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

In the first step of Dem, I wonder how $\Sigma ‘\times P^{;}Q$ is transformed into $\Sigma‘ \Sigma^;(P \overset{\downarrow}{.,})\dagger^; Q$. http://i.stack.imgur.com/ojgzn.png Thanks, 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" $✳166.1$ : $Q \times P = \Sigma‘P \underset{\overset{\textbf{.,}}{\ }}{\downarrow}^{;} Q$.

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 $x \downarrow y$.

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 :

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

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 $✳150.1$.

• Thanks, @Mauro. See below. Now I can picture this transformation. But I still don't know how the magic of ✳38.11 works. I think the precedence is $\Sigma‘(\times P^;Q)$ Aug 21 '14 at 2:17
• @GeorgeChen - But I'm still puzzled; if I follow the above comment, assuming that $Q \times P$ and $\times P^{;}Q$ are "equivalent", the formula boils down to : $\Sigma‘(Q \times P) = (\Sigma‘Q) \times P$. Does it make sense ? Aug 21 '14 at 6:28
• Almost a volume later, I'm just beginning to appreciate 38.11. The power of 38.11's magic is that it converts $\beta \times \alpha$ into a form that can take a class as arguments. When $\times \alpha$, with the help of $‘‘$, takes a class $k$ as argument, it performs $\times \alpha$ on each of $k$'s member without having to enumerate every each of them. This is one of the techniques PM uses to make infinity accessible to human knowledge. $;$ is just the ordinal counterpart of $‘‘$ Aug 24 '14 at 0:54
• $R\times P$ is analogous to $\beta \times \alpha$; $\times P ^;Q$ is analogous to $\times \alpha‘‘k$ where k is a class of classes and $Q$ is a relation of relations Aug 25 '14 at 3:28
• Sorry, @Mauro. I can't accept your answer as it is, but the final answer is totally inspired by your line of thought. Aug 28 '14 at 0:38

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

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

• I see ... But (as in the proof of $✳116.44$) why there is no need of $✳38.11$ ??? Or we have to presume that the appeal to $✳38.11$ in the proof of $✳166.44$ is mistaken ? Aug 21 '14 at 10:13
• I added another line. Yes, 38.11 is implicit in pretty much every step. Aug 21 '14 at 10:16
• $\times \alpha ‘\beta = \beta \times \alpha$ that's 38.11 Aug 21 '14 at 10:19
• Thanks agian; but how we can "reconcile" $\times \alpha ‘ \beta = \beta \times \alpha$ with $\beta \times \alpha = s‘ \alpha \downarrow ‘‘ \beta$ to get the expected transformation ? Aug 21 '14 at 10:27
• I like your way of thinking, @Mauro. I can "see" the step now. Aug 21 '14 at 10:29

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: $s‘(×α)‘‘k=s‘s‘‘α\underset{ ,,}{\downarrow}‘‘‘k$.

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, $Σ‘×P;Q=Σ‘Σ^;(P\underset{.,}{↓})†;Q$

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

$$S‘R$$-----$$S^;P$$------$$S‘T$$

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

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

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, $$\times P^;Q =\Sigma^;((P\underset{.,}{\downarrow}\dagger)^;Q)$$

$$\uparrow$$

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

||

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

$$\uparrow$$

$$\times P$$

$$\uparrow$$

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

• Sorry, I cannot grasp it. Assuming you corrected reading as : $Σ‘(\times P^{;}Q)$ if we apply $✳166.1$ we get : $\Sigma‘P^{;}Q \underset{\overset{\textbf{.,}}{\ }}{\downarrow}^{;} \Sigma‘$. In this case, we can use $✳38.11$ to "flip around" the "downarrow" but only in one way which moves it to the left, like in $\alpha \cap \beta = \cap \beta‘ \alpha$... Aug 21 '14 at 6:32
• OK I can see. We can leave out the first $\Sigma$ and work on $×P^; Q$ which is like Q by the correlater $×P$ Aug 21 '14 at 7:10
• Shown by the first step from the bottom. Aug 21 '14 at 7:15