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Autosave: 2022-12-20 07:30:07

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thomasabishop 2022-12-20 07:30:07 +00:00
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23
Logic/Propositional_logic/Boolean_function_synthesis.md
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@ -75,10 +75,27 @@ $$
\lnot z \land (\lnot(x) \lor \lnot(y))
$$
The upshot is that we now have a simpler expression that uses only NOT, OR and AND. We could therefore construct a circuit that just uses these gates to construct the conditions we specified in the first truth table.
The upshot is that we now have a simpler expression that uses only NOT, OR and AND. These are the fundamental logic gates thus we are able to construct a circuit that embodies the logic of the expression.
> This is important and is an instance of the general theorem that _any Boolean function_ can be represented using an expression containing AND, OR and NOT operations
But even this is too complex. We could get rid of the OR and just use AND and NOT, in other words, NAND:
But even this is too complex. We could get rid of the OR and just use AND and NOT:
stopped at 6:38
We can prove this theorem with the following transformation:
$$
x \lor y = \lnot(\lnot(x) \land \lnot(y))
$$
| $x$ | $y$ | $x \lor y$ | $\lnot(\lnot(x) \land \lnot(y)$ |
| --- | --- | ---------- | ------------------------------- |
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 |
Finally, we can simplify even further by doing away with AND and NOT and using a single [NAND gate](/Electronics_and_Hardware/Digital_circuits/Logic_gates.md#nand-gate) which embodies the logic of both, being true in all instances where AND would be false:
$$
\lnot (x \land y)
$$

2
Logic/Propositional_logic/Boolean_functions.md
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@ -19,7 +19,7 @@ Here is a work through where $f(1, 0, 1)$:
- The second disjunction: $x \land y$ is false because $x$ is 1 and $y$ is 1
- The overall function returns false because the main connective is disjunction and both of its disjuncts are false
We can compute all possible outputs of the function by constructing a [truth-table](/Logic/Propositional_logic/Truth-tables.md) with each possible variable as the truth conditions and the output of the function as the truth value:
We can compute all possible outputs of the function by constructing a [truth table](/Logic/Propositional_logic/Truth-tables.md) with each possible variable as the truth conditions and the output of the function as the truth value:
| $x$ | $y$ | $z$ | $f(x,y,z) = (x \land y) \lor (\lnot(x) \land z )$ |
| --- | --- | --- | ------------------------------------------------- |