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Autosave: 2022-12-28 10:30:06

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thomasabishop 2022-12-28 10:30:06 +00:00
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Logic/Propositional_logic/Boolean_function_synthesis.md
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@ -29,7 +29,7 @@ Let's say we have the following truth table:
| Line | $x$ | $y$ | $z$ | $f$ | | Line | $x$ | $y$ | $z$ | $f$ |
| ---- | --- | --- | --- | --- | | ---- | --- | --- | --- | --- |
| 1 | 0 | 0 | 0 | 1 | | 1 | 0 | 0 | 0 | 0 |
| 2 | 0 | 0 | 1 | 0 | | 2 | 0 | 0 | 1 | 0 |
| 3 | 0 | 1 | 0 | 1 | | 3 | 0 | 1 | 0 | 1 |
| 4 | 0 | 1 | 1 | 0 | | 4 | 0 | 1 | 1 | 0 |
@ -57,7 +57,7 @@ For each line we construct a Boolean expression that would result in the value i
We can now join each expression to create a complex expression that covers the entire truth table using OR: We can now join each expression to create a complex expression that covers the entire truth table using OR:
$$ $$
(\lnot(x) \land \lnot (y) \land \lnot(z)) \lor (\lnot(x) \land y \land \lnot(z)) \lor (x \land \lnot(y) \land \lnot(z)) (\lnot(x) \land \lnot (y) \land \lnot(z)) \\ \lor \\ (\lnot(x) \land y \land \lnot(z)) \\ \lor \\ (x \land \lnot(y) \land \lnot(z))
$$ $$
It's clear that we have transcribed the truth conditions accurately but that we are doing so in a rather verbose way. We can simplify by just looking at the position of the 1s in the truth table. Notice: It's clear that we have transcribed the truth conditions accurately but that we are doing so in a rather verbose way. We can simplify by just looking at the position of the 1s in the truth table. Notice: