We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^4)).
Strict Trs:
{ a__2nd(X) -> 2nd(X)
, a__2nd(cons(X, cons(Y, Z))) -> mark(Y)
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X))
, mark(2nd(X)) -> a__2nd(mark(X))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^4))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__2nd](x1) = [1] x1 + [4]
[cons](x1, x2) = [1] x1 + [1]
[mark](x1) = [0]
[a__from](x1) = [1] x1 + [0]
[from](x1) = [0]
[s](x1) = [1] x1 + [0]
[2nd](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[a__2nd(X)] = [1] X + [4]
> [1] X + [0]
= [2nd(X)]
[a__2nd(cons(X, cons(Y, Z)))] = [1] X + [5]
> [0]
= [mark(Y)]
[mark(cons(X1, X2))] = [0]
? [1]
= [cons(mark(X1), X2)]
[mark(from(X))] = [0]
>= [0]
= [a__from(mark(X))]
[mark(s(X))] = [0]
>= [0]
= [s(mark(X))]
[mark(2nd(X))] = [0]
? [4]
= [a__2nd(mark(X))]
[a__from(X)] = [1] X + [0]
? [1]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [0]
>= [0]
= [from(X)]
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^4)).
Strict Trs:
{ mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X))
, mark(2nd(X)) -> a__2nd(mark(X))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X) }
Weak Trs:
{ a__2nd(X) -> 2nd(X)
, a__2nd(cons(X, cons(Y, Z))) -> mark(Y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^4))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(s) = {1}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__2nd](x1) = [1] x1 + [0]
[cons](x1, x2) = [1] x1 + [0]
[mark](x1) = [0]
[a__from](x1) = [1] x1 + [4]
[from](x1) = [0]
[s](x1) = [1] x1 + [0]
[2nd](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[a__2nd(X)] = [1] X + [0]
>= [1] X + [0]
= [2nd(X)]
[a__2nd(cons(X, cons(Y, Z)))] = [1] X + [0]
>= [0]
= [mark(Y)]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(mark(X1), X2)]
[mark(from(X))] = [0]
? [4]
= [a__from(mark(X))]
[mark(s(X))] = [0]
>= [0]
= [s(mark(X))]
[mark(2nd(X))] = [0]
>= [0]
= [a__2nd(mark(X))]
[a__from(X)] = [1] X + [4]
> [0]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1] X + [4]
> [0]
= [from(X)]
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^4)).
Strict Trs:
{ mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X))
, mark(2nd(X)) -> a__2nd(mark(X)) }
Weak Trs:
{ a__2nd(X) -> 2nd(X)
, a__2nd(cons(X, cons(Y, Z))) -> mark(Y)
, a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^4))
We use the processor 'matrix interpretation of dimension 4' to
orient following rules strictly.
Trs: { mark(2nd(X)) -> a__2nd(mark(X)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^4)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[1 0 0 0] [1]
