YES(O(1),O(n^1))
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ a__and(X1, X2) -> and(X1, X2)
, a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(nil()) -> nil()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(X)
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(X1, X2) -> first(X1, X2)
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ mark(0()) -> 0()
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil() }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__and](x1, x2) = [1] x1 + [2] x2 + [0]
[true] = [0]
[mark](x1) = [2] x1 + [0]
[false] = [0]
[a__if](x1, x2, x3) = [1] x1 + [2] x2 + [2] x3 + [0]
[a__add](x1, x2) = [1] x1 + [2] x2 + [3]
[0] = [2]
[s](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [2]
[a__first](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[first](x1, x2) = [1] x1 + [1] x2 + [0]
[a__from](x1) = [2] x1 + [0]
[from](x1) = [1] x1 + [0]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
This order satisfies the following ordering constraints:
[a__and(X1, X2)] = [1] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [and(X1, X2)]
[a__and(true(), X)] = [2] X + [0]
>= [2] X + [0]
= [mark(X)]
[a__and(false(), Y)] = [2] Y + [0]
>= [0]
= [false()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(0())] = [4]
> [2]
= [0()]
[mark(s(X))] = [2] X + [0]
>= [1] X + [0]
= [s(X)]
[mark(add(X1, X2))] = [2] X1 + [2] X2 + [4]
> [2] X1 + [2] X2 + [3]
= [a__add(mark(X1), X2)]
[mark(nil())] = [0]
>= [0]
= [nil()]
[mark(cons(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [cons(X1, X2)]
[mark(first(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__first(mark(X1), mark(X2))]
[mark(from(X))] = [2] X + [0]
>= [2] X + [0]
= [a__from(X)]
[mark(and(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__and(mark(X1), X2)]
[mark(if(X1, X2, X3))] = [2] X1 + [2] X2 + [2] X3 + [0]
>= [2] X1 + [2] X2 + [2] X3 + [0]
= [a__if(mark(X1), X2, X3)]
[a__if(X1, X2, X3)] = [1] X1 + [2] X2 + [2] X3 + [0]
>= [1] X1 + [1] X2 + [1] X3 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [2] X + [2] Y + [0]
>= [2] X + [0]
= [mark(X)]
[a__if(false(), X, Y)] = [2] X + [2] Y + [0]
>= [2] Y + [0]
= [mark(Y)]
[a__add(X1, X2)] = [1] X1 + [2] X2 + [3]
> [1] X1 + [1] X2 + [2]
= [add(X1, X2)]
[a__add(0(), X)] = [2] X + [5]
> [2] X + [0]
= [mark(X)]
[a__add(s(X), Y)] = [1] X + [2] Y + [3]
> [1] X + [1] Y + [2]
= [s(add(X, Y))]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [first(X1, X2)]
[a__first(0(), X)] = [1] X + [2]
> [0]
= [nil()]
[a__first(s(X), cons(Y, Z))] = [1] X + [1] Y + [1] Z + [0]
>= [1] X + [1] Y + [1] Z + [0]
= [cons(Y, first(X, Z))]
[a__from(X)] = [2] X + [0]
>= [2] X + [0]
= [cons(X, from(s(X)))]
[a__from(X)] = [2] X + [0]
>= [1] X + [0]
= [from(X)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ a__and(X1, X2) -> and(X1, X2)
, a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(s(X)) -> s(X)
, mark(nil()) -> nil()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(X)
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__first(X1, X2) -> first(X1, X2)
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, a__from(X) -> from(X) }
Weak Trs:
{ mark(0()) -> 0()
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs: { mark(nil()) -> nil() }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__and](x1, x2) = [1] x1 + [2] x2 + [0]
[true] = [0]
[mark](x1) = [2] x1 + [0]
[false] = [0]
