One-bit binary subtraction — counting the distinct cases For a single-bit subtraction cell (with possible borrow), how many basic binary subtraction operations (input cases) exist?

Introduction / Context:
Understanding the truth table of a 1-bit subtractor (with or without borrow) is key to designing ripple-borrow subtractors and two’s-complement adders/subtractors. Counting the distinct input cases helps define the required logic.

Given Data / Assumptions:

• Consider basic A − B for one bit, ignoring an incoming borrow to keep the canonical “basic” set.
• Binary digits A and B can each be 0 or 1.

Concept / Approach:

There are two inputs (A, B) with 2 states each, so total combinations = 2^2 = 4. These correspond to: 0 − 0, 0 − 1, 1 − 0, and 1 − 1. Each case has a defined difference bit and potentially a borrow out. If a borrow-in is included, the truth table doubles again to 8 cases (2^3), but the term “basic binary subtraction operations” typically refers to A and B only.

Step-by-Step Solution:

Verification / Alternative check:

Compare with the half-subtractor definition which explicitly has a 2-input, 2-output truth table covering four input rows.

Why Other Options Are Wrong:

• 3, 2, or 1: undercount the combinations.
• 8: corresponds to including an extra borrow-in input (full subtractor), not the basic case.

Common Pitfalls:

• Accidentally mixing “half-subtractor” (no borrow-in) with “full subtractor” (with borrow-in).

Final Answer:

4