Identify the pattern where A < D is definitely true. Choose the expression that forces A < D.

Introduction / Context:
We must pick the single pattern that guarantees A < D in every valid assignment. Options differ in how A compares to B and how C compares to D, with equalities providing anchors.

Given Data / Assumptions:

• (a) A ≥ B = C < D
• (b) A > B = C ≥ D
• (c) A < B = C ≤ D
• (d) A > B < C ≤ D

Concept / Approach:
We need a chain that ensures A is strictly below D. If we can place A strictly below C and ensure C ≤ D, then A < D follows. Any arrangement that leaves A possibly above D is invalid.

Step-by-Step Solution:
(c) A < B = C ≤ D ⇒ A < C and C ≤ D ⇒ A < D (definite).(a) A ≥ B = C < D ⇒ A could equal or exceed C; if A is much larger than C, A < D is not guaranteed.(b) A > B = C ≥ D ⇒ A may exceed D; A < D cannot be ensured.(d) A > B < C ≤ D gives B below both A and C, but it fixes nothing about A vs D.

Verification / Alternative check:
Counterexamples for (a), (b), (d): pick values that satisfy the premises but violate A < D. For (c), any numbers respecting A < C and C ≤ D will always yield A < D.

Why Other Options Are Wrong:
They permit A ≥ D or leave the A–D relationship underdetermined.

Common Pitfalls:
Overlooking that “≤” allows equality: with A < C and C ≤ D, A remains strictly less than D regardless of whether D equals C or exceeds it.

Final Answer:
A < B = C ≤ D