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Logic/Propositional_logic/Boolean_function_synthesis.md
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@ -7,9 +7,9 @@ tags: [logic, propositional-logic, nand-to-tetris]
# Boolean function synthesis
When we looked at [boolean functions](/Logic/Propositional_logic/Boolean_functions.md) we were working in a particular direction: from a function to a truth table. When we do Boolean function synthesis we work in the opposite direction: from a function to a truth table.
When we looked at [boolean functions](/Logic/Propositional_logic/Boolean_functions.md) we were working in a particular direction: from a function to a truth table. When we do Boolean function synthesis we work in the opposite direction: from a truth table to a function.
This is an important skill that we will use when constructing [logic circuits](/Electronics_and_Hardware/Digital_circuits/Digital_circuits.md). We will go from truth conditions (i.e. what we want the circuit to do and when we want it to do it) to a function expression which is then reduced and implemented with [logic gates](/Electronics_and_Hardware/Digital_circuits/Logic_gates.md).
This is an important skill that we will use when constructing [logic circuits](/Electronics_and_Hardware/Digital_circuits/Digital_circuits.md). We will go from truth conditions (i.e. what we want the circuit to do and when we want it to do it) to a function expression which is then reduced to its simplest form and implemented with [logic gates](/Electronics_and_Hardware/Digital_circuits/Logic_gates.md).
## The process
@ -52,7 +52,7 @@ For each line we construct a Boolean expression that would result in the value i
| 3 | 0 | 1 | 0 | $ \lnot(x) \land y \land \lnot(z) $ |
| 5 | 1 | 0 | 0 | $ x \land \lnot(y) \land \lnot(z) $ |
We can now join each expression to create a complex expression that covers the entire truth table. Since 1 will be output for any one of these sub-expressions we can just join them up with OR:
We can now join each expression to create a complex expression that covers the entire truth table using OR:
$$
(\lnot(x) \land \lnot (y) \land \lnot(z)) \lor (\lnot(x) \land y \land \lnot(z)) \lor (x \land \lnot(y) \land \lnot(z))
@ -69,7 +69,7 @@ $$
(\lnot(x) \land \lnot(z)) \lor (\lnot(y) \land \lnot(z))
$$
Notice that $\lnot(z)$ is repeated so we can remove the repetition:
Notice that $\lnot(z)$ is repeated so we can remove the repetition through [idempotence](/Logic/Propositional_logic/Boolean_algebra.md#idempotent-law):
$$
\lnot z \land (\lnot(x) \lor \lnot(y))