logic - In boolean algebra, why is a+a'b = a+b? - Mathematics …
Sep 18, 2016 · In boolean algebra, why is a+a'b = a+b? [duplicate] Ask Question Asked 9 years, 2 months ago Modified 5 years, 10 months ago
If A = B, then B = A... Not Always True? Definition of
Jun 22, 2014 · So, the statement "if A = B, then B = A" is always (assumed to be) true in math, but it's not always true when applied to the meaning of words, which is what your friend is …
absolute value - Proof of $ |a-b| = |b-a| $ - Mathematics Stack …
May 22, 2015 · While working out some intriguing qualities of absolute values for my studies of calculus, I frequently used the formula below. I know that the formula below is clearly correct …
When is $A- B = B- A$? - Mathematics Stack Exchange
JFYI: $B\backslash A$ is a more common notation for $B-A$, in case you ever encounter it.
Use Algebric proof to prove the following set A- (A-B) = A ∩ B
Apr 9, 2018 · It's certainly possible, though you need additional laws such as De Morgan's law, as well the fact that $ (B^c)^c = B$.
logic - Why is "A only if B" equivalent to " (not A) or B ...
Mar 9, 2017 · The only way to contradict "$A$ only if $B$" is to find $A$ being true when $B$ is false. But that is also the only way to contradict " (not $A$) or $B$"
Showing that (A⊕B)⊕B is equal to A - Mathematics Stack Exchange
Jan 18, 2019 · Proofing associativity is one way to approach this. B⊕B cancels out, so even with A containing elements in B, it will still be A in the end.
Simplification of: AB + A'C + BC in boolean algebra
Sep 6, 2016 · I am trying to understand the simplification of the boolean expression: AB + A'C + BC I know it simplifies to A'C + BC And I understand why, but I cannot figure out how to …
Prove that (A ∩ B) ⊆ A, when A and B are sets.
Jul 3, 2019 · You are right! Straight-forward, direct from definition proof! Sometimes, when we talk about this "advanced" mathematical subjects, we expect proofs to be long, complex and …
Find all solutions of $a^b = b^a$ - Mathematics Stack Exchange
One solution is $a=2$ and $b=4$ (should be pretty easy to prove that this is the only solution where $a,b\in\mathbb {N}$ and $a<b$).