[Edit] Edits to proof-by-strong-induction.md (#3237)

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* Updated logic of strong induction

* Edit description and look over inductive step

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Co-authored-by: Daksha Deep <[email protected]>

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Author: noahpgordon

content/discrete-math/concepts/proofs/terms/proof-by-strong-induction/proof-by-strong-induction.md

@@ -1,6 +1,6 @@
 ---
 Title: 'Proof by Strong Induction'
-Description: 'An induction technique that proves a statement by providing multiple base cases, assuming the statement is true for all integers from the largest base case to some even larger integer k, and then proving the statement is true for k+1 using that assumption.'
+Description: 'Proves a universal generalization using the hypothesis that all previous elements in a series have the same property.'
 Subjects:
   - 'Computer Science'
   - 'Discrete Math'
@@ -14,43 +14,59 @@ CatalogContent:
   - 'paths/computer-science'
 ---
 
-**Proof by Strong Induction** is an induction technique that proves a statement by providing multiple base cases, assuming the statement is true for all integers from the largest base case to some even larger integer `k`, and then proving the statement is true for `k+1` using that assumption.
+**Proof by strong induction** is a mathematical technique for proving universal generalizations. It differs from ordinary mathematical induction (also known as weak mathematical induction) with respect to the inductive step.
+
+In a weak mathematical induction, the inductive step involves showing that if some element _n_ has a property, then the successor element _n + 1_ must also have that property. In a strong mathematical induction, the inductive step involves showing that if all elements up to and including _n_ have some property, then _n + 1_ has that property as well.
 
 ## Steps for Strong Induction
 
-1. _Base Cases_: Identify multiple smallest possible cases for which the statement is true.
-2. _Induction Hypothesis_: Suppose that the statement is true for some integer `k`, then the statement must be true for `k+1`.
-3. _Proof_: Prove that the induction hypothesis is true for `k+1`.
+1. _Base Step_: Prove that the first element or elements in the series have some property.
+2. _Inductive Step_: Prove that if _n_ and all elements before n have the relevant property, then _n + 1_ has that property.
 
 ## Example
 
-Statement: `2(x-1) - (x-2) = x for x >= 2`
-
-Base Cases:
+The following proposition will be proven by strong induction:
 
 ```plaintext
-At x = 2 , 2(2 - 1) - (2 - 2) = 2(1) - (0) = 2 + 0 = 2 and x = 2. Thus, the statement is true at x = 2.
-
-At x = 3, 2(3 - 1) - (3 - 2) = 2(2) - (1) = 4 - 1 = 3 and x = 3. Thus, the statement is true at x = 3.
+For all x such that x ≥ 2, 2(x-1) - (x-2) = x.
 ```
 
-Induction Hypothesis:
+Base Step:
+
+The first element is where _x = 2_. Therefore, it must be shown that _2(x - 1) - (x - 2) = x_, where _x = 2_:
 
 ```plaintext
-Suppose at x = k, 2(k-1) - (k-2) = k
-Then at x = k+1, prove that 2((k+1)-1) - ((k+1)-2) = k+1
+2(2 - 1) - (2 - 2) = 2(1) - (0) = 2 - 0 = 2
+Both sides are equal to 2. Thus, the statement is true at x = 2.
 ```
 
-Proof:
+Inductive Step:
+
+In the inductive step, it must be shown that if, for any value _x_ between _2_ and _k_ (inclusive), _2(x - 1) - (x - 2) = x_, then at _x = k + 1, 2(x - 1) - (x - 2) = x_.
+
+This conditional will be proven by assuming the antecendent (this assumption is called the inductive hypothesis) and showing the consequent:
 
 ```plaintext
-2((k+1)-1) - ((k+1)-2) = 2(k+1-1) - (k+1-2)
-                       = 2(k-1+1) - (k-2+1)
-                       = 2(k-1) + 2(1) - (k-2) -(+1)
-                       = 2(k-1) - (k-2) + 2 - 1
-                       = 2(k-1) - (k-2) + 1
-                       = k + 1  [Substituting from x = k, which states that 2(k-1) - (k-2) = k]
-2((k+1)-1) - ((k+1)-2) = k+1
-
-Thus, the statement is true at x = k+1.
+Inductive Hypothesis: Suppose that for all x such that 2 ≤ x ≤ k, 2(x - 1) - (x - 2) = x.
+To be proven: 2((k + 1) - 1) - ((k + 1) - 2) = k + 1.
+
+First, we will rearrange the left-hand of the expression above:
+ 2((k + 1) - 1) - ((k + 1) - 2) = 2(k + 1 - 1) - (k + 1 - 2) [removing extra parentheses]
+           = 2(k - 1 + 1) - (k - 2 + 1) [rearranging within parentheses]
+           = 2(k - 1) + 2 - (k - 2 + 1) [factoring 2 into the left parenthesis]
+           = 2(k - 1) + 2 - (k - 2) - 1 [rearranging the right parenthesis]
+           = 2(k - 1) - (k - 2) + 1  [adding numerical terms]
+
+Now, by the Inductive Hypothesis, 2(k - 1) - (k - 2) = k. We therefore substitute 'k' for '2(k-1) - (k-2)' in the rearranged expression above:
+ 2((k + 1) - 1) - ((k + 1) - 2) = k + 1    [substition from Inductive Hypothesis]
+
+Having shown that 2((k + 1) - 1) - ((k + 1) - 2) = k + 1, this completes the inductive step and the proof.
 ```
+
+Although the example above uses strong mathematical induction, the same strategy could be applied using weak mathematical induction, since the only value of the inductive hypothesis used was at _x = k_ (there was no need to use any values smaller than _k_).
+
+## The Logic of Strong Induction
+
+Why do the base step and the inductive step together demonstrate the truth of a universal generalization?
+
+The base step shows that the first _n_ elements in the series have the relevant property. Since all the elements prior to the element at _n + 1_ have the property, it then follows by the inductive step that the element at _n + 1_ also has the property. Similarly, since all elements prior to the element at _n + 2_ have the property, the inductive step shows that the element at _n + 2_ does as well. Since this can be done indefinitely, the entire series is shown to have the relevant property.