Disclaimer
This is not some super expert guide.
If you "know" boolean logic this will not do anything for you (I'm sorry).
It is meant to help out people playing this game, who are not too knowledgable regarding boolean logic.
Notation
Following, I will use these symbols:
∧ - AND
∨ - OR
¬ - NOT
⊼ - NAND
⊽ - NOR
[a NAND b] is equals to combining a and b with AND and negating the whole term:
a ⊼ b = ¬ (a ∧ b)
The same is true for NOR regarding OR:
a ⊽ b = ¬ (a ∨ b)
De Morgan's Law
De Morgan's Law allows replacing AND and OR with each other under specific circumstances.
¬ (a ∧ b) = ¬a ∨ ¬b
¬ (a ∨ b) = ¬a ∧ ¬b
So, if you have [a AND b] negated as a whole [NOT (a AND b)] you can rewrite it by "dragging" the negation into the term (make a ¬a and b ¬b) and replacing AND with OR.
This applies to [a OR b] as well (you just have to replace the OR with AND instead).
As you might have noticed ¬ (a ∧ b) is nothing else than a NAND b.
That means:
a ⊼ b = ¬a ∨ ¬b
a ⊽ b = ¬a ∧ ¬b
You can of course verify this by booting up Turing Complete and placing down both [a NAND b] and [(NOT a) or (NOT b)].
Reducing Gate Count Using De Morgan's Law
As you probably know double negation has no effect.
¬¬ a = a
You can use this in conjunction with De Morgan's Law to reduce your gate count, if applicable.
Just have a closer look any time you use a lot of NOT-gates - you might be able to get rid of some of them.
For example ¬(¬a ∨ ¬b) uses 4 gates.
¬(¬a ∨ ¬b) = (¬¬a ∧ ¬¬b) = (a ∧ b)
Now it only uses a single AND gate.
[¬a ∨ ¬b] uses 3 gates.
But you can replace all 3 of those gates with just a single NAND:
[¬a ∨ ¬b] = [¬ (a ∧ b)] = [a ⊼ b]
(see Morgan's Law section, this is the same term as the definition there).
Found An Error?
If you found any errors or alike, please leave a comment.
Honestly, even misspellings.
Source: https://steamcommunity.com/sharedfiles/filedetails/?id=2623974593
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