[a__2nd](x1) = [1 0 0 0] x1 + [1]
[1 0 0 0] [0]
[0 0 0 1] [1]
[1 0 0 0] [0 1 1 0] [0]
[cons](x1, x2) = [0 0 0 1] x1 + [0 0 0 0] x2 + [0]
[1 0 0 0] [0 1 1 0] [0]
[0 0 0 1] [0 1 0 0] [0]
[1 0 0 1] [0]
[mark](x1) = [1 0 0 1] x1 + [1]
[1 0 0 1] [0]
[0 0 0 1] [1]
[1 0 0 1] [0]
[a__from](x1) = [1 0 0 1] x1 + [1]
[1 0 0 1] [0]
[0 0 0 1] [1]
[1 0 0 1] [0]
[from](x1) = [0 0 0 0] x1 + [0]
[0 0 0 0] [0]
[0 0 0 1] [1]
[1 0 0 1] [0]
[s](x1) = [0 0 0 1] x1 + [1]
[0 0 0 1] [0]
[0 0 0 1] [1]
[1 0 0 0] [1]
[2nd](x1) = [0 0 0 0] x1 + [0]
[0 0 0 0] [0]
[0 0 0 1] [1]
The order satisfies the following ordering constraints:
[a__2nd(X)] = [1 0 0 0] [1]
[1 0 0 0] X + [1]
[1 0 0 0] [0]
[0 0 0 1] [1]
>= [1 0 0 0] [1]
[0 0 0 0] X + [0]
[0 0 0 0] [0]
[0 0 0 1] [1]
= [2nd(X)]
[a__2nd(cons(X, cons(Y, Z)))] = [1 0 0 0] [1 0 0 1] [0 1 1 0] [1]
[1 0 0 0] X + [1 0 0 1] Y + [0 1 1 0] Z + [1]
[1 0 0 0] [1 0 0 1] [0 1 1 0] [0]
[0 0 0 1] [0 0 0 1] [0 0 0 0] [1]
> [1 0 0 1] [0]
[1 0 0 1] Y + [1]
[1 0 0 1] [0]
[0 0 0 1] [1]
= [mark(Y)]
[mark(cons(X1, X2))] = [1 0 0 1] [0 2 1 0] [0]
[1 0 0 1] X1 + [0 2 1 0] X2 + [1]
[1 0 0 1] [0 2 1 0] [0]
[0 0 0 1] [0 1 0 0] [1]
>= [1 0 0 1] [0 1 1 0] [0]
[0 0 0 1] X1 + [0 0 0 0] X2 + [1]
[1 0 0 1] [0 1 1 0] [0]
[0 0 0 1] [0 1 0 0] [1]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1 0 0 2] [1]
[1 0 0 2] X + [2]
[1 0 0 2] [1]
[0 0 0 1] [2]
>= [1 0 0 2] [1]
[1 0 0 2] X + [2]
[1 0 0 2] [1]
[0 0 0 1] [2]
= [a__from(mark(X))]
[mark(s(X))] = [1 0 0 2] [1]
[1 0 0 2] X + [2]
[1 0 0 2] [1]
[0 0 0 1] [2]
>= [1 0 0 2] [1]
[0 0 0 1] X + [2]
[0 0 0 1] [1]
[0 0 0 1] [2]
= [s(mark(X))]
[mark(2nd(X))] = [1 0 0 1] [2]
[1 0 0 1] X + [3]
[1 0 0 1] [2]
[0 0 0 1] [2]
> [1 0 0 1] [1]
[1 0 0 1] X + [1]
[1 0 0 1] [0]
[0 0 0 1] [2]
= [a__2nd(mark(X))]
[a__from(X)] = [1 0 0 1] [0]
[1 0 0 1] X + [1]
[1 0 0 1] [0]
[0 0 0 1] [1]
>= [1 0 0 1] [0]
[0 0 0 1] X + [1]
[1 0 0 1] [0]
[0 0 0 1] [1]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1 0 0 1] [0]
[1 0 0 1] X + [1]
[1 0 0 1] [0]
[0 0 0 1] [1]
>= [1 0 0 1] [0]
[0 0 0 0] X + [0]
[0 0 0 0] [0]
[0 0 0 1] [1]
= [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^4)).
Strict Trs:
{ mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X)) }
Weak Trs:
{ a__2nd(X) -> 2nd(X)
, a__2nd(cons(X, cons(Y, Z))) -> mark(Y)
, mark(2nd(X)) -> a__2nd(mark(X))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^4))
We use the processor 'matrix interpretation of dimension 4' 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^4)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[1 0 0 1] [0]