[a__if](x1, x2, x3) = [1] x1 + [2] x2 + [2] x3 + [0]
[a__add](x1, x2) = [1] x1 + [2] x2 + [3]
[0] = [2]
[s](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [2]
[a__first](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [2]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[first](x1, x2) = [1] x1 + [1] x2 + [0]
[a__from](x1) = [2] x1 + [0]
[from](x1) = [1] x1 + [0]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
This order satisfies the following ordering constraints:
[a__and(X1, X2)] = [1] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [and(X1, X2)]
[a__and(true(), X)] = [2] X + [0]
>= [2] X + [0]
= [mark(X)]
[a__and(false(), Y)] = [2] Y + [0]
>= [0]
= [false()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(0())] = [4]
> [2]
= [0()]
[mark(s(X))] = [2] X + [0]
>= [1] X + [0]
= [s(X)]
[mark(add(X1, X2))] = [2] X1 + [2] X2 + [4]
> [2] X1 + [2] X2 + [3]
= [a__add(mark(X1), X2)]
[mark(nil())] = [4]
> [2]
= [nil()]
[mark(cons(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [cons(X1, X2)]
[mark(first(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__first(mark(X1), mark(X2))]
[mark(from(X))] = [2] X + [0]
>= [2] X + [0]
= [a__from(X)]
[mark(and(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__and(mark(X1), X2)]
[mark(if(X1, X2, X3))] = [2] X1 + [2] X2 + [2] X3 + [0]
>= [2] X1 + [2] X2 + [2] X3 + [0]
= [a__if(mark(X1), X2, X3)]
[a__if(X1, X2, X3)] = [1] X1 + [2] X2 + [2] X3 + [0]
>= [1] X1 + [1] X2 + [1] X3 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [2] X + [2] Y + [0]
>= [2] X + [0]
= [mark(X)]
[a__if(false(), X, Y)] = [2] X + [2] Y + [0]
>= [2] Y + [0]
= [mark(Y)]
[a__add(X1, X2)] = [1] X1 + [2] X2 + [3]
> [1] X1 + [1] X2 + [2]
= [add(X1, X2)]
[a__add(0(), X)] = [2] X + [5]
> [2] X + [0]
= [mark(X)]
[a__add(s(X), Y)] = [1] X + [2] Y + [3]
> [1] X + [1] Y + [2]
= [s(add(X, Y))]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [first(X1, X2)]
[a__first(0(), X)] = [1] X + [2]
>= [2]
= [nil()]
[a__first(s(X), cons(Y, Z))] = [1] X + [1] Y + [1] Z + [0]
>= [1] X + [1] Y + [1] Z + [0]
= [cons(Y, first(X, Z))]
[a__from(X)] = [2] X + [0]
>= [2] X + [0]
= [cons(X, from(s(X)))]
[a__from(X)] = [2] X + [0]
>= [1] X + [0]
= [from(X)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ a__and(X1, X2) -> and(X1, X2)
, a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(s(X)) -> s(X)
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(X)
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__first(X1, X2) -> first(X1, X2)
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, a__from(X) -> from(X) }
Weak Trs:
{ mark(0()) -> 0()
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(nil()) -> nil()
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs: { mark(from(X)) -> a__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:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__and](x1, x2) = [1] x1 + [2] x2 + [0]
[true] = [0]
[mark](x1) = [2] x1 + [0]
[false] = [0]
[a__if](x1, x2, x3) = [1] x1 + [2] x2 + [2] x3 + [0]
[a__add](x1, x2) = [1] x1 + [2] x2 + [0]
[0] = [3]
[s](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[a__first](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [3]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[first](x1, x2) = [1] x1 + [1] x2 + [0]
[a__from](x1) = [2] x1 + [2]
[from](x1) = [1] x1 + [2]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
This order satisfies the following ordering constraints:
