We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ f(x, 0()) -> x
, f(f(x, y), z) -> f(x, f(y, z))
, f(0(), y) -> y
, f(i(x), y) -> i(x)
, f(g(x, y), z) -> g(f(x, z), f(y, z))
, f(1(), g(x, y)) -> x
, f(2(), g(x, y)) -> y }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.
Trs:
{ f(1(), g(x, y)) -> x
, f(2(), g(x, y)) -> y }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are considered usable:
Uargs(f) = {2}, Uargs(g) = {1, 2}
TcT has computed the following constructor-restricted polynomial
interpretation.
[f](x1, x2) = x1 + x1*x2 + x2
[0]() = 0
[i](x1) = 0
[g](x1, x2) = 1 + x1 + x2
[1]() = 0
[2]() = 0
This order satisfies the following ordering constraints.
[f(x, 0())] = x
>= x
= [x]
[f(f(x, y), z)] = x + x*y + y + x*z + x*y*z + y*z + z
>= x + x*y + x*y*z + x*z + y + y*z + z
= [f(x, f(y, z))]
[f(0(), y)] = y
>= y
= [y]
[f(i(x), y)] = y
>=
= [i(x)]
[f(g(x, y), z)] = 1 + x + y + 2*z + x*z + y*z
>= 1 + x + x*z + 2*z + y + y*z
= [g(f(x, z), f(y, z))]
[f(1(), g(x, y))] = 1 + x + y
> x
= [x]
[f(2(), g(x, y))] = 1 + x + y
> y
= [y]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ f(x, 0()) -> x
, f(f(x, y), z) -> f(x, f(y, z))
, f(0(), y) -> y
, f(i(x), y) -> i(x)
, f(g(x, y), z) -> g(f(x, z), f(y, z)) }
Weak Trs:
{ f(1(), g(x, y)) -> x
, f(2(), g(x, y)) -> y }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.
Trs:
{ f(i(x), y) -> i(x)
, f(g(x, y), z) -> g(f(x, z), f(y, z)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are considered usable:
Uargs(f) = {2}, Uargs(g) = {1, 2}
TcT has computed the following constructor-restricted polynomial
interpretation.
[f](x1, x2) = x1 + x1*x2 + 2*x1^2 + x2
[0]() = 0
[i](x1) = 1
[g](x1, x2) = 1 + x1 + x2
[1]() = 1
[2]() = 0
This order satisfies the following ordering constraints.
[f(x, 0())] = x + 2*x^2
>= x
= [x]
[f(f(x, y), z)] = x + 3*x*y + 4*x^2 + y + x*z + x*y*z + 2*x^2*z + y*z + 8*x^2*y + 8*x^3 + 2*x^2*y^2 + 8*x^3*y + 2*x*y^2 + 8*x^4 + 2*y*x + 2*y^2*x + 4*y*x^2 + 2*y^2 + z
>= x + x*y + x*y*z + 2*x*y^2 + x*z + 2*x^2 + y + y*z + 2*y^2 + z
= [f(x, f(y, z))]
[f(0(), y)] = y
>= y
= [y]
[f(i(x), y)] = 3 + 2*y
> 1
= [i(x)]
[f(g(x, y), z)] = 3 + 5*x + 5*y + 2*z + x*z + y*z + 2*x^2 + 2*x*y + 2*y*x + 2*y^2
> 1 + x + x*z + 2*x^2 + 2*z + y + y*z + 2*y^2
= [g(f(x, z), f(y, z))]
[f(1(), g(x, y))] = 5 + 2*x + 2*y
> x
= [x]
[f(2(), g(x, y))] = 1 + x + y
> y
= [y]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ f(x, 0()) -> x
, f(f(x, y), z) -> f(x, f(y, z))
, f(0(), y) -> y }
Weak Trs:
{ f(i(x), y) -> i(x)
, f(g(x, y), z) -> g(f(x, z), f(y, z))
, f(1(), g(x, y)) -> x
, f(2(), g(x, y)) -> y }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.
Trs:
{ f(x, 0()) -> x
, f(0(), y) -> y }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are considered usable:
Uargs(f) = {2}, Uargs(g) = {1, 2}
TcT has computed the following constructor-restricted polynomial
interpretation.
[f](x1, x2) = x1 + x1*x2 + x2
[0]() = 1
[i](x1) = 0
[g](x1, x2) = 1 + x1 + x2
[1]() = 0
[2]() = 0
This order satisfies the following ordering constraints.
[f(x, 0())] = 2*x + 1
> x
= [x]
[f(f(x, y), z)] = x + x*y + y + x*z + x*y*z + y*z + z
>= x + x*y + x*y*z + x*z + y + y*z + z
= [f(x, f(y, z))]
[f(0(), y)] = 1 + 2*y
> y
= [y]
[f(i(x), y)] = y
>=
= [i(x)]
[f(g(x, y), z)] = 1 + x + y + 2*z + x*z + y*z
>= 1 + x + x*z + 2*z + y + y*z
= [g(f(x, z), f(y, z))]
[f(1(), g(x, y))] = 1 + x + y
> x
= [x]
[f(2(), g(x, y))] = 1 + x + y
> y
= [y]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs: { f(f(x, y), z) -> f(x, f(y, z)) }
Weak Trs:
{ f(x, 0()) -> x
, f(0(), y) -> y
, f(i(x), y) -> i(x)
, f(g(x, y), z) -> g(f(x, z), f(y, z))
, f(1(), g(x, y)) -> x
, f(2(), g(x, y)) -> y }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.
Trs: { f(f(x, y), z) -> f(x, f(y, z)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are considered usable:
Uargs(f) = {2}, Uargs(g) = {1, 2}
TcT has computed the following constructor-restricted polynomial
interpretation.
[f](x1, x2) = 1 + 2*x1 + 2*x1*x2 + x2
[0]() = 0
[i](x1) = 0
[g](x1, x2) = 1 + x1 + x2
[1]() = 0
[2]() = 0
This order satisfies the following ordering constraints.
[f(x, 0())] = 1 + 2*x
> x
= [x]
[f(f(x, y), z)] = 3 + 4*x + 4*x*y + 2*y + 3*z + 4*x*z + 4*x*y*z + 2*y*z
> 2 + 4*x + 4*x*y + 4*x*y*z + 2*x*z + 2*y + 2*y*z + z
= [f(x, f(y, z))]
[f(0(), y)] = 1 + y
> y
= [y]
[f(i(x), y)] = 1 + y
>
= [i(x)]
[f(g(x, y), z)] = 3 + 2*x + 2*y + 3*z + 2*x*z + 2*y*z
>= 3 + 2*x + 2*x*z + 2*z + 2*y + 2*y*z
= [g(f(x, z), f(y, z))]
[f(1(), g(x, y))] = 2 + x + y
> x
= [x]
[f(2(), g(x, y))] = 2 + x + y
> y
= [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(x, 0()) -> x
, f(f(x, y), z) -> f(x, f(y, z))
, f(0(), y) -> y
, f(i(x), y) -> i(x)
, f(g(x, y), z) -> g(f(x, z), f(y, z))
, f(1(), g(x, y)) -> x
, f(2(), g(x, y)) -> y }
Obligation:
runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^2))