2 Boolean Algebra cont The digital abstraction Theorem
מבוא למחשבים ספרתיים + מבנה המחשב 2# תרגול Boolean Algebra cont’ The digital abstraction
Theorem: Absorption Law For every pair of elements a , b B, 1. a + a · b = a 2. a · ( a + b ) = a Proof: (1) Identity Distributivity Commutativity Theorem: For any a B, a+1=1 Identity (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
Proof: (1) Let Distributivity Commutativity Distributivity Idempotent Law Absorption Law
Commutativity Distributivity Idempotent Law Absorption Law Commutativity Absorption Law
Same transitions Putting it all together: · before +
Also, (2) Duality
Theorem 11: De. Morgan’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) (2) Duality (a+b)(a’b’) = 0. (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: → • Closure of (∪) and (∩) over B (functions |B| ≥ 2. ).
A 1. 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,
A 2. Distributivity of ( ∪ ) and ( ∩ ). Let and or If , We have and . 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.
A 3. Existence of additive and multiplicative identity element. A 4. Existence of the complement. All axioms are satisfied Algebra of sets is Boolean algebra.
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 ) 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) ]
Let fd be the dual of a function f ( x 1 , x 2 , … , xn ) Lemma: In switching algebra, fd = f’ ( x 1’ , x 2’ , … , xn’ ) Proof: Let f ( x 1 , x 2 , … , xn ) 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. 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 If then, E 1’ ( E 1 + E 2 ) ( E 1 · E 2 )
If then,
Definition: A function f is called self-dual if f = fd Lemma: For any function f and any two-valued variable A, the function g = Af + A’fd is a self-dual. Proof: (holds for any Boolean algebra) Dual definition Distributivity Commutativity
Distributivity Commutativity A’ is the complement of A Identity Commutativity Notice that the above expression has the form: ab + a’c +bc where “a” =A, “b”=f, “c” = fd.
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 algebra Identity OR
Easier proof (2) for switching algebra only: (case analysis) A=0 0’ = 1 Identity Commutativity Theorem: For any a B, a· 0=0 Identity Absorption Law
A=1
Example of a transfer function for an inverter
slope = -1
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: f (x) = x slope < -1
f (x) = x slope < -1
slope = -1 f (x) = x slope = -1 true if: slope < -1
- Slides: 35