We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ f(0()) -> true()
, f(1()) -> false()
, f(s(x)) -> f(x)
, if(true(), x, y) -> x
, if(false(), x, y) -> y
, g(x, c(y)) -> g(x, g(s(c(y)), y))
, g(s(x), s(y)) -> if(f(x), s(x), s(y)) }
Obligation:
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(if) = {1}, Uargs(g) = {2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[f](x1) = [0]
[0] = [0]
[true] = [0]
[1] = [0]
[false] = [4]
[s](x1) = [0]
[if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [1]
[g](x1, x2) = [1] x1 + [1] x2 + [0]
[c](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[f(0())] = [0]
>= [0]
= [true()]
[f(1())] = [0]
? [4]
= [false()]
[f(s(x))] = [0]
>= [0]
= [f(x)]
[if(true(), x, y)] = [1] x + [1] y + [1]
> [1] x + [0]
= [x]
[if(false(), x, y)] = [1] x + [1] y + [5]
> [1] y + [0]
= [y]
[g(x, c(y))] = [1] x + [1] y + [0]
>= [1] x + [1] y + [0]
= [g(x, g(s(c(y)), y))]
[g(s(x), s(y))] = [0]
? [1]
= [if(f(x), s(x), s(y))]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ f(0()) -> true()
, f(1()) -> false()
, f(s(x)) -> f(x)
, g(x, c(y)) -> g(x, g(s(c(y)), y))
, g(s(x), s(y)) -> if(f(x), s(x), s(y)) }
Weak Trs:
{ if(true(), x, y) -> x
, if(false(), x, y) -> y }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ f(0()) -> true()
, f(1()) -> false() }
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(if) = {1}, Uargs(g) = {2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[f](x1) = [1]
[0] = [0]
[true] = [0]
[1] = [0]
[false] = [0]
[s](x1) = [0]
[if](x1, x2, x3) = [2] x1 + [3] x2 + [2] x3 + [0]
[g](x1, x2) = [1] x2 + [2]
[c](x1) = [1] x1 + [2]
The order satisfies the following ordering constraints:
[f(0())] = [1]
> [0]
= [true()]
[f(1())] = [1]
> [0]
= [false()]
[f(s(x))] = [1]
>= [1]
= [f(x)]
[if(true(), x, y)] = [3] x + [2] y + [0]
>= [1] x + [0]
= [x]
[if(false(), x, y)] = [3] x + [2] y + [0]
>= [1] y + [0]
= [y]
[g(x, c(y))] = [1] y + [4]
>= [1] y + [4]
= [g(x, g(s(c(y)), y))]
[g(s(x), s(y))] = [2]
>= [2]
= [if(f(x), s(x), s(y))]
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:
{ f(s(x)) -> f(x)
, g(x, c(y)) -> g(x, g(s(c(y)), y))
, g(s(x), s(y)) -> if(f(x), s(x), s(y)) }
Weak Trs:
{ f(0()) -> true()
, f(1()) -> false()
, if(true(), x, y) -> x
, if(false(), x, y) -> y }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
Trs:
{ g(x, c(y)) -> g(x, g(s(c(y)), y))
, g(s(x), s(y)) -> if(f(x), s(x), s(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:
----------
The following argument positions are usable:
Uargs(if) = {1}, Uargs(g) = {2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[f](x1) = [0]
[0] = [0]
[true] = [0]
[1] = [0]
[false] = [0]
[s](x1) = [1]
[if](x1, x2, x3) = [4] x1 + [1] x2 + [3] x3 + [0]
[g](x1, x2) = [1] x1 + [1] x2 + [3]
[c](x1) = [1] x1 + [7]
The order satisfies the following ordering constraints:
[f(0())] = [0]
>= [0]
= [true()]
[f(1())] = [0]
>= [0]
= [false()]
[f(s(x))] = [0]
>= [0]
= [f(x)]
[if(true(), x, y)] = [1] x + [3] y + [0]
>= [1] x + [0]
= [x]
[if(false(), x, y)] = [1] x + [3] y + [0]
>= [1] y + [0]
= [y]
[g(x, c(y))] = [1] x + [1] y + [10]
> [1] x + [1] y + [7]
= [g(x, g(s(c(y)), y))]
[g(s(x), s(y))] = [5]
> [4]
= [if(f(x), s(x), s(y))]
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: { f(s(x)) -> f(x) }
Weak Trs:
{ f(0()) -> true()
, f(1()) -> false()
, if(true(), x, y) -> x
, if(false(), x, y) -> y
, g(x, c(y)) -> g(x, g(s(c(y)), y))
, g(s(x), s(y)) -> if(f(x), s(x), s(y)) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs: { f(s(x)) -> f(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(if) = {1}, Uargs(g) = {2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[f](x1) = [0 1] x1 + [0]
[0 5] [4]
[0] = [0]
[0]
[true] = [0]
[0]
[1] = [0]
[0]
[false] = [0]
[0]
[s](x1) = [0 4] x1 + [0]
[0 1] [1]
[if](x1, x2, x3) = [2 0] x1 + [2 0] x2 + [1 0] x3 + [0]
[0 2] [0 1] [0 4] [0]
[g](x1, x2) = [3 1] x1 + [1 0] x2 + [0]
[1 7] [1 0] [6]
[c](x1) = [1 0] x1 + [7]
[0 0] [0]
The order satisfies the following ordering constraints:
[f(0())] = [0]
[4]
>= [0]
[0]
= [true()]
[f(1())] = [0]
[4]
>= [0]
[0]
= [false()]
[f(s(x))] = [0 1] x + [1]
[0 5] [9]
> [0 1] x + [0]
[0 5] [4]
= [f(x)]
[if(true(), x, y)] = [2 0] x + [1 0] y + [0]
[0 1] [0 4] [0]
>= [1 0] x + [0]
[0 1] [0]
= [x]
[if(false(), x, y)] = [2 0] x + [1 0] y + [0]
[0 1] [0 4] [0]
>= [1 0] y + [0]
[0 1] [0]
= [y]
[g(x, c(y))] = [3 1] x + [1 0] y + [7]
[1 7] [1 0] [13]
> [3 1] x + [1 0] y + [1]
[1 7] [1 0] [7]
= [g(x, g(s(c(y)), y))]
[g(s(x), s(y))] = [0 13] x + [0 4] y + [1]
[0 11] [0 4] [13]
> [0 10] x + [0 4] y + [0]
[0 11] [0 4] [13]
= [if(f(x), s(x), s(y))]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ f(0()) -> true()
, f(1()) -> false()
, f(s(x)) -> f(x)
, if(true(), x, y) -> x
, if(false(), x, y) -> y
, g(x, c(y)) -> g(x, g(s(c(y)), y))
, g(s(x), s(y)) -> if(f(x), s(x), s(y)) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^1))