Results for query "  AB  x C   "

Boolean algebra: (A'+B)(A+C)

I have a little problem with this expression:

x = (A'+B)(A+C)

I know it can be simplified to:


since ive used some software to simplify it, but i simply can't see how it is done.This is what i've done so far:

(A'+B)(A+C) =>
A'A + AB + A'C + BC =>
0 + AB + A'C + BC =>
AB + A'C + BC

I just fail to see how i can do this differently and get to the correct result.

Examples of enriched categories which are (co)powered or (co)tensored

We have for each $G: C \to \textbf{Ab}$ an isomorphism $\textbf{Ab}^C(G, F^A) \cong \textbf{Ab}(A, \textbf{Ab}^C(G, F))$ since

\textbf{Ab}^C(G, F^A) & \cong & \int_c \textbf{Ab}(Gc, F^A(c)) \\
& \cong & \int_c \textbf{Ab}(Gc, \textbf{Ab}(A, Fc)) \\
& \cong & \int_c \textbf{Ab}(Gc \otimes A, Fc) \\
& \cong & \int_c \textbf{Ab}(A, \textbf{Ab}(Gc, Fc)) \\
& \cong & \textbf{Ab}(A, \int_c \textbf{Ab}(Gc, Fc) \\
& \cong & \textbf{Ab}(A, \textbf{Ab}^C(G, F)).For we have a universal element $\mathbb{Z} \to \textbf{Ab}(A, \textbf{Ab}^C(F^A, F))$ and a map

$$\textbf{Ab}^C(F^A, F) \to \textbf{Ab}^C(\sigma(F^A), \sigma(F)).

An example for a construction on monads/operads?

A monad $(X, A)$ in $C$ is defined as a monoid object $(A : X \rightarrow X, p^A : AA \rightarrow A, e^A: 1_X \rightarrow A)$ in $C(X, X)$.Given two monads $(X, A)$ and $(X, B)$, a distibutive law is a 2-cell $d : BA \rightarrow AB$ satisfying few axioms.

An Entropy Inequality

As I mentioned in the comment, the key to the solution is the inequality
(1+a)\log(1+b)+(1+b)\log(1+a)\ge 2(1+c)\log(1+c)
where $a,b,c>0, ab=c^2$.Dividing by $XY$ and putting $X^{-1}=1+x, Y^{-1}=1+y$, so that $\frac{1-X}X=x,\frac{1-Y}Y=y$, we see that this case reduces exactly to the above inequality.

power log distance between matrices

Then their singular values satisfy the majorization
\log \sigma(AB) - \log \sigma(B) \prec \log\sigma(A),
where $s(X)$ denotes the vector of singular values of an operator $X$.Then,
d(A,B) \le d(A,C) + d(B,C).

Injective objects in Mor(Ab)

Consider the abelian (Grothendieck) category $\mathcal{C} := \mathrm{Fun}(\{0<1\},\mathrm{Ab}) = \mathrm{Mor}(\mathrm{Ab})$.I will use notation $A_0 \to A_1$ for objects of $\mathrm{Mor}(\mathrm{Ab})$.

Why are ring actions much harder to find than group actions?

And of course, a group $G$ acting on $X$ is just a group homomorphism $G \to \text{Aut }X$.In fact, any ordinary category $C$ has a free $\text{Ab}$-enriched category $\mathbb{Z}[C]$ equipped with the universal functor from $C$ to an $\text{Ab}$-enriched category.

c^2 = a^2 + b^2 + ab and its solutions

The equation $c^2 = a^2 + b^2 + ab$ is the law of cosines for a triangle with integer sides $a$, $b$, and $c$, and a 120 degree angle opposite side $c$.By the substitution $x = (a-b)/2$, $y = (a+b)/2$ it can be transformed to $x^2 + 3y^2 = 4z^2$, which is a more familiar equation, whose solutions are given in parametric form on page 353 (Corollary 6.

Cohomological functor from triangulated category

Say we have a cohomological functor F from a triangulated category $C$ to the category $Ab$ of abelian groups, e.Given two objects $x, y$ in $C$, can we find a complex that's like $RHom(x,y)$ whose cohomology computes $Ext_C^i(x,y):=Hom_C(x,y[i])$?