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Extended Normal Form

Edited: Late Feb 2016 to align to changes to my blog on Fundamental Sequences.

This blog extends Cantor's Normal Form to provide precise definitions for ordinals with arbitrary complexity. It follows from my earlier blog on Fundamental Sequences and uses the Aristo Sequence definition from that blog. In turn this blog will be referred to in my other blogs for the J Function.

Constructing Ordinals with Arbitrary Complexity

The following construction procedure can be used to construct any ordinal of arbitrary complexity (up to the size of the Small Veblen Ordinal (SVO) which is defined as follows:

$$SVO = \varphi(1,0_{[\omega]})$$

The construction procedure enforces Extended Normal Form and will be consistent to the Aristo Sequence which is defined on this blog.

Let $$\gamma$$ be an arbitrary transfinite ordinal, $$\lambda$$ is an arbitrary limit ordinal, and $$t, q$$ are finite integers.

Let $$\gamma$$ be an arbitrary ordinal to be constructed:

Step 1: Define $$\gamma = \gamma_b + \gamma_a$$ where $$\gamma_b > \gamma_a >= 0$$

Step 2: Define $$\gamma_b = \gamma_d.\gamma_c$$ where $$\gamma_d > \gamma_c > 0$$

Step 3: Define $$\gamma_d = (\lambda\uparrow\uparrow t)^{\gamma_e}$$ where $$\lambda^{\lambda} > \gamma_e > 0$$

Step 4: Define $$\lambda = \varphi(\gamma_{[q]})$$ where $$q > 1$$ or $$\lambda = \omega$$ when $$q = 1$$

Step 5: Is a special case. When $$q = 0$$ then $$\lambda = \gamma_d = \gamma_b =$$ null and $$\gamma_a = n$$ i.e. a finite integer.

Applying all the steps for $$\gamma$$ we get:

$$\gamma = (\lambda\uparrow\uparrow t)^{\gamma_e}.\gamma_c + \gamma_a = (\varphi(\gamma_{[q]})\uparrow\uparrow t)^{\gamma_e}.\gamma_c + \gamma_a$$

The construction procedure is heavily recursive. Every instance of $$\gamma$$ is recursively constructed using the same procedure.

Step 3 where $$\lambda^{\lambda} > \gamma_e$$

Step 3 is not obvious but follows from this logic:

Let $$\gamma_d = (\lambda\uparrow\uparrow t)^{\gamma_e} = (\delta^{\lambda})^{\gamma_e}$$

When $$\gamma_e = 2$$

$$\gamma_d = (\delta^{\lambda})^2 = \delta^{\lambda.2}$$

When $$\gamma_e = \lambda$$

$$\gamma_d = (\delta^{\lambda})^{\lambda} = \delta^{\lambda.\lambda} = \delta^{\lambda^2}$$

When $$\gamma_e = \lambda^2$$

$$\gamma_d = (\delta^{\lambda})^{\lambda^2} = \delta^{\lambda.\lambda^2} = \delta^{\lambda^3}$$

When $$\gamma_e = \lambda^{\lambda}$$

$$\gamma_d = (\delta^{\lambda})^{\lambda^{\lambda}} = \delta^{\lambda.\lambda^{\lambda}}$$

$$= \delta^{\lambda^{1 + \lambda}} = \delta^{\lambda^{\lambda}} = \delta^{\lambda^{\lambda}} = \lambda\uparrow\uparrow (t+1)$$

Therefore $$\gamma_e$$ has the upper limit of $$\lambda^{\lambda}$$ because this is equivalent to increasing $$t$$ by 1.

Step 4 where $$\lambda = \varphi(\gamma_{[q]})$$

Step 4 requires a construction procedure for $$\lambda$$ with additional rules. The procedure begins as follows:

$$\lambda = \varphi(\gamma_{[q]})$$ with these initial rules:

• $$q = 0$$ then $$\gamma = \gamma_a = n$$ (i.e. a finite integer) or
• $$q = 1$$ and $$\gamma_{[q]} = \gamma_{[1]} = \gamma_1 = 1$$ and $$\lambda = \varphi(1) = \omega$$

When $$q > 1$$ we begin constructing $$\lambda$$ with:

$$\lambda_0 = \varphi(\mu_{[q]})$$ where $$\mu$$ is a special case of a $$\gamma$$ ordinal but recursively defined with $$q < 2$$

and

$$\lambda_1 = \varphi^{m_0}(\mu_{[1]}*,\mu_{[q-1]})$$

then

$$\lambda_2 = \varphi^{m_1}(\gamma_{[1]},(\lambda_1 + a_1)_*,0_{[q-2]})$$ where

• $$m_1$$ is a finite integer
• $$\gamma_1$$ is a recursively defined $$\gamma$$ ordinal, and
• If $$\gamma_1 = \mu_1$$ then $$a_1 = 0$$
• If $$0 < \gamma_1 < \mu_1$$ then $$a_1 = 1$$
• If $$0 = \gamma_1$$ then $$m_1 = 0$$ and $$\lambda_2$$ collapses to $$\lambda_1$$

