Discrete Math : proving of predicate logic

Question

Assume $P(n)$ is a statement about integers $n\ge 1$ such that $∀n,P(n) \implies P (n + 4)$. Is the following statement $(∀n,P(n))$ always true, or always false, or can be true for some values and false for some other values?

Answer 1

If $P(1)$ is true, by induction principle and the hypothesis $\forall n(P(n) \Rightarrow P(n+4))$, then you can assure that $P(n)$ is true for all $n \geq 5$. However, you don't know if $P(n)$ is true for any $n \geq 1$ and you can't do much. Also, even if you knew $P(1)$ is true, to assure $\forall nP(n)$, you would still have to prove that $P(n)$ is true for $n \in \{2,3,4\}$, and you can't infer that also just by the hypothesis given. So, by the current state of your question, I think this can be true for some values and false for others.

Answer 2

The formula $\forall n [P(n) \to P(n+4)]$ alone does not imply that $\forall n P(n)$ is true. $P(n)$ could be true for all $n$; it could be false for all $n$; it could true for some $n$ and false for others. My point is there's no way to prove $\forall n P(n)$ is necessarily true when all you have to work with is the formula $\forall n [P(n) \to P(n+4)]$.

In order to utilize the power of induction and make inferences about specific integers, you must go one step further and prove a base case. In other words, you must demonstrate that a formula such as $P(1)$ is true. However, even proving $P(1)$ alone would not imply $\forall n P(n)$; it would merely imply the formulas $P(1), P(5), P(9), P(13), ...$ are all true. In order to infer $\forall n P(n)$, you would have to prove the following four formulas: $P(1), P(2), P(3), P(4)$.

By induction, proving $P(1)$ implies $P(1), P(5), P(9), P(13), ...$ are all true.

By induction, proving $P(2)$ implies $P(2), P(6), P(10), P(14), ...$ are all true.

By induction, proving $P(3)$ implies $P(3), P(7), P(11), P(15), ...$ are all true.

By induction, proving $P(4)$ implies $P(4), P(8), P(12), P(16), ...$ are all true.

So, after combining the above results you can see that proving $P(1), P(2), P(3), P(4)$ implies $P(n)$ is true for all $n \ge 1$.