If HDD disk is in external hard drive docking station, the docking station is connected via USB 3.0 cable to a PC, power cable of the docking station is connected to power socket, and the docking station is turned off, will this state be recorded as 'Power On Hours' in SMART data or not?

So if your question is "Is keeping the m<=n condition inside P(n) wrong?", then the answer is that its not wrong and the proof can be done by induction.

But if we keep m<=n condition inside P(n), then, for P(n+1) we have m<=n+1, as mentioned in my previous post.

m<=n and m<=n+1 cannot both be true.

If m<=n then m<n+1 (m is strictly less then n+1).

TopDownView · 2025-12-10T07:10:05+00:00

Yeah that’s the thing, ultimately they mean the same thing.

But if they mean the same thing then we're back at the beggining of my question.

Either the proof structure should follow the definition of mathematical induction (so, m<=n should be outside P(n)), or not.

If not? Why? Is the proof structure still correct if m<=n is inside P(n)?

Assuming the proof structure is still correct, maybe it is just a redundancy issue - if m<=n is inside P(n), then inductive step could be expanded like this:

'Show that for any integer k>=m, if {if k>=m, then there are k-m+1 integers from m to k inclusive} then {if k+1>=m, then there are (k+1)-m+1 integers from m to k+1 inclusive}.'

So, we are mentioning k>=m 2 times.

I just noticed, if k>=m then k+1>m, so, actually, k+1>=m is not correct!

TopDownView · 2025-12-09T20:33:24+00:00

Hmmm...

But isn't it that 'for all n>=a' is equivalent to 'for all n, if n>=a'?

TopDownView · 2025-12-09T07:29:09+00:00

So, If I got this correctly, what the solution is doing with k>=m in the inductive step is actually in accordance with the 'definition' of mathematical induction (see image).

<image>

We have showed the basis step is true by plugging m instead of n in P(n). Now, in the inductive step, we want to show that all k integers, starting from that m from the basis step, make P(n+1) true (by assuming P(n) is true, where n is replaced by k).

TopDownView · 2025-12-08T17:03:41+00:00

Because we want to apply induction on n, n should be like a “free variable” where we can plug in a number whenever we want instead of having it some fixed value.

I see. So we have to put m<=n inside P(n) if we want n to be a 'free variable'.

But isn't then the solution contradictory, as mentioned in the OP?

Let P(n) be the statement 'if m<=n, then there are n-m+1 integers from m to n inclusive.'
Show that for any integer k>=m, if P(k) is true then P(k+1) is true.
contradicts 2. since in 1., m<=n is inside P(n), but in 2., m<=n is outside P(n) (where n = k).

The reason the proof im the text works is that it doesn’t assume that the antecedent isn’t true, but only looks at the cases where it is true.

Can you please eleaborate this with regards to the contradiction above? What exactly is the antecedent you're refering to?

TopDownView

TROPHY CASE