## FANDOM

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According to Knuth Arrows Theorem by Sbiis Saibian,

for all $$a≥2, b≥1, c≥1, k≥2: (a\uparrow^kb)\uparrow^kc<a\uparrow^k(b+c)$$

$$(a\uparrow^k)^bc=\underbrace{a\uparrow^ka\uparrow^k...a\uparrow^k}_{b}(c)$$

prove that: for all $$a≥2, b≥1, c≥1, k≥3: (a\uparrow^kb)\uparrow^kc<(a\uparrow^{k-1})^{b+c-2}(1+a)$$

$$(a\uparrow^k)^{b+c-2}(1+a)$$

$$=(a\uparrow^k)^{b+c-3}(a\uparrow^k(1+a))$$

$$=(a\uparrow^k)^{b+c-3}(a\uparrow^{k-1}a\uparrow^ka)$$

$$>(a\uparrow^k)^{b+c-3}(a+a\uparrow^{k+1}2)$$

Let $$M(1)=a+a\uparrow^{k+1}2$$

$$M(n+1)=M(n)+a\uparrow^{k+1}(n+2)$$

1. for all $$n,m∈N,m<n:a\uparrow^{k+1}m<a\uparrow^{k+1}n$$

Therefore $$M(n)<(n+1)\times a\uparrow^{k+1}(n+1)$$

2.$$M(1)=a+a\uparrow^{k+1}2≥2+4>4=1+3$$

also for all $$n∈N:a\uparrow^{k+1}(n+2)>n+2>1$$

Therefore $$M(n)>n+3$$

$$a\uparrow^kM(n)$$

$$=a\uparrow^k(M(n-1)+a\uparrow^{k+1}(n+1))$$

$$>(a\uparrow^ka\uparrow^{k+1}(n+1))\uparrow^kM(n-1)$$

$$=(a\uparrow^{k+1}(n+2))\uparrow^kM(n-1)$$

$$>a\uparrow^{k+1}(n+2)\times M(n-1)$$

$$>a\uparrow^{k+1}(n+2)\times (n+2)$$

$$>M(n+1)$$

Therefore,

$$(a\uparrow^k)^{b+c-3}M(1)$$

$$>(a\uparrow^k)^{c-1}M(b-1)$$

$$=(a\uparrow^k)^{c-1}(M(b-2)+a\uparrow^{k+1}b)$$

lemma:

for all $$n∈N:a\uparrow^k(M(b-2)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)>M(b-2)+(a\uparrow^{k+1}b)\uparrow^{k+1}(n+1)$$

proof:

$$a\uparrow^k(M(b-2)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)$$

$$=a\uparrow^k(M(b-3)+a\uparrow^{k+1}(b-1)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)$$

$$=(a\uparrow^{k-1})^{a\uparrow^{k+1} (b-1)+(a\uparrow^{k+1} b)\uparrow^{k+1} n}(a\uparrow^kM(b-3))$$

$$>a\uparrow^k(a\uparrow^{k+1}(b-1)+(a\uparrow^{k+1}b)\uparrow^{k+1}n)+a\uparrow^kM(b-3)$$

$$>(a\uparrow^{k+1}b)\uparrow^k((a\uparrow^{k+1}b)\uparrow^{k+1}n)+a\uparrow^kM(b-3)$$

$$=(a\uparrow^{k+1}b)\uparrow^{k+1}(n+1)+a\uparrow^kM(b-3)$$

$$>M(b-2)+(a\uparrow^{k+1}b)\uparrow^{k+1}(n+1)$$

Therefore,

$$(a\uparrow^k)^{c-1}(M(b-2)+a\uparrow^{k+1}b)$$

$$>M(b-2)+(a\uparrow^{k+1}b)\uparrow^{k+1}c$$

$$>(a\uparrow^{k+1}b)\uparrow^{k+1}c$$