Hyperoperators [n] basics for large n

[dyitto]

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.