Study Boolean Algebra & Logic Minimization — complete the practice problems in your notebook.
Which Boolean identity is De Morgan's first theorem?
De Morgan's laws come in a pair: NOT(A∧B) = ¬A∨¬B, and NOT(A∨B) = ¬A∧¬B. The first is about negating AND.
The distributive law expands AND over OR. De Morgan's laws relate negation of AND/OR to OR/AND of negations.
A ∨ ¬A = 1 is the complement/tautology law. De Morgan's laws involve negating compound expressions.
A Karnaugh map is used to:
K-maps minimize Boolean expressions. RAM design involves addressing, decoders, and latches — different topic.
BCD (Binary-Coded Decimal) represents each decimal digit in 4 bits. K-maps minimize logic functions.
IEEE 754 defines sign/exponent/mantissa fields. K-maps are for combinational logic optimization.
The Boolean expression A·(A+B) simplifies to:
Absorption law: A·(A+B) = A. Think: if A is 0 the whole expression is 0; if A is 1 the whole expression is 1. Result = A.
Test with A=1,B=0: original = 1·(1+0)=1·1=1. A·B=1·0=0. Not equal. Use the absorption law: A·(A+B)=A.
If A=0: 0·(0+B)=0. So the result is not always 1. Correct simplification is A.
In sum-of-products (SOP) form, each term is:
SOP = Σ(products). Each minterm is an AND of literals. The minterms are then summed (OR'd).
SOP uses AND within terms and OR between terms. XOR (⊕) is its own operation.
NAND = NOT AND. SOP terms are standard AND of literals without additional inversion.
Which law allows replacing A + A·B with A?
Idempotent: A+A=A or A·A=A. The expression A+A·B uses the absorption law: A+A·B=A.
Commutativity reorders operands. Absorption removes a redundant term: A+AB=A.
Associativity changes grouping, not simplification. Absorption: A+AB=A.