# Boolean Algebra cont’ The digital abstraction מבנה המחשב + מבוא למחשבים ספרתיים תרגול 2#

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Boolean Algebra cont’ The digital abstraction מבנה המחשב + מבוא למחשבים ספרתיים תרגול 2#

Theorem: Absorption Law For every pair of elements a, b  B, 1. a + a · b = a 2. a · ( a + b ) = a Proof: (1) Identity Commutativity Distributivity Identity Theorem: For any a  B, a + 1 = 1 (2) duality.

Theorem: Associative Law In a Boolean algebra, each of the binary operations ( + ) and ( · ) is associative. That is, for every a, b, c  B, 1. a + ( b + c ) = ( a + b ) + c 2. a · ( b · c ) = ( a · b ) · c

Distributivity Commutativity Distributivity Absorption Law Idempotent Law Proof: (1) Let

Commutativity Distributivity Idempotent Law Absorption Law Commutativity Absorption Law

Putting it all together: Same transitions · before +

(2) Duality Also,

Theorem 11: DeMorgan’s Law For every pair of elements a, b  B, 1. ( a + b )’ = a’ · b’ 2. ( a · b )’ = a’ + b’ Proof: (1) We first prove that (a+b) is the complement of a’·b’. Thus, (a+b)’ = a’·b’ By the definition of the complement and its uniqueness, it suffices to show: (i) (a+b)+(a’b’) = 1 and (ii) (a+b)(a’b’) = 0. (2) Duality(a·b)’ = a’+b’

Distributivity Commutativity Associativity a’ and b’ are the complements of a and b respectively Theorem: For any a  B, a + 1 = 1 Idempotent Law

Commutativity Distributivity Commutativity Associativity Commutativity a’ and b’ are the complements of a and b respectively Theorem: For any a  B, a · 0 = 0 Idempotent Law

Algebra of Sets Consider a set S. B = all the subsets of S (denoted by P(S)). “plus”  set-union ∪ “times”  set-intersection ∩ Additive identity element – empty set Ø Multiplicative identity element – the set S. Complement of X  B:

Theorem: The algebra of sets is a Boolean algebra. Proof: By satisfying the axioms of Boolean algebra: B is a set of at least two elements For every non empty set S: →|B| ≥ 2. Closure of ( ∪ ) and (∩) over B (functions ).

A1. Cummutativity of ( ∪ ) and ( ∩ ). An element lies in the union precisely when it lies in one of the two sets X and Y. Equally an element lies in the union precisely when it lies in one of the two sets X and Y. Hence,

A2. Distributivity of ( ∪ ) and ( ∩ ). Let and or If,We haveand. Hence, If,We haveand. Hence, or

This can be conducted in the same manner as ⊆. We present an alternative way: Definition of intersection and Also, definition of intersection definition of union Similarly, * **

Taking (*) and (**) we get, Distributivity of union over intersection can be conducted in the same manner.

A3. Existence of additive and multiplicative identity element. A4. Existence of the complement. Algebra of sets is Boolean algebra.All axioms are satisfied

Dual transformation - Recursive definition: Dual: expressions → expressions base: 0 → 1 1 → 0 a → a, a  B\{0,1} recursion step: Let E 1 and E 2 be Boolean expressions. Then, E 1 ’ → [dual(E 1 )]’ ( E 1 + E 2 ) → [ dual(E 1 ) · dual(E 2 ) ] ( E 1 · E 2 ) → [ dual(E 1 ) + dual(E 2 ) ] Boolean expression - Recursive definition: base: 0, 1, a  B – expressions. recursion step: Let E 1 and E 2 be Boolean expressions. Then, E 1 ’ ( E 1 + E 2 ) ( E 1 · E 2 )

Proof: Let f ( x 1, x 2, …, x n ) be a Boolean expression. We show that applying the complement on the whole expression together with replacing each variable by it’s complement, yields the dual transformation definition. Let f d be the dual of a function f ( x 1, x 2, …, x n ) Lemma: In switching algebra, f d = f’ ( x 1 ’, x 2 ’, …, x n ’ ) Induction basis: 0, 1 – expressions.

Induction hypothesis: Lemma holds for Boolean expressions: E 1 and E 2. That is: Induction step: show that it is true for E 1 ’ ( E 1 + E 2 ) ( E 1 · E 2 ) If then,

If then, If then,

Definition: A function f is called self-dual if f = f d Lemma: For any function f and any two-valued variable A, the function g = Af + A’f d is a self-dual. Proof: (holds for any Boolean algebra) Dual definition Distributivity Commutativity

Notice that the above expression has the form: ab + a’c +bc where “a” =A, “b”=f, “c” = f d. Distributivity Commutativity A’ is the complement of A Identity Commutativity

We now prove a stronger claim: Identity a’ is the complement of a Distributivity Commutativity Distributivity Theorem: For any a  B, a + 1 = 1 Identity

For example: self-dual

Easier proof (1) for switching algebra only: (using dual properties) Switching algebraOR Identity

A = 0 0’ = 1 Identity Commutativity Absorption Law Theorem: For any a  B, a · 0 = 0 Identity Easier proof (2) for switching algebra only: (case analysis)

A = 1

Example of a transfer function for an inverter

slope = -1

true only if:

BUT, this is not always the case. For example: slope = -1 Moreover, in this example it can be proved that no threshold values exist, which are consistent with definition 3 from lecture notes.

Using the assumption: slope < -1 f (x) = x

slope < -1

true if: f (x) = x slope < -1 slope = -1

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