then

$$\lambda_3 = \varphi^{m_2}(\gamma_{[2]},(\lambda_2 + a_2)_*,0_{[q-3]})$$ where

• $$m_2$$ is a finite integer
• $$\gamma_{[2]}$$ are recursively defined $$\gamma$$ ordinals, and
• If $$\gamma_1 = \gamma_1$$ (value within $$\lambda_2$$) then $$a_2 = 0$$
• If $$0 < \gamma_1 < \gamma_1$$ then $$a_2 = 1$$
• If $$0 = \gamma_1$$ then $$q$$ is decremented to $$q - 1$$ and $$0 < \gamma_2$$

then ... until

$$\lambda_q = \varphi^{m_{q-1}}(\gamma_{[q-1]},(\lambda_{q-1} + a_{q-1})_*)$$ where

• $$m_{q-1}$$ is a finite integer
• $$\gamma_{[q-1]}$$ are recursively defined $$\gamma$$ ordinals, and
• If $$\gamma_1 = \gamma_1$$ (value within $$\lambda_{q-1}$$) then $$a_{q-1} = 0$$
• If $$0 < \gamma_1 < \gamma_1$$ then $$a_{q-1} = 1$$
• If $$0 = \gamma_1$$ then $$q$$ is decremented to $$q - 1$$ and $$0 < \gamma_2$$ or
• If $$0 = \gamma_2$$ then $$q = q - 1$$ and $$0 < \gamma_3$$ and
• For successive $$0 = \gamma_i$$ then $$q = q - 1$$ until $$0 < \gamma_{i+1}$$

The construction procedure continues. So $$\lambda = \lambda_{q+j}$$ where $$j$$ is the number of iterations of the following:

$$\lambda_{q+k} = \varphi^{m_{q+k-1}}(\gamma_{[q-1]},(\lambda_{q+k-1} + a_{q+k-1})_*)$$ where

• $$m_{q+k-1}$$ is a finite integer
• $$\gamma_{[q-1]}$$ are recursively defined $$\gamma$$ ordinals, and
• If $$\gamma_1 = \gamma_1$$ (value within $$\lambda_{q+k-1}$$) then $$a_{q+k-1} = 0$$
• If $$0 < \gamma_1 < \gamma_1$$ then $$a_{q+k-1} = 1$$
• If $$0 = \gamma_1$$ then $$q$$ is decremented to $$q - 1$$ and $$0 < \gamma_2$$ or
• If $$0 = \gamma_2$$ then $$q = q - 1$$ and $$0 < \gamma_3$$ and
• For successive $$0 = \gamma_i$$ then $$q = q - 1$$ until $$0 < \gamma_{i+1}$$

These steps are identical to those for the construction of $$\lambda_q$$ but will continue until the construction of $$\lambda_{q+j+1}$$ when $$0 = \gamma_1$$ has occurred sufficiently to decrement $$q = 1$$ and

$$\lambda_{q+j+1}$$ collapses to $$= \lambda_{q+j} = \lambda$$

Construction Example for $$\omega^{\omega} + \omega.3 + 2$$

Here is a construction procedure for this arbitrary ordinal:

$$\gamma = \omega^{\omega} + \omega.3 + 2 = (\lambda\uparrow\uparrow t)^{\gamma_e}.\gamma_c + \gamma_a = (\varphi(\gamma_{[q]})\uparrow\uparrow t)^{\gamma_e}.\gamma_c + \gamma_a$$

Step 1: Let $$\gamma_a = \omega.3 + 2$$

Step 2: Let $$\gamma_b = \omega^{\omega}.\gamma_c$$ where $$\gamma_c = 1$$

Step 3: Let $$\gamma_d = (\lambda\uparrow\uparrow 2)^{\gamma_e}$$ where $$\gamma_e = 1$$

Step 4: Let $$\lambda = \varphi(\gamma_{[q]})$$ and $$\lambda = \omega$$ for $$q = 1$$ and $$\gamma_1 = 1$$

Then repeat these steps to define $$\gamma_a = \omega.3 + 2$$ according to the rules.

Step 5: Let $$\gamma_a = 2$$

Step 6: Let $$\gamma_b = \omega.\gamma_c$$ where $$\gamma_c = 3$$

Step 7: Let $$\gamma_d = (\lambda\uparrow\uparrow 1)^{\gamma_e}$$ where $$\gamma_e = 1$$

Step 8: Let $$\lambda = \varphi(\gamma_{[q]})$$ and $$\lambda = \omega$$ for $$q = 1$$ and $$\gamma_1 = 1$$

Setting $$\gamma_x = n$$ (i.e. any finite number) follows the special case rule

Step 9: Let $$\gamma_a = n$$ (i.e. any finite number) where $$q = 0$$

Resulting in:

$$\gamma = \omega^{\omega} + \omega.3 + 2$$