[a__and(X1, X2)] = [1] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [and(X1, X2)]
[a__and(true(), X)] = [2] X + [0]
>= [2] X + [0]
= [mark(X)]
[a__and(false(), Y)] = [2] Y + [0]
>= [0]
= [false()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(0())] = [6]
> [3]
= [0()]
[mark(s(X))] = [2] X + [0]
>= [1] X + [0]
= [s(X)]
[mark(add(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__add(mark(X1), X2)]
[mark(nil())] = [6]
> [3]
= [nil()]
[mark(cons(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [cons(X1, X2)]
[mark(first(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__first(mark(X1), mark(X2))]
[mark(from(X))] = [2] X + [4]
> [2] X + [2]
= [a__from(X)]
[mark(and(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__and(mark(X1), X2)]
[mark(if(X1, X2, X3))] = [2] X1 + [2] X2 + [2] X3 + [0]
>= [2] X1 + [2] X2 + [2] X3 + [0]
= [a__if(mark(X1), X2, X3)]
[a__if(X1, X2, X3)] = [1] X1 + [2] X2 + [2] X3 + [0]
>= [1] X1 + [1] X2 + [1] X3 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [2] X + [2] Y + [0]
>= [2] X + [0]
= [mark(X)]
[a__if(false(), X, Y)] = [2] X + [2] Y + [0]
>= [2] Y + [0]
= [mark(Y)]
[a__add(X1, X2)] = [1] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [2] X + [3]
> [2] X + [0]
= [mark(X)]
[a__add(s(X), Y)] = [1] X + [2] Y + [0]
>= [1] X + [1] Y + [0]
= [s(add(X, Y))]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [first(X1, X2)]
[a__first(0(), X)] = [1] X + [3]
>= [3]
= [nil()]
[a__first(s(X), cons(Y, Z))] = [1] X + [1] Y + [1] Z + [0]
>= [1] X + [1] Y + [1] Z + [0]
= [cons(Y, first(X, Z))]
[a__from(X)] = [2] X + [2]
>= [2] X + [2]
= [cons(X, from(s(X)))]
[a__from(X)] = [2] X + [2]
>= [1] X + [2]
= [from(X)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ a__and(X1, X2) -> and(X1, X2)
, a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(s(X)) -> s(X)
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__first(X1, X2) -> first(X1, X2)
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, a__from(X) -> from(X) }
Weak Trs:
{ mark(0()) -> 0()
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(nil()) -> nil()
, mark(from(X)) -> a__from(X)
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ a__from(X) -> cons(X, from(s(X)))
, a__from(X) -> 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:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__and](x1, x2) = [1] x1 + [2] x2 + [0]
[true] = [0]
[mark](x1) = [2] x1 + [0]
[false] = [0]
[a__if](x1, x2, x3) = [1] x1 + [2] x2 + [2] x3 + [0]
[a__add](x1, x2) = [1] x1 + [2] x2 + [0]
[0] = [3]
[s](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[a__first](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [3]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[first](x1, x2) = [1] x1 + [1] x2 + [0]
[a__from](x1) = [2] x1 + [3]
[from](x1) = [1] x1 + [2]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
This order satisfies the following ordering constraints:
[a__and(X1, X2)] = [1] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [and(X1, X2)]
[a__and(true(), X)] = [2] X + [0]
>= [2] X + [0]
= [mark(X)]
[a__and(false(), Y)] = [2] Y + [0]
>= [0]
= [false()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(0())] = [6]
> [3]
= [0()]
[mark(s(X))] = [2] X + [0]
>= [1] X + [0]
= [s(X)]
[mark(add(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__add(mark(X1), X2)]
[mark(nil())] = [6]
> [3]
= [nil()]
