User:Polygon/Better lower bound for BB(4,3)

Better lower bound for BB(4,3)

Definitions

$f(n)=2^{2^{n+1}}$

$g(n)=\frac{5\times 2^{2^{f^n(0)+1}+2}-8}{9}$

$n0=\frac{5\times 2^{2^{2^{32}+1}+1}-4}{9}$

$n1=2^{2^{32}+1}-4$

Σ = $5\times 2^{2^{f^{g^{n1}(n0)}(0)+1}+2}+7$

Lower bound on $f^n(0)$

$f(n)=2^{2^{n+1}}$

$f(0)=2^{2^1};f(f(0))=2^{2^{(2^{2^1}+1)}}>(2\uparrow {)}^{4}1;f^3(0)=2^{2^{(2^{2^{(2^{2^1}+1)}}+1)}}>(2\uparrow {)}^{6}1$

$f^k(0)=2^{2^{f^{k-1}(0)+1}}>2^{2^{f^{k-1}(0)}}=>(2\uparrow {)}^2{f}^{k-1}(0)=>(2\uparrow {)}^{2a}{f}^{k-a}(0)$

$=>f^k(0)>(2\uparrow {)}^{2k}1=2\uparrow \uparrow 2k$

$=>f^n(0)>2\uparrow \uparrow 2n$

Lower bound on $g^k(n)$

$g(n)=\frac{5\times 2^{2^{f^n(0)+1}+2}-8}{9}>\frac{5\times 2^{2^{2\uparrow \uparrow 2n}+2}-8}{9}>2^{2^{2\uparrow \uparrow 2n}}=2\uparrow \uparrow (2n+2)$

$g^2(n)>\frac{5\times 2^{2^{f^{2\uparrow \uparrow (2n+2)}(0)+1}+2}-8}{9}>\frac{5\times 2^{2^{2\uparrow \uparrow 2\uparrow \uparrow (2n+2)}+2}-8}{9}>2\uparrow \uparrow 2\uparrow \uparrow (2n+2)$

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

$g^k(n)>(2\uparrow \uparrow {)}^k(2n+2)>(2\uparrow \uparrow {)}^{k}2n$

Lower bound on $g^{n1}(n0)$

$g^{n1}(n0)=g^{2^{2^{32}+1}-4}(\frac{5\times 2^{2^{2^{32}+1}+1}-4}{9})>g^{2^{2^{32}+1}-4}(2^{2^{2^{32}}})>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}{2}^{2^{2^{32}}}$ ; Note that $2\uparrow \uparrow 6<2^{2^{2^{32}}}<2\uparrow \uparrow 7$

$(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}{2}^{2^{2^{32}}}>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-4}2\uparrow \uparrow 6=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}6>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}4=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-3}2\uparrow \uparrow 2=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2$

$g^{n1}(n0)>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2$

Lower bound on $f^{g^{n1}(n0)}(0)$

$g^{n1}(n0)>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2$ and $f^n(0)>2\uparrow \uparrow 2n$

$f^{g^{n1}(n0)}(0)>f^{(2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2}(0)>2\uparrow \uparrow (2\times (2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2)>2\uparrow \uparrow (2\uparrow \uparrow {)}^{2^{2^{32}+1}-2}2=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-1}2$

$(2\uparrow \uparrow {)}^{2^{2^{32}+1}-1}2=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}-1}2\uparrow \uparrow 1=>(2\uparrow \uparrow {)}^{2^{2^{32}+1}}1=>2\uparrow \uparrow \uparrow 2^{2^{32}+1}$

$f^{g^{n1}(n0)}(0)>2\uparrow \uparrow \uparrow 2^{2^{32}+1}$

Lower bound on Σ

Σ = $5\times 2^{2^{f^{g^{n1}(n0)}(0)+1}+2}+7>5\times 2^{2^{(2\uparrow \uparrow \uparrow 2^{2^{32}+1})+1}+2}+7>2\uparrow \uparrow \uparrow 2^{2^{32}+1}$

Σ > $2\uparrow \uparrow \uparrow 2^{2^{32}+1}$