Match the Boolean-algebra laws to their relational forms: (A) Distributive law, (B) De Morgan’s law, (C) Idempotent law, (D) Involution law — with (1) x + x = x, (2) x(y + z) = xy + xz, (3) (x)̄̄ = x, (4) (x + y)̄ = x̄ ȳ.

Difficulty: Easy

Correct Answer: A-2, B-4, C-1, D-3

Explanation:


Introduction:
Boolean algebra laws provide the algebraic toolkit for simplifying logic expressions and designing digital circuits. This item tests recognition of four cornerstone identities and their canonical forms.


Given Data / Assumptions:

  • A: Distributive law.
  • B: De Morgan’s law.
  • C: Idempotent law.
  • D: Involution (double-complement) law.
  • Relational forms: (1) x + x = x, (2) x(y + z) = xy + xz, (3) (x)̄̄ = x, (4) (x + y)̄ = x̄ ȳ.


Concept / Approach:

Each law has a standard, widely used expression. The distributive law expands products over sums. De Morgan’s laws transform complements of sums/products. Idempotence states redundancy of combining a variable with itself. Involution states that complementing twice restores the original variable.


Step-by-Step Solution:

A → (2): x(y + z) = xy + xz (distribution of product over sum).B → (4): (x + y)̄ = x̄ ȳ (the dual is (xy)̄ = x̄ + ȳ).C → (1): x + x = x (and x·x = x).D → (3): (x)̄̄ = x (double negation).


Verification / Alternative check:

Truth tables or Venn diagrams confirm each identity; Karnaugh maps also illustrate idempotence and distributivity.


Why Other Options Are Wrong:

Any permutation that mismatches these canonical forms contradicts standard Boolean identities taught in digital logic.


Common Pitfalls:

Confusing De Morgan’s transformations or mixing the two De Morgan forms; forgetting that idempotence applies to both + and · operations.


Final Answer:

A-2, B-4, C-1, D-3

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