[mark(cons(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [cons(X1, X2)]
[mark(first(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__first(mark(X1), mark(X2))]
[mark(from(X))] = [2] X + [4]
> [2] X + [3]
= [a__from(X)]
[mark(and(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__and(mark(X1), X2)]
[mark(if(X1, X2, X3))] = [2] X1 + [2] X2 + [2] X3 + [0]
>= [2] X1 + [2] X2 + [2] X3 + [0]
= [a__if(mark(X1), X2, X3)]
[a__if(X1, X2, X3)] = [1] X1 + [2] X2 + [2] X3 + [0]
>= [1] X1 + [1] X2 + [1] X3 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [2] X + [2] Y + [0]
>= [2] X + [0]
= [mark(X)]
[a__if(false(), X, Y)] = [2] X + [2] Y + [0]
>= [2] Y + [0]
= [mark(Y)]
[a__add(X1, X2)] = [1] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [2] X + [3]
> [2] X + [0]
= [mark(X)]
[a__add(s(X), Y)] = [1] X + [2] Y + [0]
>= [1] X + [1] Y + [0]
= [s(add(X, Y))]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [first(X1, X2)]
[a__first(0(), X)] = [1] X + [3]
>= [3]
= [nil()]
[a__first(s(X), cons(Y, Z))] = [1] X + [1] Y + [1] Z + [0]
>= [1] X + [1] Y + [1] Z + [0]
= [cons(Y, first(X, Z))]
[a__from(X)] = [2] X + [3]
> [2] X + [2]
= [cons(X, from(s(X)))]
[a__from(X)] = [2] X + [3]
> [1] X + [2]
= [from(X)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ a__and(X1, X2) -> and(X1, X2)
, a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(s(X)) -> s(X)
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__first(X1, X2) -> first(X1, X2)
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z)) }
Weak Trs:
{ mark(0()) -> 0()
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(nil()) -> nil()
, mark(from(X)) -> a__from(X)
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, a__from(X) -> cons(X, from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, a__first(X1, X2) -> first(X1, X2)
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z)) }
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:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__and](x1, x2) = [1] x1 + [2] x2 + [0]
[true] = [0]
[mark](x1) = [2] x1 + [0]
[false] = [0]
[a__if](x1, x2, x3) = [1] x1 + [2] x2 + [2] x3 + [0]
[a__add](x1, x2) = [1] x1 + [2] x2 + [0]
[0] = [3]
[s](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[a__first](x1, x2) = [1] x1 + [1] x2 + [3]
[nil] = [3]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[first](x1, x2) = [1] x1 + [1] x2 + [2]
[a__from](x1) = [2] x1 + [0]
[from](x1) = [1] x1 + [0]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
This order satisfies the following ordering constraints:
[a__and(X1, X2)] = [1] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [and(X1, X2)]
[a__and(true(), X)] = [2] X + [0]
>= [2] X + [0]
= [mark(X)]
[a__and(false(), Y)] = [2] Y + [0]
>= [0]
= [false()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(0())] = [6]
> [3]
= [0()]
[mark(s(X))] = [2] X + [0]
>= [1] X + [0]
= [s(X)]
[mark(add(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__add(mark(X1), X2)]
[mark(nil())] = [6]
> [3]
= [nil()]
[mark(cons(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [cons(X1, X2)]
[mark(first(X1, X2))] = [2] X1 + [2] X2 + [4]
> [2] X1 + [2] X2 + [3]
= [a__first(mark(X1), mark(X2))]
[mark(from(X))] = [2] X + [0]
>= [2] X + [0]
= [a__from(X)]
[mark(and(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__and(mark(X1), X2)]
[mark(if(X1, X2, X3))] = [2] X1 + [2] X2 + [2] X3 + [0]
>= [2] X1 + [2] X2 + [2] X3 + [0]
= [a__if(mark(X1), X2, X3)]
