FANDOM


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,

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

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

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

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