[a__2nd](x1) = [0 1 0 1] x1 + [0]
[0 0 1 0] [1]
[0 0 0 1] [1]
[1 0 0 0] [0 1 0 0] [0]
[cons](x1, x2) = [1 0 0 0] x1 + [0 1 0 0] x2 + [0]
[0 0 0 1] [0 0 1 0] [0]
[0 0 0 1] [0 0 1 0] [0]
[1 0 0 1] [0]
[mark](x1) = [1 0 0 1] x1 + [0]
[0 0 0 1] [1]
[0 0 0 1] [1]
[1 0 0 1] [1]
[a__from](x1) = [1 0 0 1] x1 + [0]
[0 0 0 1] [1]
[0 0 0 1] [1]
[1 0 0 1] [1]
[from](x1) = [0 0 0 0] x1 + [0]
[0 0 0 0] [0]
[0 0 0 1] [1]
[1 0 0 0] [0]
[s](x1) = [0 0 0 1] x1 + [0]
[0 0 0 0] [1]
[0 0 0 1] [1]
[1 0 0 1] [0]
[2nd](x1) = [0 0 0 1] x1 + [0]
[0 0 1 0] [0]
[0 0 0 1] [1]
The order satisfies the following ordering constraints:
[a__2nd(X)] = [1 0 0 1] [0]
[0 1 0 1] X + [0]
[0 0 1 0] [1]
[0 0 0 1] [1]
>= [1 0 0 1] [0]
[0 0 0 1] X + [0]
[0 0 1 0] [0]
[0 0 0 1] [1]
= [2nd(X)]
[a__2nd(cons(X, cons(Y, Z)))] = [1 0 0 1] [1 0 0 1] [0 1 1 0] [0]
[1 0 0 1] X + [1 0 0 1] Y + [0 1 1 0] Z + [0]
[0 0 0 1] [0 0 0 1] [0 0 1 0] [1]
[0 0 0 1] [0 0 0 1] [0 0 1 0] [1]
>= [1 0 0 1] [0]
[1 0 0 1] Y + [0]
[0 0 0 1] [1]
[0 0 0 1] [1]
= [mark(Y)]
[mark(cons(X1, X2))] = [1 0 0 1] [0 1 1 0] [0]
[1 0 0 1] X1 + [0 1 1 0] X2 + [0]
[0 0 0 1] [0 0 1 0] [1]
[0 0 0 1] [0 0 1 0] [1]
>= [1 0 0 1] [0 1 0 0] [0]
[1 0 0 1] X1 + [0 1 0 0] X2 + [0]
[0 0 0 1] [0 0 1 0] [1]
[0 0 0 1] [0 0 1 0] [1]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1 0 0 2] [2]
[1 0 0 2] X + [2]
[0 0 0 1] [2]
[0 0 0 1] [2]
>= [1 0 0 2] [2]
[1 0 0 2] X + [1]
[0 0 0 1] [2]
[0 0 0 1] [2]
= [a__from(mark(X))]
[mark(s(X))] = [1 0 0 1] [1]
[1 0 0 1] X + [1]
[0 0 0 1] [2]
[0 0 0 1] [2]
> [1 0 0 1] [0]
[0 0 0 1] X + [1]
[0 0 0 0] [1]
[0 0 0 1] [2]
= [s(mark(X))]
[mark(2nd(X))] = [1 0 0 2] [1]
[1 0 0 2] X + [1]
[0 0 0 1] [2]
[0 0 0 1] [2]
>= [1 0 0 2] [1]
[1 0 0 2] X + [1]
[0 0 0 1] [2]
[0 0 0 1] [2]
= [a__2nd(mark(X))]
[a__from(X)] = [1 0 0 1] [1]
[1 0 0 1] X + [0]
[0 0 0 1] [1]
[0 0 0 1] [1]
> [1 0 0 1] [0]
[1 0 0 1] X + [0]
[0 0 0 1] [1]
[0 0 0 1] [1]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1 0 0 1] [1]
[1 0 0 1] X + [0]
[0 0 0 1] [1]
[0 0 0 1] [1]
>= [1 0 0 1] [1]
[0 0 0 0] X + [0]
[0 0 0 0] [0]
[0 0 0 1] [1]
= [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^4)).
Strict Trs:
{ mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(from(X)) -> a__from(mark(X)) }
Weak Trs:
{ a__2nd(X) -> 2nd(X)
, a__2nd(cons(X, cons(Y, Z))) -> mark(Y)
, mark(s(X)) -> s(mark(X))
, mark(2nd(X)) -> a__2nd(mark(X))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^4))
We use the processor 'matrix interpretation of dimension 4' to
orient following rules strictly.
