We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ *(x, *(y, z)) -> *(otimes(x, y), z)
, *(x, oplus(y, z)) -> oplus(*(x, y), *(x, z))
, *(1(), y) -> y
, *(+(x, y), z) -> oplus(*(x, z), *(y, z)) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
We add the following weak dependency pairs:
Strict DPs:
{ *^#(x, *(y, z)) -> c_1(*^#(otimes(x, y), z))
, *^#(x, oplus(y, z)) -> c_2(*^#(x, y), *^#(x, z))
, *^#(1(), y) -> c_3(y)
, *^#(+(x, y), z) -> c_4(*^#(x, z), *^#(y, z)) }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ *^#(x, *(y, z)) -> c_1(*^#(otimes(x, y), z))
, *^#(x, oplus(y, z)) -> c_2(*^#(x, y), *^#(x, z))
, *^#(1(), y) -> c_3(y)
, *^#(+(x, y), z) -> c_4(*^#(x, z), *^#(y, z)) }
Strict Trs:
{ *(x, *(y, z)) -> *(otimes(x, y), z)
, *(x, oplus(y, z)) -> oplus(*(x, y), *(x, z))
, *(1(), y) -> y
, *(+(x, y), z) -> oplus(*(x, z), *(y, z)) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
No rule is usable, rules are removed from the input problem.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ *^#(x, *(y, z)) -> c_1(*^#(otimes(x, y), z))
, *^#(x, oplus(y, z)) -> c_2(*^#(x, y), *^#(x, z))
, *^#(1(), y) -> c_3(y)
, *^#(+(x, y), z) -> c_4(*^#(x, z), *^#(y, z)) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following constant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}, Uargs(c_4) = {1, 2}
TcT has computed the following constructor-restricted matrix
interpretation.
[*](x1, x2) = [0]
[0]
[otimes](x1, x2) = [0]
[0]
[1] = [2]
[0]
[+](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[oplus](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 1] [0 1] [0]
[*^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2]
[1 0] [1 0] [0]
[c_1](x1) = [1 0] x1 + [2]
[0 1] [0]
[c_2](x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 1] [0 1] [2]
[c_3](x1) = [0]
[0]
[c_4](x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 1] [0 1] [2]
The order satisfies the following ordering constraints:
[*^#(x, *(y, z))] = [0 0] x + [2]
[1 0] [0]
? [0 0] z + [4]
[1 0] [0]
= [c_1(*^#(otimes(x, y), z))]
[*^#(x, oplus(y, z))] = [0 0] x + [0 0] y + [0 0] z + [2]
[1 0] [1 0] [1 0] [0]
? [0 0] x + [0 0] y + [0 0] z + [6]
[2 0] [1 0] [1 0] [2]
= [c_2(*^#(x, y), *^#(x, z))]
[*^#(1(), y)] = [0 0] y + [2]
[1 0] [2]
> [0]
[0]
= [c_3(y)]
[*^#(+(x, y), z)] = [0 0] x + [0 0] y + [0 0] z + [2]
[1 0] [1 0] [1 0] [0]
? [0 0] x + [0 0] y + [0 0] z + [6]
[1 0] [1 0] [2 0] [2]
= [c_4(*^#(x, z), *^#(y, z))]
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^2)).
Strict DPs:
{ *^#(x, *(y, z)) -> c_1(*^#(otimes(x, y), z))
, *^#(x, oplus(y, z)) -> c_2(*^#(x, y), *^#(x, z))
, *^#(+(x, y), z) -> c_4(*^#(x, z), *^#(y, z)) }
Weak DPs: { *^#(1(), y) -> c_3(y) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.
DPs:
{ 3: *^#(+(x, y), z) -> c_4(*^#(x, z), *^#(y, z))
, 4: *^#(1(), y) -> c_3(y) }
Sub-proof:
----------
The following argument positions are considered usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}, Uargs(c_4) = {1, 2}
TcT has computed the following constructor-restricted polynomial
interpretation.
[*](x1, x2) = 1 + 2*x2
[otimes](x1, x2) = 0
[1]() = 0
[+](x1, x2) = 1 + x1 + x2
[oplus](x1, x2) = 1 + x1 + x2
[*^#](x1, x2) = 1 + 2*x1 + 2*x1*x2 + x2
[c_1](x1) = 2*x1
[c_2](x1, x2) = x1 + x2
[c_3](x1) = 0
[c_4](x1, x2) = x1 + x2
This order satisfies the following ordering constraints.
[*^#(x, *(y, z))] = 2 + 4*x + 4*x*z + 2*z
>= 2 + 2*z
= [c_1(*^#(otimes(x, y), z))]
[*^#(x, oplus(y, z))] = 2 + 4*x + 2*x*y + 2*x*z + y + z
>= 2 + 4*x + 2*x*y + y + 2*x*z + z
= [c_2(*^#(x, y), *^#(x, z))]
[*^#(1(), y)] = 1 + y
>
= [c_3(y)]
[*^#(+(x, y), z)] = 3 + 2*x + 2*y + 3*z + 2*x*z + 2*y*z
> 2 + 2*x + 2*x*z + 2*z + 2*y + 2*y*z
= [c_4(*^#(x, z), *^#(y, z))]
The strictly oriented rules are moved into the weak component.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ *^#(x, *(y, z)) -> c_1(*^#(otimes(x, y), z))
, *^#(x, oplus(y, z)) -> c_2(*^#(x, y), *^#(x, z)) }
Weak DPs:
{ *^#(1(), y) -> c_3(y)
, *^#(+(x, y), z) -> c_4(*^#(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.