[a__if(X1, X2, X3)] = [1] X1 + [2] X2 + [2] X3 + [0]
>= [1] X1 + [1] X2 + [1] X3 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [2] X + [2] Y + [0]
>= [2] X + [0]
= [mark(X)]
[a__if(false(), X, Y)] = [2] X + [2] Y + [0]
>= [2] Y + [0]
= [mark(Y)]
[a__add(X1, X2)] = [1] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [2] X + [3]
> [2] X + [0]
= [mark(X)]
[a__add(s(X), Y)] = [1] X + [2] Y + [0]
>= [1] X + [1] Y + [0]
= [s(add(X, Y))]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [3]
> [1] X1 + [1] X2 + [2]
= [first(X1, X2)]
[a__first(0(), X)] = [1] X + [6]
> [3]
= [nil()]
[a__first(s(X), cons(Y, Z))] = [1] X + [1] Y + [1] Z + [3]
> [1] X + [1] Y + [1] Z + [2]
= [cons(Y, first(X, Z))]
[a__from(X)] = [2] X + [0]
>= [2] X + [0]
= [cons(X, from(s(X)))]
[a__from(X)] = [2] X + [0]
>= [1] X + [0]
= [from(X)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ a__and(X1, X2) -> and(X1, X2)
, a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(s(X)) -> s(X)
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y) }
Weak Trs:
{ mark(0()) -> 0()
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(nil()) -> nil()
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(X)
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(X1, X2) -> first(X1, X2)
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y) }
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:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__and](x1, x2) = [1] x1 + [2] x2 + [0]
[true] = [0]
[mark](x1) = [2] x1 + [0]
[false] = [0]
[a__if](x1, x2, x3) = [1] x1 + [2] x2 + [2] x3 + [3]
[a__add](x1, x2) = [1] x1 + [2] x2 + [0]
[0] = [3]
[s](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[a__first](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [3]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[first](x1, x2) = [1] x1 + [1] x2 + [0]
[a__from](x1) = [2] x1 + [0]
[from](x1) = [1] x1 + [0]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [2]
This order satisfies the following ordering constraints:
[a__and(X1, X2)] = [1] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [and(X1, X2)]
[a__and(true(), X)] = [2] X + [0]
>= [2] X + [0]
= [mark(X)]
[a__and(false(), Y)] = [2] Y + [0]
>= [0]
= [false()]
[mark(true())] = [0]
>= [0]
= [true()]
[mark(false())] = [0]
>= [0]
= [false()]
[mark(0())] = [6]
> [3]
= [0()]
[mark(s(X))] = [2] X + [0]
>= [1] X + [0]
= [s(X)]
[mark(add(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__add(mark(X1), X2)]
[mark(nil())] = [6]
> [3]
= [nil()]
[mark(cons(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [cons(X1, X2)]
[mark(first(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__first(mark(X1), mark(X2))]
[mark(from(X))] = [2] X + [0]
>= [2] X + [0]
= [a__from(X)]
[mark(and(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__and(mark(X1), X2)]
[mark(if(X1, X2, X3))] = [2] X1 + [2] X2 + [2] X3 + [4]
> [2] X1 + [2] X2 + [2] X3 + [3]
= [a__if(mark(X1), X2, X3)]
[a__if(X1, X2, X3)] = [1] X1 + [2] X2 + [2] X3 + [3]
> [1] X1 + [1] X2 + [1] X3 + [2]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [2] X + [2] Y + [3]
> [2] X + [0]
= [mark(X)]
[a__if(false(), X, Y)] = [2] X + [2] Y + [3]
> [2] Y + [0]
= [mark(Y)]
[a__add(X1, X2)] = [1] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [2] X + [3]
> [2] X + [0]
= [mark(X)]
[a__add(s(X), Y)] = [1] X + [2] Y + [0]
>= [1] X + [1] Y + [0]