Trs: { mark(from(X)) -> a__from(mark(X)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^4)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[1 0 1 0] [0]
[a__2nd](x1) = [1 0 1 0] x1 + [0]
[0 0 1 0] [0]
[0 0 1 1] [0]
[1 0 0 0] [0 1 0 1] [0]
[cons](x1, x2) = [0 0 1 0] x1 + [0 0 0 0] x2 + [0]
[0 0 1 0] [0 1 0 0] [0]
[1 0 0 0] [0 0 0 1] [0]
[1 0 1 0] [0]
[mark](x1) = [1 0 1 0] x1 + [0]
[0 0 1 0] [0]
[1 0 1 0] [0]
[1 0 1 0] [1]
[a__from](x1) = [0 0 1 1] x1 + [1]
[0 0 1 0] [1]
[1 0 1 0] [1]
[1 0 1 0] [1]
[from](x1) = [0 0 0 0] x1 + [1]
[0 0 1 0] [1]
[0 0 0 0] [0]
[1 0 0 0] [0]
[s](x1) = [0 0 0 0] x1 + [0]
[0 0 1 0] [1]
[0 0 0 0] [0]
[1 0 1 0] [0]
[2nd](x1) = [0 0 1 0] x1 + [0]
[0 0 1 0] [0]
[0 0 1 0] [0]
The order satisfies the following ordering constraints:
[a__2nd(X)] = [1 0 1 0] [0]
[1 0 1 0] X + [0]
[0 0 1 0] [0]
[0 0 1 1] [0]
>= [1 0 1 0] [0]
[0 0 1 0] X + [0]
[0 0 1 0] [0]
[0 0 1 0] [0]
= [2nd(X)]
[a__2nd(cons(X, cons(Y, Z)))] = [1 0 1 0] [1 0 2 0] [0 0 0 1] [0]
[1 0 1 0] X + [1 0 2 0] Y + [0 0 0 1] Z + [0]
[0 0 1 0] [0 0 1 0] [0 0 0 0] [0]
[1 0 1 0] [1 0 1 0] [0 0 0 1] [0]
>= [1 0 1 0] [0]
[1 0 1 0] Y + [0]
[0 0 1 0] [0]
[1 0 1 0] [0]
= [mark(Y)]
[mark(cons(X1, X2))] = [1 0 1 0] [0 2 0 1] [0]
[1 0 1 0] X1 + [0 2 0 1] X2 + [0]
[0 0 1 0] [0 1 0 0] [0]
[1 0 1 0] [0 2 0 1] [0]
>= [1 0 1 0] [0 1 0 1] [0]
[0 0 1 0] X1 + [0 0 0 0] X2 + [0]
[0 0 1 0] [0 1 0 0] [0]
[1 0 1 0] [0 0 0 1] [0]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1 0 2 0] [2]
[1 0 2 0] X + [2]
[0 0 1 0] [1]
[1 0 2 0] [2]
> [1 0 2 0] [1]
[1 0 2 0] X + [1]
[0 0 1 0] [1]
[1 0 2 0] [1]
= [a__from(mark(X))]
[mark(s(X))] = [1 0 1 0] [1]
[1 0 1 0] X + [1]
[0 0 1 0] [1]
[1 0 1 0] [1]
> [1 0 1 0] [0]
[0 0 0 0] X + [0]
[0 0 1 0] [1]
[0 0 0 0] [0]
= [s(mark(X))]
[mark(2nd(X))] = [1 0 2 0] [0]
[1 0 2 0] X + [0]
[0 0 1 0] [0]
[1 0 2 0] [0]
>= [1 0 2 0] [0]
[1 0 2 0] X + [0]
[0 0 1 0] [0]
[1 0 2 0] [0]
= [a__2nd(mark(X))]
[a__from(X)] = [1 0 1 0] [1]
[0 0 1 1] X + [1]
[0 0 1 0] [1]
[1 0 1 0] [1]
>= [1 0 1 0] [1]
[0 0 1 0] X + [0]
[0 0 1 0] [1]
[1 0 1 0] [0]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1 0 1 0] [1]
[0 0 1 1] X + [1]
[0 0 1 0] [1]
[1 0 1 0] [1]
>= [1 0 1 0] [1]
[0 0 0 0] X + [1]
[0 0 1 0] [1]
[0 0 0 0] [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^4)).
Strict Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) }
Weak Trs:
{ a__2nd(X) -> 2nd(X)
, a__2nd(cons(X, cons(Y, Z))) -> mark(Y)
, mark(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X))
, mark(2nd(X)) -> a__2nd(mark(X))
, a__from(X) -> cons(mark(X), from(s(X)))
, a__from(X) -> from(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^4))
We use the processor 'matrix interpretation of dimension 4' to
orient following rules strictly.
