Which statement about predicate P(x) over natural numbers is TRUE?

The correct option is:
( P(0) ∧ (∀x [ P(x+1) ]) ) ⇒ (∀x P(x))

Let's analyze why this statement is logically true for any predicate $P\left(x\right)$ defined over the set of natural numbers (which typically starts at 0, i.e., $\mathbb{N}=\{0,1,2,...\}$).

1. Understanding the Premise (Antecedent):
The premise of the implication is:
$P\left(0\right)\wedge (\forall x[P(x+1)\left]\right)$
This premise is a conjunction of two distinct parts:
• $P\left(0\right)$: The predicate holds true for $x=0$.
• $\forall x\left[P\right(x+1\left)\right]$: The predicate holds true for the successor of every natural number $x$.

2. Evaluating the Second Conjunct:
Let's look closely at the statement $\forall x\left[P\right(x+1\left)\right]$.
Since $x$ ranges over all natural numbers $\{0,1,2,...\}$:
• For $x=0$, this asserts $P\left(1\right)$ is true.
• For $x=1$, this asserts $P\left(2\right)$ is true.
• For $x=2$, this asserts $P\left(3\right)$ is true, and so on.
In general, this conjunct guarantees that $P\left(k\right)$ is true for every positive integer $k\ge 1$.

3. Combining the Parts:
By combining the first conjunct $P\left(0\right)$ (which covers $x=0$) and the second conjunct $\forall x\left[P\right(x+1\left)\right]$ (which covers all $x\ge 1$), we establish that $P\left(x\right)$ is true for every single element in the domain of natural numbers.
Therefore, the conclusion $\forall xP\left(x\right)$ must be true.

4. Why the Other Options are Incorrect:
• Option 2: $\left(P\right(0)\wedge (\forall x\left[P\right(x)\Rightarrow P(x-1\left)\right]\left)\right)\Rightarrow (\forall xP(x\left)\right)$
The implication runs backward (from $x$ to $x-1$). Knowing $P\left(0\right)$ is true does not allow us to propagate the truth upward to larger natural numbers.
• Option 3: $\left(P\right(1000)\wedge (\forall x\left[P\right(x)\Rightarrow P(x-1\left)\right]\left)\right)\Rightarrow (\forall xP(x\left)\right)$
This allows us to prove that $P\left(x\right)$ holds for values less than or equal to 1000, but it does not tell us anything about any $x>1000$.
• Option 4: $\left(P\right(1000)\wedge (\forall x\left[P\right(x)\Rightarrow P(x+1\left)\right]\left)\right)\Rightarrow (\forall xP(x\left)\right)$
This allows us to prove that $P\left(x\right)$ holds for all $x\ge 1000$, but it does not guarantee that the predicate holds for any elements smaller than 1000 (e.g., $x=0$ to $999$).