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A student is trying to prove a statement ๐‘„(๐‘›)๐‘„(๐‘›) for all integers ๐‘›โ‰ฅ2๐‘›โ‰ฅ2 by strong induction. As an induction step they have proved that, for all integers ๐‘˜โ‰ฅ3๐‘˜โ‰ฅ3, if ๐‘„(๐‘˜โˆ’1)๐‘„(๐‘˜โˆ’1) then ๐‘„(๐‘˜+1)๐‘„(๐‘˜+1). What is the minimum number of base steps they will need to finish their proof?

Question

A student is trying to prove a statement ๐‘„(๐‘›)๐‘„(๐‘›) for all integers ๐‘›โ‰ฅ2๐‘›โ‰ฅ2 by strong induction. As an induction step they have proved that, for all integers ๐‘˜โ‰ฅ3๐‘˜โ‰ฅ3, if ๐‘„(๐‘˜โˆ’1)๐‘„(๐‘˜โˆ’1) then ๐‘„(๐‘˜+1)๐‘„(๐‘˜+1). What is the minimum number of base steps they will need to finish their proof?

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Solution

The student will need to prove two base cases.

The first base case is Q(2) because the statement needs to be proven for all integers nโ‰ฅ2.

The second base case is Q(3) because the induction step starts from k=3.

So, the minimum number of base steps needed to finish the proof is 2.

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