DPs:
{ 2: *^#(x, oplus(y, z)) -> c_2(*^#(x, y), *^#(x, z))
, 3: *^#(1(), y) -> c_3(y)
, 4: *^#(+(x, y), z) -> c_4(*^#(x, z), *^#(y, z)) }
Sub-proof:
----------
The following argument positions are considered usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}, Uargs(c_4) = {1, 2}
TcT has computed the following constructor-restricted polynomial
interpretation.
[*](x1, x2) = x2
[otimes](x1, x2) = 0
[1]() = 0
[+](x1, x2) = 2 + x1 + x2
[oplus](x1, x2) = 2 + x1 + x2
[*^#](x1, x2) = 1 + x1 + x1*x2 + x2
[c_1](x1) = x1
[c_2](x1, x2) = x1 + x2
[c_3](x1) = 0
[c_4](x1, x2) = x1 + x2
This order satisfies the following ordering constraints.
[*^#(x, *(y, z))] = 1 + x + x*z + z
>= 1 + z
= [c_1(*^#(otimes(x, y), z))]
[*^#(x, oplus(y, z))] = 3 + 3*x + x*y + x*z + y + z
> 2 + 2*x + x*y + y + x*z + z
= [c_2(*^#(x, y), *^#(x, z))]
[*^#(1(), y)] = 1 + y
>
= [c_3(y)]
[*^#(+(x, y), z)] = 3 + x + y + 3*z + x*z + y*z
> 2 + x + x*z + 2*z + y + y*z
= [c_4(*^#(x, z), *^#(y, z))]
The strictly oriented rules are moved into the weak component.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs: { *^#(x, *(y, z)) -> c_1(*^#(otimes(x, y), z)) }
Weak DPs:
{ *^#(x, oplus(y, z)) -> c_2(*^#(x, y), *^#(x, z))
, *^#(1(), y) -> c_3(y)
, *^#(+(x, y), z) -> c_4(*^#(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.
DPs:
{ 1: *^#(x, *(y, z)) -> c_1(*^#(otimes(x, y), z))
, 2: *^#(x, oplus(y, z)) -> c_2(*^#(x, y), *^#(x, z))
, 3: *^#(1(), y) -> c_3(y)
, 4: *^#(+(x, y), z) -> c_4(*^#(x, z), *^#(y, z)) }
Sub-proof:
----------
The following argument positions are considered usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}, Uargs(c_4) = {1, 2}
TcT has computed the following constructor-restricted polynomial
interpretation.
[*](x1, x2) = 2 + 2*x1 + 2*x1*x2 + 2*x1^2 + 2*x2 + 2*x2^2
[otimes](x1, x2) = 1
[1]() = 0
[+](x1, x2) = 2 + x1 + x2
[oplus](x1, x2) = 1 + x1 + x2
[*^#](x1, x2) = 1 + x1 + 2*x1*x2 + 2*x2
[c_1](x1) = 2 + x1
[c_2](x1, x2) = x1 + x2
[c_3](x1) = 0
[c_4](x1, x2) = x1 + x2
This order satisfies the following ordering constraints.
[*^#(x, *(y, z))] = 5 + 5*x + 4*x*y + 4*x*y*z + 4*x*y^2 + 4*x*z + 4*x*z^2 + 4*y + 4*y*z + 4*y^2 + 4*z + 4*z^2
> 4 + 4*z
= [c_1(*^#(otimes(x, y), z))]
[*^#(x, oplus(y, z))] = 3 + 3*x + 2*x*y + 2*x*z + 2*y + 2*z
> 2 + 2*x + 2*x*y + 2*y + 2*x*z + 2*z
= [c_2(*^#(x, y), *^#(x, z))]
[*^#(1(), y)] = 1 + 2*y
>
= [c_3(y)]
[*^#(+(x, y), z)] = 3 + x + y + 6*z + 2*x*z + 2*y*z
> 2 + x + 2*x*z + 4*z + y + 2*y*z
= [c_4(*^#(x, z), *^#(y, z))]
The strictly oriented rules are moved into the weak component.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak DPs:
{ *^#(x, *(y, z)) -> c_1(*^#(otimes(x, y), z))
, *^#(x, oplus(y, z)) -> c_2(*^#(x, y), *^#(x, z))
, *^#(1(), y) -> c_3(y)
, *^#(+(x, y), z) -> c_4(*^#(x, z), *^#(y, z)) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(1))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ *^#(x, *(y, z)) -> c_1(*^#(otimes(x, y), z))
, *^#(x, oplus(y, z)) -> c_2(*^#(x, y), *^#(x, z))
, *^#(1(), y) -> c_3(y)
, *^#(+(x, y), z) -> c_4(*^#(x, z), *^#(y, z)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Rules: Empty
Obligation:
runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^2))