Correct : a

The correct answer is Option A.
This question directly tests the principle of mathematical induction expressed in predicate logic. The standard induction principle over natural numbers states:
If P(0) is true (base case), and for every natural number x, P(x) implies P(x+1) (inductive step), then P(x) holds for all natural numbers.
This is exactly what Option A says: (P(0) ∧ ∀x[P(x) ⇒ P(x+1)]) ⇒ ∀x P(x). Correct.
Option B - (P(0) ∧ ∀x[P(x) ⇒ P(x−1)]) ⇒ ∀x P(x): This goes downward - from x to x−1. Starting at P(0) and going to P(−1) leaves the domain of natural numbers. This doesn''t establish P for any x > 0. Incorrect.
Option C - (P(1000) ∧ ∀x[P(x) ⇒ P(x−1)]) ⇒ ∀x P(x): Starting at 1000 and going downward proves P for 0 through 1000 only. Natural numbers above 1000 are never covered. Incorrect.
Option D - (P(1000) ∧ ∀x[P(x) ⇒ P(x+1)]) ⇒ ∀x P(x): Going upward from 1000 proves P for all x ≥ 1000, but says nothing about 0 through 999. Incorrect.
The key insight - valid induction must start from the smallest element (0 for natural numbers) and go upward. Only Option A does both correctly.