Claim:Weak induction principle starting at k is equivalent to weak induction

From CS2800 wiki

Suppose you have an inductive proof in the following style:

To prove [math]\href{/cs2800/wiki/index.php/%5Cforall}{\forall} n \geq k, P(n) [/math] using weak induction, you can prove [math]P(k) [/math] and prove [math]P(n+1) [/math] for an arbitrary [math]n \geq k [/math], assuming [math]P(n) [/math].

But you are only willing to accept the basic weak induction principle:

To prove "[math]\href{/cs2800/wiki/index.php/%5Cforall}{\forall} n \href{/cs2800/wiki/index.php/%5Cin}{\in} \mathbb{N},P(n) [/math]" using weak induction, you must prove [math]P(0) [/math] (this is often called the base case), and then you must prove [math]P(n+1) [/math] for an arbitrary [math]n [/math], assuming [math]P(n) [/math] (this is called the inductive step).

You can systematically convert from the first style to the second:

Suppose you know the following:
  1. [math]P(k) [/math]
  2. [math]\href{/cs2800/wiki/index.php/%5Cforall}{\forall} n \geq k, P(n) [/math] implies [math]P(n+1) [/math]
Then you can conclude [math]\href{/cs2800/wiki/index.php/%5Cforall}{\forall} n \geq k, P(n) [/math], using only the basic weak induction principle.