= [s(add(X, Y))]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [first(X1, X2)]
[a__first(0(), X)] = [1] X + [3]
>= [3]
= [nil()]
[a__first(s(X), cons(Y, Z))] = [1] X + [1] Y + [1] Z + [0]
>= [1] X + [1] Y + [1] Z + [0]
= [cons(Y, first(X, Z))]
[a__from(X)] = [2] X + [0]
>= [2] X + [0]
= [cons(X, from(s(X)))]
[a__from(X)] = [2] X + [0]
>= [1] X + [0]
= [from(X)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ a__and(X1, X2) -> and(X1, X2)
, a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(s(X)) -> s(X)
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(and(X1, X2)) -> a__and(mark(X1), X2) }
Weak Trs:
{ mark(0()) -> 0()
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(nil()) -> nil()
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(X)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(X1, X2) -> first(X1, X2)
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ mark(true()) -> true()
, mark(false()) -> false()
, mark(s(X)) -> s(X)
, mark(cons(X1, X2)) -> cons(X1, 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:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__and](x1, x2) = [1] x1 + [3] x2 + [0]
[true] = [1]
[mark](x1) = [3] x1 + [1]
[false] = [0]
[a__if](x1, x2, x3) = [1] x1 + [3] x2 + [3] x3 + [1]
[a__add](x1, x2) = [1] x1 + [3] x2 + [0]
[0] = [2]
[s](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[a__first](x1, x2) = [1] x1 + [1] x2 + [1]
[nil] = [1]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[first](x1, x2) = [1] x1 + [1] x2 + [1]
[a__from](x1) = [3] x1 + [0]
[from](x1) = [1] x1 + [0]
[and](x1, x2) = [1] x1 + [1] x2 + [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1]
This order satisfies the following ordering constraints:
[a__and(X1, X2)] = [1] X1 + [3] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [and(X1, X2)]
[a__and(true(), X)] = [3] X + [1]
>= [3] X + [1]
= [mark(X)]
[a__and(false(), Y)] = [3] Y + [0]
>= [0]
= [false()]
[mark(true())] = [4]
> [1]
= [true()]
[mark(false())] = [1]
> [0]
= [false()]
[mark(0())] = [7]
> [2]
= [0()]
[mark(s(X))] = [3] X + [1]
> [1] X + [0]
= [s(X)]
[mark(add(X1, X2))] = [3] X1 + [3] X2 + [1]
>= [3] X1 + [3] X2 + [1]
= [a__add(mark(X1), X2)]
[mark(nil())] = [4]
> [1]
= [nil()]
[mark(cons(X1, X2))] = [3] X1 + [3] X2 + [1]
> [1] X1 + [1] X2 + [0]
= [cons(X1, X2)]
[mark(first(X1, X2))] = [3] X1 + [3] X2 + [4]
> [3] X1 + [3] X2 + [3]
= [a__first(mark(X1), mark(X2))]
[mark(from(X))] = [3] X + [1]
> [3] X + [0]
= [a__from(X)]
[mark(and(X1, X2))] = [3] X1 + [3] X2 + [1]
>= [3] X1 + [3] X2 + [1]
= [a__and(mark(X1), X2)]
[mark(if(X1, X2, X3))] = [3] X1 + [3] X2 + [3] X3 + [4]
> [3] X1 + [3] X2 + [3] X3 + [2]
= [a__if(mark(X1), X2, X3)]
[a__if(X1, X2, X3)] = [1] X1 + [3] X2 + [3] X3 + [1]
>= [1] X1 + [1] X2 + [1] X3 + [1]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [3] X + [3] Y + [2]
> [3] X + [1]
= [mark(X)]
[a__if(false(), X, Y)] = [3] X + [3] Y + [1]
>= [3] Y + [1]
= [mark(Y)]
[a__add(X1, X2)] = [1] X1 + [3] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [3] X + [2]
> [3] X + [1]
= [mark(X)]
[a__add(s(X), Y)] = [1] X + [3] Y + [0]
>= [1] X + [1] Y + [0]
= [s(add(X, Y))]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [1]
>= [1] X1 + [1] X2 + [1]
= [first(X1, X2)]
[a__first(0(), X)] = [1] X + [3]
> [1]
= [nil()]
[a__first(s(X), cons(Y, Z))] = [1] X + [1] Y + [1] Z + [1]
>= [1] X + [1] Y + [1] Z + [1]
= [cons(Y, first(X, Z))]
[a__from(X)] = [3] X + [0]
>= [2] X + [0]
