boolean-algebra
In boolean algebra I found the rules like the redundancy theorem and de morgan's law, a little unintuitive.
Although the truth table shows it all, I wonder if the rules were made by experimenting like this in the first place, and if not, how do I prove such rules without using the truth table?
Use the identities: \$a+\bar{a} = 1\$ and \$a+1 = 1\$
\$\bar{x}\bar{y}\bar{z}+x\bar{z}+xy\$
\$\bar{x}\bar{y}\bar{z}+x\bar{z}(y+\bar{y})+xy\$
\$\bar{x}\bar{y}\bar{z}+x\bar{z}y+x\bar{z}\bar{y}+xy\$
\$\bar{x}\bar{y}\bar{z}+x\bar{z}\bar{y}+x\bar{z}y+xy\$
\$(\bar{x}+x)\bar{y}\bar{z}+(\bar{z}+1)xy\$
\$\bar{y}\bar{z}+xy\$
Actally, it is very simple:
f = (y XOR z) XOR x
simulate this circuit – Schematic created using CircuitLab
This just by looking at the thruth table and realzing that the bottom half is the inverse of thetop half.
Here's the math using boolean calculus.
We know x ^ y = x'y + xy' (^ is the XOR operator). Then:
g = a'b'c + abc + ab'c' + a'bc'
g = (a'b' + ab)c + (ab' + a'b)c'
^using a for X and b' for Y
g = (a ^ b')c + (a ^ b)c'
Now we have (a ^ b') and (a ^ b), which may seem not be possible to match, unless you realise that due to the nature of the XOR, (a ^ b') = (a ^ b)' [I'll leave that as an exercise to the reader]. Thus
g = (a ^ b)'c + (a ^ b)c'
g = (a ^ b) ^ c
This is only possible due to the special nature of the original equation. If you need an AND or OR you will find it impossible to reduce the function like this.
Best Answer
Since de Morgan did his work at a time when "logic" was in upheaval, we might consider what he used: Venn Diagrams.