A question regarding ❋166.44 in Whitehead & Russell's Principia Mathematica

Question

In the first step of Dem, I wonder how $\Sigma ‘\times P^{;}Q$ is transformed into $\Sigma‘ \Sigma^;(P \overset{\downarrow}{.,})\dagger^; Q$. https://i.stack.imgur.com/ojgzn.png Thanks,

Answer 1

This answer is inspired by a previous answer in which @Mauro discovered the similarities between ✳113.44 and ✳166.44 and suggested that their dems are analogous to each other.

$×α‘β$ is convenient because it is viewed as the descriptive function $×α$ taking $β$ as its input.

By ✳38.11, $ ×α‘\beta=α×β $

By ✳113.1, $ α×β=s‘α\underset{,,}{\downarrow}‘‘β $

By ✳37.101, $s‘α\underset{,,}{\downarrow}‘‘β=s‘α\underset{,,}{\downarrow}ϵ‘β$. In other words, $s‘α\underset{,,}{\downarrow}ϵ$ is also a descriptive function.

Now we can see that the two descriptive functions are equivalent: $×α=s‘α\underset{,,}{\downarrow}ϵ$

Given $k$ as a class of $β$'s, when we want the descriptive function $×α$ to go inside $k$ and perform $×α$ on every each of $k$'s members, we write $×α‘‘k$. (See ✳37.01)

Thus, $(\times \alpha)‘‘k = s‘‘\alpha \underset{,,}{\downarrow} \epsilon‘‘k=s‘‘ \alpha \underset{\overset{\textbf{,,}}{\ }}{\downarrow}‘‘‘k$.

I.e. First let $α\underset{,,}{\downarrow}ϵ$ go in and take each of $k$'s member as input, then let $s$ go into the resulting class and operate on each of the newly transformed members.

This explains the first step of 113.44's Dem: $s‘(×α)‘‘k=s‘s‘‘α\underset{ ,,}{\downarrow}‘‘‘k$.

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Now, cardinal $s$ is analogous to ordinal $Σ$, and cardinal $R‘‘$, $Rϵ$, $‘‘$ are analogous to ordinal $Σ;$, $Σ†$ and $;$ respectively. Like $×α‘β$, $×P‘R$ is a descriptive function of $R$.

By ✳38.11, $×P‘R=R×P$

By ✳166.1, $R×P=Σ‘P\underset{.,}{\downarrow}^;R$

By ✳150.1, $Σ‘P\underset{.,}{↓}^;R=Σ‘(P\underset{.,}{↓}†)‘R$, in other words, $Σ‘(P\underset{.,}{↓}†)‘R$ is a descriptive function of R.

Now we can see, the two descriptive functions are equivalent:

$×P=Σ‘(P\underset{.,}{↓}†)$ ------- (1)

Given $Q$ as a relation of relations,$ ×P;Q$ performs $×P$ on every term of $Q$'s field. In virtue of (1), $×P;Q$ is the same as $(Σ‘(P\underset{.,}{↓}†));Q$, which is to perform $Σ‘(P\underset{.,}{↓}†)$ on every each of Q's field, which is equivalent to perform $P\underset{.,}{\downarrow}†$ on each of Q's field, then perform $Σ $on each of the newly transformed field members, i.e. $Σ^;(P\underset{.,}{↓})†^;Q$. This explains the first step of 166.44 Dem, $Σ‘×P;Q=Σ‘Σ^;(P\underset{.,}{↓})†;Q$

Answer 2

It is very hard to "disentangle" this formula.

We have to start with the LHS of $✳166.44$ :

$\Sigma‘ \times P^{;} Q$.

The first step is to apply the "transformation" $✳166.1$ : $Q \times P = \Sigma‘P \underset{\overset{\textbf{.,}}{\ }}{\downarrow}^{;} Q$.

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Note

In the following steps I'll use "$\downarrow$" in place of "down-arrow with dot & comma". Alas! "downarrow" is already used in PM for the couple $x \downarrow y$.

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Applying $✳166.1$ to $\Sigma‘ \times P$ we get :

$\Sigma‘ P \downarrow^{;} \Sigma‘^{;} Q$.

Now we need $✳38.11$ to "play with" $\downarrow$ [see $✳38.11$ in vol.I, page 313 :

$x \downarrow‘ y = \downarrow y‘x = x \downarrow y$

and the example with "$\cap$" : $\alpha \cap \beta =\cap \beta‘ \alpha$. This is so because we can "read" the intersection of $\alpha$ and $\beta$ as a function $\cap \beta‘$ "applied to" $\alpha$.]

Assuming that we have correctly applied it (see the Addendum for details), we have :

$[\Sigma‘(\Sigma‘ P \downarrow)^{;}]^{;} Q$.

We finally apply $✳150.1$, which again is "quite simple" : $S^{;}Q = S \dagger Q$, to get :

$[\Sigma‘(\Sigma‘ P \downarrow) \dagger]^{;} Q$.

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Addendum

We have to consider (thanks to George...) $✳116.44$, which is the cardinal counterpart of $✳166.44$ (please, note the same number after the dot ...) :

$(s‘k) \times \alpha = s‘(\times \alpha)‘‘k$

Forgetting the "switch" of LHS with RHS, it has clearly the same "form" of :

$\Sigma‘ \times P^{;} Q = (\Sigma‘ Q) \times P$.

