7. ((a > 2 or b > 2) and a > 1 and b > 1) -> a[n+1]b > a[n]b

Proof:

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The case n = 1: we wish to prove that

a > 1 and b > 1 and (a > 2 or b > 2) -> a * b > a + b

2 * 3 > 2 + 3

a > 2 -> a * 2 = a + a > a + 2

Now suppose a * b > a + b for any a, b > 1

a * (b + 1) = a * b + a > a + b + a > a + (b + 1)

Now let's consider lemma 7 true for some n > 0.

2 [n+2] 3 = 2 [n+1] (2 [n+2] 2) = 2 [n+1] 4 > 2 [n+1] 3

a > 2 -> a [n+2] 2 = a [n+1] (a [n+2] 1) = a [n+1] a > a [n+1] 2

Now suppose for some a, b: a > 1 and b > 1 and (a > 2 or b > 2) and a [n+2] b > a [n+1] b

a [n+2] (b + 1) = a [n+1] (a [n+2] b)

> a [n+1] (a [n+1] b)

> a [n] (a [n+1] b) (mind that a [n+1] b > b >= 2)

= a [n+1] (b + 1)

This proves lemma 7 to be true for n + 1.

So lemma 7 also has been proved for all n by induction applied to n.

Proof:

======

The case n = 1: we wish to prove that

a > 1 and b > 1 and (a > 2 or b > 2) -> a * b > a + b

2 * 3 > 2 + 3

a > 2 -> a * 2 = a + a > a + 2

Now suppose a * b > a + b for any a, b > 1

a * (b + 1) = a * b + a > a + b + a > a + (b + 1)

Now let's consider lemma 7 true for some n > 0.

2 [n+2] 3 = 2 [n+1] (2 [n+2] 2) = 2 [n+1] 4 > 2 [n+1] 3

a > 2 -> a [n+2] 2 = a [n+1] (a [n+2] 1) = a [n+1] a > a [n+1] 2

Now suppose for some a, b: a > 1 and b > 1 and (a > 2 or b > 2) and a [n+2] b > a [n+1] b

a [n+2] (b + 1) = a [n+1] (a [n+2] b)

> a [n+1] (a [n+1] b)

> a [n] (a [n+1] b) (mind that a [n+1] b > b >= 2)

= a [n+1] (b + 1)

This proves lemma 7 to be true for n + 1.

So lemma 7 also has been proved for all n by induction applied to n.