Trs: { mark(cons(X1, X2)) -> cons(mark(X1), X2) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^4)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(a__2nd) = {1}, Uargs(cons) = {1}, Uargs(a__from) = {1},
Uargs(s) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[1 0 0 1] [0]
[a__2nd](x1) = [0 1 0 0] x1 + [0]
[1 1 0 0] [0]
[0 0 0 1] [0]
[1 0 0 0] [0 0 1 0] [0]
[cons](x1, x2) = [0 0 0 1] x1 + [0 1 0 0] x2 + [0]
[1 0 0 0] [0 0 0 0] [1]
[0 0 0 1] [0 1 0 0] [1]
[1 0 0 1] [0]
[mark](x1) = [0 0 0 1] x1 + [0]
[1 0 0 1] [1]
[0 0 0 1] [0]
[1 0 0 1] [1]
[a__from](x1) = [0 0 0 1] x1 + [1]
[1 0 0 1] [1]
[0 0 0 1] [1]
[1 0 0 1] [0]
[from](x1) = [0 0 0 0] x1 + [0]
[0 0 0 0] [0]
[0 0 0 1] [1]
[1 0 0 1] [0]
[s](x1) = [0 0 0 0] x1 + [0]
[0 0 1 1] [0]
[0 0 0 1] [1]
[1 0 0 1] [0]
[2nd](x1) = [0 1 0 0] x1 + [0]
[0 0 0 0] [0]
[0 0 0 1] [0]
The order satisfies the following ordering constraints:
[a__2nd(X)] = [1 0 0 1] [0]
[0 1 0 0] X + [0]
[1 1 0 0] [0]
[0 0 0 1] [0]
>= [1 0 0 1] [0]
[0 1 0 0] X + [0]
[0 0 0 0] [0]
[0 0 0 1] [0]
= [2nd(X)]
[a__2nd(cons(X, cons(Y, Z)))] = [1 0 0 1] [1 0 0 1] [0 1 0 0] [2]
[0 0 0 1] X + [0 0 0 1] Y + [0 1 0 0] Z + [0]
[1 0 0 1] [1 0 0 1] [0 1 0 0] [1]
[0 0 0 1] [0 0 0 1] [0 1 0 0] [1]
> [1 0 0 1] [0]
[0 0 0 1] Y + [0]
[1 0 0 1] [1]
[0 0 0 1] [0]
= [mark(Y)]
[mark(cons(X1, X2))] = [1 0 0 1] [0 1 1 0] [1]
[0 0 0 1] X1 + [0 1 0 0] X2 + [1]
[1 0 0 1] [0 1 1 0] [2]
[0 0 0 1] [0 1 0 0] [1]
> [1 0 0 1] [0 0 1 0] [0]
[0 0 0 1] X1 + [0 1 0 0] X2 + [0]
[1 0 0 1] [0 0 0 0] [1]
[0 0 0 1] [0 1 0 0] [1]
= [cons(mark(X1), X2)]
[mark(from(X))] = [1 0 0 2] [1]
[0 0 0 1] X + [1]
[1 0 0 2] [2]
[0 0 0 1] [1]
>= [1 0 0 2] [1]
[0 0 0 1] X + [1]
[1 0 0 2] [1]
[0 0 0 1] [1]
= [a__from(mark(X))]
[mark(s(X))] = [1 0 0 2] [1]
[0 0 0 1] X + [1]
[1 0 0 2] [2]
[0 0 0 1] [1]
> [1 0 0 2] [0]
[0 0 0 0] X + [0]
[1 0 0 2] [1]
[0 0 0 1] [1]
= [s(mark(X))]
[mark(2nd(X))] = [1 0 0 2] [0]
[0 0 0 1] X + [0]
[1 0 0 2] [1]
[0 0 0 1] [0]
>= [1 0 0 2] [0]
[0 0 0 1] X + [0]
[1 0 0 2] [0]
[0 0 0 1] [0]
= [a__2nd(mark(X))]
[a__from(X)] = [1 0 0 1] [1]
[0 0 0 1] X + [1]
[1 0 0 1] [1]
[0 0 0 1] [1]
> [1 0 0 1] [0]
[0 0 0 1] X + [0]
[1 0 0 1] [1]
[0 0 0 1] [1]
= [cons(mark(X), from(s(X)))]
[a__from(X)] = [1 0 0 1] [1]
[0 0 0 1] X + [1]
[1 0 0 1] [1]
[0 0 0 1] [1]
> [1 0 0 1] [0]
[0 0 0 0] X + [0]
[0 0 0 0] [0]
[0 0 0 1] [1]
= [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__2nd(X) -> 2nd(X)
, a__2nd(cons(X, cons(Y, Z))) -> mark(Y)
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(from(X)) -> a__from(mark(X))
, mark(s(X)) -> s(mark(X))
, mark(2nd(X)) -> a__2nd(mark(X))
, a__from(X) -> cons(mark(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^4))