The first step in the proof of $✳116.44$ is an application of $✳113.1$ :

$\beta \times \alpha = s‘ \alpha \underset{\overset{\textbf{,,}}{\ }}{\downarrow} ‘‘\beta$ Def ["downarrow with two commas"].

This, in turn, is the correlate of $✳166.1$ :

$Q \times P = \Sigma‘P \underset{\overset{\textbf{.,}}{\ }}{\downarrow}^{;} Q$ Def.

Now, back again to the proof of $✳116.44$ : we apply $✳113.1$ to the RHS [where $s‘$ is the "$\beta$" and $\alpha$ is the "$\alpha$"], to get :

$s‘(\times \alpha)‘‘k = s‘ \alpha \underset{\overset{\textbf{,,}}{\ }}{\downarrow}‘‘s‘ ‘‘k$.

Here we need some transformation of the above RHS [see the text of PM page 114 and George's answer] :

$s‘s‘‘ \alpha \underset{\overset{\textbf{,,}}{\ }}{\downarrow}‘ ‘‘k$

Now, we can exploit the analogy between $✳166.44$ and $✳116.44$, "mimicking" the steps of $✳116.44$ :

$s‘(\times \alpha)‘‘k = s‘ \alpha \underset{\overset{\textbf{,,}}{\ }}{\downarrow}‘‘s‘ ‘‘k = s‘s‘‘ \alpha \underset{\overset{\textbf{,,}}{\ }}{\downarrow}‘ ‘‘k$

with something like :

$\Sigma‘ \times P^{;} Q = \Sigma‘ P \underset{\overset{\textbf{.,}}{\ }}{\downarrow}^{;} \Sigma‘^{;} Q = \Sigma‘ \Sigma‘ P \underset{\overset{\textbf{.,}}{\ }}{\downarrow}^{;;}Q$

and then replace one of the two "$;$" with the "dagger" ($\dagger$), according to $✳150.1$.

Answer 3

If $S^;P$ is like $S‘‘C‘P$ in an order similar to $P$, then in picture:

$S‘R$-----$S^;P$------$S‘T$

$S\uparrow$------------------$\overset{\smile}{S} \downarrow$

$R$----------$P$---------$T$

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Given $R=Q‘T$,When the correlator is $ \times P$, then $\times P^;Q$ can be pictured like this (from bottom up):

$R$----------------$\huge{Q}$--------------$T$

$\uparrow$

Take $(P\underset{.,}{\downarrow}\dagger) $ out of $(P\underset{.,}{\downarrow}\dagger)^;Q$'s field we have $Q$'s field.

$\uparrow$

$P\underset{.,}{\downarrow}^;R$-----$\huge{(P\underset{.,}{\downarrow}\dagger)^;Q}$---------$P\underset{.,}{\downarrow}^;T $

$\uparrow$

Take $\Sigma $ out of $\times P ^;Q$'s field, we have $(P\underset{.,}{\downarrow}\dagger)^;Q$'s field. Thus, $\times P^;Q =\Sigma^;((P\underset{.,}{\downarrow}\dagger)^;Q) $

$\uparrow$

$\Sigma‘P\underset{.,}{\downarrow}^;R$-----$\huge{\times P^;Q}$------$\Sigma‘P\underset{.,}{\downarrow}^;T $

||

$R\times P$--------$\huge{\times P^;Q}$----------$T \times P $

$\uparrow$

$ \times P$

$\uparrow$

$R$----------------$\huge{Q}$--------------$T$

Answer 4

This is just an illustration for 113.44:

Suppose $k=\{ \beta, \gamma, \delta, ... \}$

Then

$(\times \alpha)‘‘k$

$ = \{\times \alpha ‘\beta, \times \alpha‘\gamma, \times \alpha‘\delta, ...\} $

$=\{\beta \times \alpha, \gamma \times \alpha, \delta \times \alpha, ...\}$

$=\{s‘\alpha \underset{,,}{\downarrow}‘‘\beta, s‘\alpha \underset{,,}{\downarrow}‘‘\gamma,s‘\alpha \underset{,,}{\downarrow}‘‘\delta,... \}$

$=s‘‘\{\alpha \underset{,,}{\downarrow}‘‘\beta, \alpha \underset{,,}{\downarrow}‘‘\gamma,\alpha \underset{,,}{\downarrow}‘‘\delta,... \} $

$=s‘‘\alpha\underset{,,}{\downarrow}‘‘‘\{\beta, \gamma,\delta,... \} $

$=s‘‘\alpha\underset{,,}{\downarrow}‘‘‘k$

Or, For any $\beta\in k$

$\times \alpha‘\beta = \beta \times \alpha = s‘\alpha \underset{,,} {\downarrow}‘‘\beta= s‘(\alpha \underset{,,}{\downarrow})_{\epsilon}‘\beta$

Then

$(\times \alpha)‘‘k=s‘‘(\alpha\underset{,,}{\downarrow})_\epsilon‘‘k=s‘‘\alpha\underset{,,}{\downarrow}‘‘‘k$