Single-bit subtraction coverage: There are four basic input combinations for subtracting two single-bit binary numbers (minuend, subtrahend) when considered case-by-case. Evaluate this statement.

Introduction / Context:Designing half- and full-subtractors starts with enumerating all 1-bit cases for the minuend (M) and subtrahend (S). Recognizing the complete set of cases avoids mistakes when deriving logic for the difference and borrow outputs.

Given Data / Assumptions:

• M and S are single bits: each ∈ {0, 1}.
• No initial borrow-in considered (half-subtractor perspective).
• We classify cases by the (M, S) pair.

Concept / Approach:With two binary inputs, there are 2^2 = 4 possible combinations. For subtraction, these are: 0−0, 0−1, 1−0, 1−1. Each case yields a specific difference bit and borrow-out pattern, forming the canonical truth table for the half-subtractor.

Step-by-Step Solution:

Verification / Alternative check:The complete table leads to the well-known relations: D = M ⊕ S (for half-subtractor) and Bout = ~M * S. Extending to full-subtractor adds a borrow-in term but preserves the base four cases per (M,S).

Why Other Options Are Wrong:

• Incorrect: Understates or overstates the number of base cases for two inputs.
• Ambiguous / Cannot be determined: The enumeration is straightforward from combinatorics.

Common Pitfalls:Forgetting to treat the borrow-in separately for full-subtractors; the base (M,S) cases remain four, with additional combinations when Bin is included.

Final Answer:Correct