[SOLVED] Boolean Algebra

**starman_dx** · February 27th 2009, 01:10 AM

Assume that B is a Boolean algebra with operations + and ·.
For each true statement below, give a detailed proof. For each false statement below, write
out its negation, then give a proof of the negation.

(∀a, b ∈ B) (a + b = 1 ↔ b · ā = ā)
(∀a, b ∈ B) (a · b = 0 ↔ b + ā = ā)
(∀a, b ∈ B) (a + b = a + c → b = c)

for the last one, i had the same proof as http://www.mathhelpforum.com/math-he...b-b-c-b-c.html
but it just doesn't seem right to me...
completely stumped on the first two

.

please help!!!!
thanks!!!

**Grandad** · February 27th 2009, 02:51 AM

Hello starman_dx

Originally Posted by starman_dx

Assume that B is a Boolean algebra with operations + and ·.
For each true statement below, give a detailed proof. For each false statement below, write
out its negation, then give a proof of the negation.

(∀a, b ∈ B) (a + b = 1 ↔ b · ā = ā)
(∀a, b ∈ B) (a · b = 0 ↔ b + ā = ā)
(∀a, b ∈ B) (a + b = a + c → b = c)

for the last one, i had the same proof as http://www.mathhelpforum.com/math-he...b-b-c-b-c.html
but it just doesn't seem right to me...
completely stumped on the first two

.

please help!!!!
thanks!!!

Here are the proofs you need. I'll leave you to supply the names of the Laws I'm using at each stage.

(1) $a+b= 1$

$\rightarrow (a+b)\cdot \bar{a} = 1\cdot\bar{a}$

$\rightarrow a\cdot\bar{a} +b\cdot\bar{a} = \bar{a}$

$\rightarrow 0+b\cdot\bar{a} = \bar{a}$

$\rightarrow b\cdot\bar{a} = \bar{a}$

And $b\cdot\bar{a} = \bar{a}$

$\rightarrow b\cdot\bar{a} + a = \bar{a}+a$

$\rightarrow (b+a)\cdot(\bar{a}+a) = 1$

$\rightarrow (a+b).1 = 1$

$\rightarrow a+b = 1$

(2) $a\cdot b=0$

$\rightarrow a\cdot b + \bar{a} = 0 + \bar{a}$

$\rightarrow (a+\bar{a})\cdot(b+\bar{a}) = \bar{a}$

$\rightarrow 1\cdot(b+\bar{a}) = \bar{a}$

$\rightarrow b+\bar{a} = \bar{a}$

And $b+\bar{a} = \bar{a}$

$\rightarrow a\cdot(b+\bar{a}) = a\cdot\bar{a}$

$\rightarrow a\cdot b + a\cdot \bar{a} = 0$

$\rightarrow a\cdot b + 0 = 0$

$\rightarrow a\cdot b = 0$

(3) is not true. The negation is $\exists a, b, c \in B, (a+b = a+c) \wedge (b \ne c)$

As a counter-example, consider sets $A = \{1\}, B = \{2\}$ and $C = \{1, 2\}$ . Then $A \cup B = A \cup C$ , but $B \ne C$ .

Grandad

**starman_dx** · February 27th 2009, 06:19 AM

you are a life saver
thank you!!!!

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