jdihdib BM4isZ2
动物界脊索动物门哺乳纲灵长目人科人属智人种一员
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【转】一个据称极强的序数记号 作者是@test_alpha0 ,名字是Lavar Table Yarn 我们称一个(∗,∘)表达式,是二元运算∗,∘组合出的一个一元表达式,包括id。 设R定义在N*×N*上,输出(∗,∘)表达式。如果没有特别标示R,我们默认R(k,n)=id,对一切k,n。 我们递归地定义一族(∗,∘)表达式: L(1,a,b,c,d)(x)=L(1,1,b,1,d)(x) L(1,1,b,1,b)(x)=x L(1,1,b,1,d+1)(x)=L(1,1,b,1,d)(R(1,d)(L(1,1,1,1,d)(x))∗x) L(k,a,b,c+1,d)(x)=L(k,a,b,c,d)(R(k,c)(L(k,1,c+1,c,d)(x))∗x) T(k-1,a,b)=L(k-1,a,a,b,b) L(k,a,b,a,d)(x)=T(k-1,b,d)(x) 其中,这些正整数变量满足k>1,a≤b,c,d≥b,c。 现在,定义k-R-序关系是N*×N上的一个序关系,满足: (a,0)<(b,0),若a<b。 (a,0)<(b,c),若c>0。 (a,b)<(a,c),若b<c。 若a<b,m,n>0,则(a,m)>(b,n)当且仅当在Laver table A_n里,(T(k,a,b)(1))^2^(m-1)=2^n。这里laver table里项的指数a^n指从a出发,∗a迭代n-1次。 不难证明以下结论: (ZFC+I2) 1-R-序关系是良序关系。 (ZFC+I1) 对任意正整数n,n-R-序关系是良序关系。 这些序关系把多个Laver table编织在了一起。我们把它们称为Laver Table Yarn,简称LTY。记LTY(n,R)为n-R-序关系,LTY(n)=LTY(n,id)。 我们略微分析LTY(k,R)。不难看出,(n,0)对应序数n-1。(a,n)对应后继序数,当且仅当a>1,且[lbk]n=0,或者存在自然数m,使得在A_n里,(T(k,a-1,a)(1))^2^m=2^(n-1)[rbk]。对于后者的情况,(a-1,m+1)即为(a,n)的前继。这也给出了后继的计算方法。 若(a,n)对应极限序数,则它有个标准的基本列:(a,n)[lbk]s[rbk]=(a+s,m-1),其中(a+s,m)对应(a,n)的第s个后继。 现在,可以定义一个四元函数f(a,b,c,d),其中b是自然数,a,c,d是正整数,满足: f(1,0,c,d)=10^d f(a,b,c,d)=(λx.f(a',b',c,x))^d (d),若(a,b)是(a',b')的LTY(c)后继。 f(a,b,c,d)=f(a',b',c,d),若在LTY(c)中,(a',b')=(a,b)[lbk]d[rbk]。 然后我们定义LTYF(n)=f(1,n,n,n)。
PSS VS BOCF 因为希腊字母难以打出,我用p代替ψ,W代替Ω,w代替ω (0) 1 (0)(0) 2 (0)(1) w (0)(1)(0) w+1 (0)(1)(0)(1) w*2 (0)(1)(0)(1)(0)(1) w*3 (0)(1)(1) w^2 (0)(1)(1)(0)(1)(1) w^2*2 (0)(1)(1)(1) w^3 (0)(1)(1)(1)(1) w^4 (0)(1)(2) w^w (0)(1)(2)(1) w^(w+1) (0)(1)(2)(1)(2) w^(w*2) (0)(1)(2)(2) w^w^2 (0)(1)(2)(2)(2) w^w^3 (0)(1)(2)(3) w^w^w (0)(1)(2)(3)(4) w^w^w^w (0)(1,1) p(W) (0)(1,1)(0)(1,1) p(W)*2 (0)(1,1)(1) p(W+1) (0)(1,1)(1)(2) p(W+w) (0)(1,1)(1)(2)(3) p(W+w^w) (0)(1,1)(1)(2,1) p(W+p(W)) (0)(1,1)(1)(2,1)(1)(2,1) p(W+p(W)*2) (0)(1,1)(1)(2,1)(2) p(W+p(W+1)) (0)(1,1)(1)(2,1)(2)(3) p(W+p(W+w)) (0)(1,1)(1)(2,1)(2)(3,1) p(W+p(W+p(W))) (0)(1,1)(1,1) p(W*2) (0)(1,1)(1,1)(1)(2,1)(2,1) p(W*2+p(W*2)) (0)(1,1)(1,1)(1,1) p(W*3) (0)(1,1)(2) p(W*w) (0)(1,1)(2)(1,1)(2) p(W*w*2) (0)(1,1)(2)(3) p(W*w^w) (0)(1,1)(2)(3,1) p(W*p(W)) (0)(1,1)(2)(3,1)(3)(4,1) p(W*p(W*p(W))) (0)(1,1)(2,1) p(W^2) (0)(1,1)(2,1)(1,1)(2)(3,1)(4,1) p(W^2+W*p(W^2+W)) (0)(1,1)(2,1)(1,1)(2,1) p(W^2*2) (0)(1,1)(2,1)(2) p(W^2*w) (0)(1,1)(2,1)(2)(3,1)(4,1) p(W^2*p(W^2)) (0)(1,1)(2,1)(2,1) p(W^3) (0)(1,1)(2,1)(2,1)(2,1) p(W^4) (0)(1,1)(2,1)(3) p(W^w) (0)(1,1)(2,1)(3)(4,1) p(W^p(W)) (0)(1,1)(2,1)(3,1) p(W^W) (0)(1,1)(2,1)(3,1)(2) p(W^W*w) (0)(1,1)(2,1)(3,1)(2)(3,1)(4,1)(5,1) p(W^W*p(W^W)) (0)(1,1)(2,1)(3,1)(2,1) p(W^(W+1)) (0)(1,1)(2,1)(3,1)(2,1)(2,1) p(W^(W+2)) (0)(1,1)(2,1)(3,1)(2,1)(3)(4,1)(5,1)(6,1) p(W^(W+p(W^W))) (0)(1,1)(2,1)(3,1)(2,1)(3,1) p(W^(W*2)) (0)(1,1)(2,1)(3,1)(2,1)(3,1)(2,1)(3,1) p(W^(W*3)) (0)(1,1)(2,1)(3,1)(3) p(W^(W*w)) (0)(1,1)(2,1)(3,1)(3)(4,1)(5,1)(6,1) p(W^(W*p(W^W)) (0)(1,1)(2,1)(3,1)(3,1) p(W^W^2) (0)(1,1)(2,1)(3,1)(3,1)(3,1) p(W^W^3) (0)(1,1)(2,1)(3,1)(4) p(W^W^w) (0)(1,1)(2,1)(3,1)(4)(5,1)(6,1)(7,1) p(W^W^p(W^W)) (0)(1,1)(2,1)(3,1)(4,1) p(W^W^W) (0)(1,1)(2,1)(3,1)(4,1)(1,1) p(W^W^W+W) (0)(1,1)(2,1)(3,1)(4,1)(2,1) p(W^(W^W+1)) (0)(1,1)(2,1)(3,1)(4,1)(3,1) p(W^W^(W+1)) (0)(1,1)(2,1)(3,1)(4,1)(4,1) p(W^W^W^2) (0)(1,1)(2,1)(3,1)(4,1)(5,1) p(W^W^W^W) (0)(1,1)(2,1)(3,1)(4,1)(5,1)(6,1) p(W^W^W^W^W) (0)(1,1)(2,2) p(W_2)
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