Talk:High School Mathematics Extensions/Logic/Problem Set/Solutions

That 203.88.166.38 is me, Tom Lam or lemontea(it keeps kicking me out while I'm editing)

Why is there no boarder with table?

Problem on the solution
On the questions about NAND, I've find the answer that's simplier,using only six NAND operators.

{[(x NAND x) NAND (y NAND y)] NAND w} NAND (z NAND z)

Solution: Let (x+y)w+z = a NAND b, where a and b can either be one of x,y,w,z or another NAND operator.

$$(x+y)w+z=(ab)'$$

$$(x+y)w+z=a'+b'$$

Therefore $$a=[(x+y)w]'$$ and $$b=z'$$, both need further NAND operators. Let a = c NAND d, and let b = e NAND f.

$$(x+y)w+z={(cd)'}^{'}+{(ef)'}^{'}$$

$$(x+y)w+z=cd+ef$$

Therefore d=w, e=f=z,c=x+y.Let c = g NAND h.

$$x+y=(gh)'$$

$$x+y=g'+h'$$

Now g=x' and h=y', we still need more NAND operators. Let g = i NAND j and let h = k NAND l.

$$(ij)'=x'$$ $$x=ij$$ $$(kl)'=y'$$ $$kl=y$$

Therefore i=j=x and k=l=y.

Now substitute all the variables back, you should get:

(x+y)w+z={[(x NAND x) NAND (y NAND y)] NAND w} NAND (z NAND z)

Thanks.


 * Hey, tops! Did you figure this out yourself? Really good. Xiaodai 09:08, 20 Nov 2004 (UTC)
 * Actually i didn't read the solutions. The one in there is a shocker. Xiaodai 09:14, 20 Nov 2004 (UTC)
 * Hi!Yes, I did figure this out myself, but I just want to ask: what do you mean by "shocker" and "i didn't read the solutions." in your last comment? I'm completely confused.Lemontea 19:25, 20 Nov 2004 (GMT+8)