 + [3 2] X3 + [0]
[0 0] [0 1] [0 1] [0]
>= [1 4] X1 + [1 0] X2 + [1 0] X3 + [0]
[0 0] [0 1] [0 1] [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [3 2] Y + [3 2] X + [4]
[0 1] [0 1] [0]
>= [3 2] X + [4]
[0 1] [0]
= [mark(X)]
[a__if(false(), X, Y)] = [3 2] Y + [3 2] X + [4]
[0 1] [0 1] [0]
>= [3 2] Y + [4]
[0 1] [0]
= [mark(Y)]
[mark(0())] = [4]
[0]
> [0]
[0]
= [0()]
[mark(s(X))] = [3 3] X + [6]
[0 0] [1]
> [3 3] X + [4]
[0 0] [1]
= [s(mark(X))]
[mark(true())] = [6]
[1]
> [0]
[1]
= [true()]
[mark(false())] = [6]
[1]
> [0]
[1]
= [false()]
[mark(div(X1, X2))] = [3 9] X1 + [0 3] X2 + [9]
[0 0] [0 0] [1]
>= [3 9] X1 + [0 3] X2 + [9]
[0 0] [0 0] [1]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [4]
[0]
> [0]
[0]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [3 0] X1 + [6]
[0 0] [1]
> [1 0] X1 + [0]
[0 0] [1]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [3 12] X1 + [3 2] X2 + [3 2] X3 + [4]
[0 0] [0 1] [0 1] [0]
>= [3 6] X1 + [3 2] X2 + [3 2] X3 + [4]
[0 0] [0 1] [0 1] [0]
= [a__if(mark(X1), X2, X3)]
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:
{ mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Weak Trs:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(s(X)) -> s(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2) }
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:
{ mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
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(s) = {1}, Uargs(a__div) = {1}, Uargs(a__if) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[a__minus](x1, x2) = [0]
[0]
[0] = [0]
[0]
[s](x1) = [1 0] x1 + [0]
[0 0] [7]
[a__geq](x1, x2) = [0 0] x2 + [4]
[1 0] [4]
[true] = [4]
[4]
[false] = [2]
[0]
[a__div](x1, x2) = [1 1] x1 + [0 0] x2 + [7]
[0 0] [0 1] [1]
[a__if](x1, x2, x3) = [1 0] x1 + [3 0] x2 + [3 0] x3 + [1]
[0 0] [0 1] [0 1] [0]
[div](x1, x2) = [1 1] x1 + [0 0] x2 + [3]
[0 0] [0 1] [1]
[minus](x1, x2) = [0]
[0]
[mark](x1) = [3 0] x1 + [3]
[0 1] [0]
[geq](x1, x2) = [0 0] x2 + [2]
[1 0] [4]
[if](x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1]
[0 0] [0 1] [0 1] [0]
The order satisfies the following ordering constraints:
[a__minus(X1, X2)] = [0]
[0]
>= [0]
[0]
= [minus(X1, X2)]
[a__minus(0(), Y)] = [0]
[0]
>= [0]
[0]
= [0()]
[a__minus(s(X), s(Y))] = [0]
[0]
>= [0]
[0]
= [a__minus(X, Y)]
[a__geq(X1, X2)] = [0 0] X2 + [4]
[1 0] [4]
> [0 0] X2 + [2]
[1 0] [4]
= [geq(X1, X2)]
[a__geq(X, 0())] = [4]
[4]
>= [4]
[4]
= [true()]
[a__geq(0(), s(Y))] = [0 0] Y + [4]
[1 0] [4]
> [2]
[0]
= [false()]
[a__geq(s(X), s(Y))] = [0 0] Y + [4]
[1 0] [4]
>= [0 0] Y + [4]
[1 0] [4]
= [a__geq(X, Y)]
[a__div(X1, X2)] = [1 1] X1 + [0 0] X2 + [7]
[0 0] [0 1] [1]
> [1 1] X1 + [0 0] X2 + [3]
[0 0] [0 1] [1]
= [div(X1, X2)]
[a__div(0(), s(Y))] = [7]
[8]
> [0]
[0]
= [0()]
[a__div(s(X), s(Y))] = [1 0] X + [14]
[0 0] [8]
>= [14]
[7]
= [a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())]
[a__if(X1, X2, X3)] = [1 0] X1 + [3 0] X2 + [3 0] X3 + [1]
[0 0] [0 1] [0 1] [0]
>= [1 0] X1 + [1 0] X2 + [1 0] X3 + [1]
[0 0] [0 1] [0 1] [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [3 0] Y + [3 0] X + [5]
[0 1] [0 1] [0]
> [3 0] X + [3]
[0 1] [0]
= [mark(X)]
[a__if(false(), X, Y)] = [3 0] Y + [3 0] X + [3]
[0 1] [0 1] [0]
>= [3 0] Y + [3]
[0 1] [0]
= [mark(Y)]
[mark(0())] = [3]
[0]
> [0]
[0]
= [0()]
[mark(s(X))] = [3 0] X + [3]
[0 0] [7]
>= [3 0] X + [3]
[0 0] [7]
= [s(mark(X))]
[mark(true())] = [15]
[4]
> [4]
[4]
= [true()]
[mark(false())] = [9]
[0]
> [2]
[0]
= [false()]
[mark(div(X1, X2))] = [3 3] X1 + [0 0] X2 + [12]
[0 0] [0 1] [1]
> [3 1] X1 + [0 0] X2 + [10]
[0 0] [0 1] [1]
= [a__div(mark(X1), X2)]
[mark(minus(X1, X2))] = [3]
[0]
> [0]
[0]
= [a__minus(X1, X2)]
[mark(geq(X1, X2))] = [0 0] X2 + [9]
[1 0] [4]
> [0 0] X2 + [4]
[1 0] [4]
= [a__geq(X1, X2)]
[mark(if(X1, X2, X3))] = [3 0] X1 + [3 0] X2 + [3 0] X3 + [6]
[0 0] [0 1] [0 1] [0]
> [3 0] X1 + [3 0] X2 + [3 0] X3 + [4]
[0 0] [0 1] [0 1] [0]
= [a__if(mark(X1), X2, X3)]
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:
{ a__minus(X1, X2) -> minus(X1, X2)
, a__minus(0(), Y) -> 0()
, a__minus(s(X), s(Y)) -> a__minus(X, Y)
, a__geq(X1, X2) -> geq(X1, X2)
, a__geq(X, 0()) -> true()
, a__geq(0(), s(Y)) -> false()
, a__geq(s(X), s(Y)) -> a__geq(X, Y)
, a__div(X1, X2) -> div(X1, X2)
, a__div(0(), s(Y)) -> 0()
, a__div(s(X), s(Y)) ->
a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, mark(0()) -> 0()
, mark(s(X)) -> s(mark(X))
, mark(true()) -> true()
, mark(false()) -> false()
, mark(div(X1, X2)) -> a__div(mark(X1), X2)
, mark(minus(X1, X2)) -> a__minus(X1, X2)
, mark(geq(X1, X2)) -> a__geq(X1, X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))