162 lines
3.6 KiB
Markdown
162 lines
3.6 KiB
Markdown
---
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tags: [propositional-logic, algebra, nand-to-tetris]
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---
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# Boolean algebra
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## Algebraic laws
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Many of the laws that obtain in the mathematical realm of algebra also obtain
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for Boolean expressions.
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### The Commutative Law
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$$
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x \land y = y \land x \\
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$$
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$$
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x \lor y = y \lor x
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$$
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Compare the
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[Commutative Law](Whole_numbers.md#the-commutative-property)
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in the context of arithmetic.
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### The Associative Law
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$$
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x \land (y \land z) = (x \land y) \land z
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$$
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$$
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x \lor (y \lor z) = (x \lor y) \lor z
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$$
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Compare the
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[Associative Law](Whole_numbers.md#the-associative-property)
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in the context of arithmetic.
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### The Distributive Law
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$$
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x \land (y \lor z) = (x \land y) \lor (x \land z)
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$$
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$$
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x \lor (y \land z) = (x \lor y) \land (x \lor z)
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$$
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Compare how the
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[Distributive Law applies in the case of algebra based on arithmetic](Distributivity.md):
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$$
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a \cdot (b + c) = a \cdot b + a \cdot c
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$$
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### Double Negation Elimination
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$$
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\lnot \lnot x = x
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$$
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### Idempotent Law
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$$
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x \land x = x
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$$
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> Combining a quantity with itself either by logical addition or logical
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> multiplication will result in a logical sum or product that is the equivalent
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> of the quantity
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### DeMorgan's Laws
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In addition we have
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[DeMorgan's Laws](DeMorgan's_Laws.md) which express
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the relationship that obtains between the negations of conjunctive and
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disjunctive expressions:
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$$
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\lnot(x \land y) = \lnot x \lor \lnot y
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$$
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$$
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\lnot (x \lor y) = \lnot x \land \lnot y
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$$
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## Applying the laws to simplify complex Boolean expressions
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Say we have the following expression:
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$$
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\lnot(\lnot(x) \land \lnot (x \lor y))
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$$
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We can employ DeMorgan's Laws to convert the second conjunct to a different
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form:
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$$
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\lnot (x \lor y) = \lnot x \land \lnot y
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$$
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So now we have:
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$$
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\lnot(\lnot(x) \land (\lnot x \land \lnot y ))
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$$
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As we have now have an expression of the form _P and (Q and R)_ we can apply the
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Distributive Law to simplify the brackets (_P and Q and R_):
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$$
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\lnot( \lnot(x) \land \lnot(x) \land \lnot(y))
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$$
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Notice that we are repeating ourselves in this reformulation. We have
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$\lnot(x) \land \lnot(x)$ but this is just the same $\lnot(x)$ by the principle
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of **idempotence**. So we can reduce to:
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$$
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\lnot(\lnot(x) \land \lnot(y))
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$$
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This gives our expression the form of the first DeMorgan Law
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($\lnot (P \land Q)$), thus we can apply the law ($\lnot P \lor \lnot Q$) to
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get:
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$$
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\lnot(\lnot(x)) \lor \lnot(\lnot(y))
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$$
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Of course now we have two double negatives. We can apply the double negation law
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to get:
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$$
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x \lor y
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$$
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### Truth table
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Whenever we simplify an algebraic expression the value of the resulting
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expression should match that of the complex expression. We can demonstrate this
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with a truth table:
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| $x$ | $y$ | $\lnot(\lnot(x) \land \lnot (x \lor y))$ | $x \lor y$ |
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| --- | --- | ---------------------------------------- | ---------- |
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| 0 | 0 | 0 | 0 |
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| 0 | 1 | 1 | 1 |
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| 1 | 0 | 1 | 1 |
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| 1 | 1 | 1 | 1 |
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### Significance for computer architecture
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The fact that we can take a complex Boolean function and reduce it to a simpler
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formulation has great significance for the development of computer
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architectures, specifically
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[logic gates](Logic_gates.md). It
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would be rather resource intensive and inefficient to create a gate that is
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representative of the complex function. Whereas the simplified version only
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requires a single
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[OR gate](Logic_gates.md#or-gate).
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