We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ :(z, +(x, f(y))) -> :(g(z, y), +(x, a()))
, :(:(x, y), z) -> :(x, :(y, z))
, :(+(x, y), z) -> +(:(x, z), :(y, z)) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.
Trs: { :(+(x, y), z) -> +(:(x, z), :(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(:) = {2}, Uargs(+) = {1, 2}
TcT has computed the following constructor-restricted polynomial
interpretation.
[:](x1, x2) = 2*x1 + 2*x1*x2 + x2
[+](x1, x2) = 1 + x1 + x2
[f](x1) = x1
[g](x1, x2) = 0
[a]() = 0
This order satisfies the following ordering constraints.
[:(z, +(x, f(y)))] = 4*z + 2*z*x + 2*z*y + 1 + x + y
>= 1 + x
= [:(g(z, y), +(x, a()))]
[:(:(x, y), z)] = 4*x + 4*x*y + 2*y + 4*x*z + 4*x*y*z + 2*y*z + z
>= 2*x + 4*x*y + 4*x*y*z + 2*x*z + 2*y + 2*y*z + z
= [:(x, :(y, z))]
[:(+(x, y), z)] = 2 + 2*x + 2*y + 3*z + 2*x*z + 2*y*z
> 1 + 2*x + 2*x*z + 2*z + 2*y + 2*y*z
= [+(:(x, z), :(y, z))]
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:
{ :(z, +(x, f(y))) -> :(g(z, y), +(x, a()))
, :(:(x, y), z) -> :(x, :(y, z)) }
Weak Trs: { :(+(x, y), z) -> +(:(x, z), :(y, z)) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.
Trs: { :(:(x, y), z) -> :(x, :(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(:) = {2}, Uargs(+) = {1, 2}
TcT has computed the following constructor-restricted polynomial
interpretation.
[:](x1, x2) = 1 + x1 + x1*x2 + x1^2 + x2
[+](x1, x2) = 2 + x1 + x2
[f](x1) = 2 + x1
[g](x1, x2) = 0
[a]() = 2
This order satisfies the following ordering constraints.
[:(z, +(x, f(y)))] = 5 + 5*z + z*x + z*y + z^2 + x + y
>= 5 + x
= [:(g(z, y), +(x, a()))]
[:(:(x, y), z)] = 3 + 3*x + 4*x*y + 4*x^2 + 3*y + 2*z + x*z + x*y*z + x^2*z + y*z + 3*x^2*y + 2*x^3 + x^2*y^2 + 2*x^3*y + x*y^2 + x^4 + y*x + y^2*x + y*x^2 + y^2
> 2 + 2*x + x*y + x*y*z + x*y^2 + x*z + x^2 + y + y*z + y^2 + z
= [:(x, :(y, z))]
[:(+(x, y), z)] = 7 + 5*x + 5*y + 3*z + x*z + y*z + x^2 + x*y + y*x + y^2
> 4 + x + x*z + x^2 + 2*z + y + y*z + y^2
= [+(:(x, z), :(y, z))]
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: { :(z, +(x, f(y))) -> :(g(z, y), +(x, a())) }
Weak Trs:
{ :(:(x, y), z) -> :(x, :(y, z))
, :(+(x, y), z) -> +(:(x, z), :(y, z)) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.
Trs: { :(z, +(x, f(y))) -> :(g(z, y), +(x, a())) }
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(:) = {2}, Uargs(+) = {1, 2}
TcT has computed the following constructor-restricted polynomial
interpretation.
[:](x1, x2) = 1 + x1 + x1*x2 + x1^2 + x2
[+](x1, x2) = 1 + x1 + x2
[f](x1) = 2 + x1
[g](x1, x2) = 0
[a]() = 0
This order satisfies the following ordering constraints.
[:(z, +(x, f(y)))] = 4 + 4*z + z*x + z*y + z^2 + x + y
> 2 + x
= [:(g(z, y), +(x, a()))]
[:(:(x, y), z)] = 3 + 3*x + 4*x*y + 4*x^2 + 3*y + 2*z + x*z + x*y*z + x^2*z + y*z + 3*x^2*y + 2*x^3 + x^2*y^2 + 2*x^3*y + x*y^2 + x^4 + y*x + y^2*x + y*x^2 + y^2
> 2 + 2*x + x*y + x*y*z + x*y^2 + x*z + x^2 + y + y*z + y^2 + z
= [:(x, :(y, z))]
[:(+(x, y), z)] = 3 + 3*x + 3*y + 2*z + x*z + y*z + x^2 + x*y + y*x + y^2
>= 3 + x + x*z + x^2 + 2*z + y + y*z + y^2
= [+(:(x, z), :(y, z))]
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:
{ :(z, +(x, f(y))) -> :(g(z, y), +(x, a()))
, :(:(x, y), z) -> :(x, :(y, z))
, :(+(x, y), z) -> +(:(x, z), :(y, z)) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^2))