= [cons(X, from(s(X)))]
[a__from(X)] = [3] X + [0]
>= [1] X + [0]
= [from(X)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ a__and(X1, X2) -> and(X1, X2)
, a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, mark(and(X1, X2)) -> a__and(mark(X1), X2) }
Weak Trs:
{ mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(nil()) -> nil()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(X)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(X1, X2) -> first(X1, X2)
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, mark(and(X1, X2)) -> a__and(mark(X1), 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:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__and](x1, x2) = [1] x1 + [2] x2 + [2]
[true] = [2]
[mark](x1) = [2] x1 + [0]
[false] = [2]
[a__if](x1, x2, x3) = [1] x1 + [2] x2 + [2] x3 + [0]
[a__add](x1, x2) = [1] x1 + [2] x2 + [0]
[0] = [3]
[s](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[a__first](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [3]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[first](x1, x2) = [1] x1 + [1] x2 + [0]
[a__from](x1) = [2] x1 + [0]
[from](x1) = [1] x1 + [0]
[and](x1, x2) = [1] x1 + [1] x2 + [2]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
This order satisfies the following ordering constraints:
[a__and(X1, X2)] = [1] X1 + [2] X2 + [2]
>= [1] X1 + [1] X2 + [2]
= [and(X1, X2)]
[a__and(true(), X)] = [2] X + [4]
> [2] X + [0]
= [mark(X)]
[a__and(false(), Y)] = [2] Y + [4]
> [2]
= [false()]
[mark(true())] = [4]
> [2]
= [true()]
[mark(false())] = [4]
> [2]
= [false()]
[mark(0())] = [6]
> [3]
= [0()]
[mark(s(X))] = [2] X + [0]
>= [1] X + [0]
= [s(X)]
[mark(add(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__add(mark(X1), X2)]
[mark(nil())] = [6]
> [3]
= [nil()]
[mark(cons(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [cons(X1, X2)]
[mark(first(X1, X2))] = [2] X1 + [2] X2 + [0]
>= [2] X1 + [2] X2 + [0]
= [a__first(mark(X1), mark(X2))]
[mark(from(X))] = [2] X + [0]
>= [2] X + [0]
= [a__from(X)]
[mark(and(X1, X2))] = [2] X1 + [2] X2 + [4]
> [2] X1 + [2] X2 + [2]
= [a__and(mark(X1), X2)]
[mark(if(X1, X2, X3))] = [2] X1 + [2] X2 + [2] X3 + [0]
>= [2] X1 + [2] X2 + [2] X3 + [0]
= [a__if(mark(X1), X2, X3)]
[a__if(X1, X2, X3)] = [1] X1 + [2] X2 + [2] X3 + [0]
>= [1] X1 + [1] X2 + [1] X3 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [2] X + [2] Y + [2]
> [2] X + [0]
= [mark(X)]
[a__if(false(), X, Y)] = [2] X + [2] Y + [2]
> [2] Y + [0]
= [mark(Y)]
[a__add(X1, X2)] = [1] X1 + [2] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [2] X + [3]
> [2] X + [0]
= [mark(X)]
[a__add(s(X), Y)] = [1] X + [2] Y + [0]
>= [1] X + [1] Y + [0]
= [s(add(X, Y))]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [first(X1, X2)]
[a__first(0(), X)] = [1] X + [3]
>= [3]
= [nil()]
[a__first(s(X), cons(Y, Z))] = [1] X + [1] Y + [1] Z + [0]
>= [1] X + [1] Y + [1] Z + [0]
= [cons(Y, first(X, Z))]
[a__from(X)] = [2] X + [0]
>= [2] X + [0]
= [cons(X, from(s(X)))]
[a__from(X)] = [2] X + [0]
>= [1] X + [0]
= [from(X)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs: { a__and(X1, X2) -> and(X1, X2) }
Weak Trs:
{ a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(nil()) -> nil()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(X)
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(X1, X2) -> first(X1, X2)
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs: { a__and(X1, X2) -> and(X1, 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:
