

As far as BA + BC + C = BA + C(B + 1), that's really simple factoring.
Look at it as 1BC + 1C. Factor out the C.
C(1B + 1).
And clearly the coefficient 1 before the B is extraneous, so it's not written.

1) Yes
2) See above post, just factoring.
3) So this is an AND gate with inversions on all inputs and outputs?
Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'
Then you have a NOR with an inverted output which is the same as OR.

China7058 Posts
I really don't see how BA + BC + C = BA + C(B + 1) as factoring, I mean..
x^2 + 4x + 4 would be (x+2)(x+2) ...
Can you explain further, I really don't see it.

In general C(A+B)=CA+CB for any three numbers A,B,C. This is called the distributive law of multiplication.
Plug in A=1, and you get C(1+B)=C(1)+CB=C+CB

Forget the BA for a moment.
BC + C = C(B + 1)

China7058 Posts
On April 21 2008 11:38 fight_or_flight wrote: 1) Yes
2) See above post, just factoring.
3) So this is an AND gate with inversions on all inputs and outputs?
Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'
Then you have a NOR with an inverted output which is the same as OR. For number 3, so it's still (AB)'? Because there's an inverter in the input which makes it:
A' B' = AB? It doesn't work that way?

On April 21 2008 11:38 Raithed wrote: I really don't see how BA + BC + C = BA + C(B + 1) as factoring, I mean..
x^2 + 4x + 4 would be (x+2)(x+2) ...
Can you explain further, I really don't see it. x^2 + 4x + 4 = (x+2)(x+2) x^3 + x^2 + 4x + 4 = x^3+(x+2)(x+2)
BC + C = C(B+1) BA + BC + C = BA + C(B+1)

China7058 Posts
On April 21 2008 11:40 Muirhead wrote: In general C(A+B)=CA+CB for any three numbers A,B,C. This is called the distributive law of multiplication.
Plug in A=1, and you get C(1+B)=C(1)+CB=C+CB Oh, wow, nice explanation. But what if you don't know the answer, would you always plug A = 1?

On April 21 2008 11:46 Raithed wrote:Show nested quote +On April 21 2008 11:40 Muirhead wrote: In general C(A+B)=CA+CB for any three numbers A,B,C. This is called the distributive law of multiplication.
Plug in A=1, and you get C(1+B)=C(1)+CB=C+CB Oh, wow, nice explanation. But what if you don't know the answer, would you always plug A = 1? Lol, you kinda missed the point.
He was using an example to demonstrate the factorization of BC + C = C(B + 1).
He pretended that the 1 in the above example was an A, which would have been written as:
BC + CA = C(B + A).
Exact same problem, just with another variable.

On April 21 2008 11:43 Raithed wrote:Show nested quote +On April 21 2008 11:38 fight_or_flight wrote: 1) Yes
2) See above post, just factoring.
3) So this is an AND gate with inversions on all inputs and outputs?
Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'
Then you have a NOR with an inverted output which is the same as OR. For number 3, so it's still (AB)'? Because there's an inverter in the input which makes it: A' B' = AB? It doesn't work that way? Does not work that way. Look at the truth tables:
out = A'B'
A B out 0 0 1 0 1 0 1 0 0 1 1 0

out = AB
A B out 0 0 0 0 1 0 1 0 0 1 1 1

Therefore you conclude that A'B' = NOR. (look up demorgan's theorem)

On April 21 2008 11:53 fight_or_flight wrote:Show nested quote +On April 21 2008 11:43 Raithed wrote:On April 21 2008 11:38 fight_or_flight wrote: 1) Yes
2) See above post, just factoring.
3) So this is an AND gate with inversions on all inputs and outputs?
Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'
Then you have a NOR with an inverted output which is the same as OR. For number 3, so it's still (AB)'? Because there's an inverter in the input which makes it: A' B' = AB? It doesn't work that way? Does not work that way. Look at the truth tables: out = A'B' A B out 0 0 1 0 1 0 1 0 0 1 1 0  out = AB A B out 0 0 0 0 1 0 1 0 0 1 1 1  Therefore you conclude that A'B' = NOR. (look up demorgan's theorem) Yeah, this is exactly accurate. Not A AND Not B evaluates as true in very different circumstances than A AND B evaluates as true.

China7058 Posts
Without looking at the truth table, I thought not from input, with a not in output would cancel, this just confirms that it's NOR. I didn't mean exactly as (AB)' == AB, I just meant that the ' cancels.

No, they don't cancel. If you have 2 bubbles on the input of an AND, you can make it into an OR with a bubble on the output.
Same thing if the inputs on an OR have bubbles...it becomes an AND with a bubble on the output.

China7058 Posts
So basically, the bubbles DON'T cancel, whatsoever?

China7058 Posts
But wait, the gate is a NAND gate...


China7058 Posts
So if it's
Bubbled input NAND  inverter  bubbles cancel? Only this way?

China7058 Posts
Which is basically
NAND + inverter = comes out as AB? Instead of (AB)' ?




