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