----------
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__and](x1, x2) = [1] x1 + [3] x2 + [3]
[true] = [1]
[mark](x1) = [3] x1 + [1]
[false] = [2]
[a__if](x1, x2, x3) = [1] x1 + [3] x2 + [3] x3 + [0]
[a__add](x1, x2) = [1] x1 + [3] x2 + [0]
[0] = [2]
[s](x1) = [1] x1 + [0]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[a__first](x1, x2) = [1] x1 + [1] x2 + [2]
[nil] = [0]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[first](x1, x2) = [1] x1 + [1] x2 + [1]
[a__from](x1) = [3] x1 + [0]
[from](x1) = [1] x1 + [0]
[and](x1, x2) = [1] x1 + [1] x2 + [1]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
This order satisfies the following ordering constraints:
[a__and(X1, X2)] = [1] X1 + [3] X2 + [3]
> [1] X1 + [1] X2 + [1]
= [and(X1, X2)]
[a__and(true(), X)] = [3] X + [4]
> [3] X + [1]
= [mark(X)]
[a__and(false(), Y)] = [3] Y + [5]
> [2]
= [false()]
[mark(true())] = [4]
> [1]
= [true()]
[mark(false())] = [7]
> [2]
= [false()]
[mark(0())] = [7]
> [2]
= [0()]
[mark(s(X))] = [3] X + [1]
> [1] X + [0]
= [s(X)]
[mark(add(X1, X2))] = [3] X1 + [3] X2 + [1]
>= [3] X1 + [3] X2 + [1]
= [a__add(mark(X1), X2)]
[mark(nil())] = [1]
> [0]
= [nil()]
[mark(cons(X1, X2))] = [3] X1 + [3] X2 + [1]
> [1] X1 + [1] X2 + [0]
= [cons(X1, X2)]
[mark(first(X1, X2))] = [3] X1 + [3] X2 + [4]
>= [3] X1 + [3] X2 + [4]
= [a__first(mark(X1), mark(X2))]
[mark(from(X))] = [3] X + [1]
> [3] X + [0]
= [a__from(X)]
[mark(and(X1, X2))] = [3] X1 + [3] X2 + [4]
>= [3] X1 + [3] X2 + [4]
= [a__and(mark(X1), X2)]
[mark(if(X1, X2, X3))] = [3] X1 + [3] X2 + [3] X3 + [1]
>= [3] X1 + [3] X2 + [3] X3 + [1]
= [a__if(mark(X1), X2, X3)]
[a__if(X1, X2, X3)] = [1] X1 + [3] X2 + [3] X3 + [0]
>= [1] X1 + [1] X2 + [1] X3 + [0]
= [if(X1, X2, X3)]
[a__if(true(), X, Y)] = [3] X + [3] Y + [1]
>= [3] X + [1]
= [mark(X)]
[a__if(false(), X, Y)] = [3] X + [3] Y + [2]
> [3] Y + [1]
= [mark(Y)]
[a__add(X1, X2)] = [1] X1 + [3] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(X1, X2)]
[a__add(0(), X)] = [3] X + [2]
> [3] X + [1]
= [mark(X)]
[a__add(s(X), Y)] = [1] X + [3] Y + [0]
>= [1] X + [1] Y + [0]
= [s(add(X, Y))]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [2]
> [1] X1 + [1] X2 + [1]
= [first(X1, X2)]
[a__first(0(), X)] = [1] X + [4]
> [0]
= [nil()]
[a__first(s(X), cons(Y, Z))] = [1] X + [1] Y + [1] Z + [2]
> [1] X + [1] Y + [1] Z + [1]
= [cons(Y, first(X, Z))]
[a__from(X)] = [3] X + [0]
>= [2] X + [0]
= [cons(X, from(s(X)))]
[a__from(X)] = [3] X + [0]
>= [1] X + [0]
= [from(X)]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ a__and(X1, X2) -> and(X1, X2)
, a__and(true(), X) -> mark(X)
, a__and(false(), Y) -> false()
, mark(true()) -> true()
, mark(false()) -> false()
, mark(0()) -> 0()
, mark(s(X)) -> s(X)
, mark(add(X1, X2)) -> a__add(mark(X1), X2)
, mark(nil()) -> nil()
, mark(cons(X1, X2)) -> cons(X1, X2)
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(from(X)) -> a__from(X)
, mark(and(X1, X2)) -> a__and(mark(X1), X2)
, mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
, a__if(X1, X2, X3) -> if(X1, X2, X3)
, a__if(true(), X, Y) -> mark(X)
, a__if(false(), X, Y) -> mark(Y)
, a__add(X1, X2) -> add(X1, X2)
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(X1, X2) -> first(X1, X2)
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
, a__from(X) -> cons(X, from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))