Turing 1936 Skeleton Tables Procedure

I am reading Turing 1936 to learn about the halting problem from its origin. However, I encountered a roadblock upon reaching section four, in which Turing demonstrates that his m-configuration tables can be expressed as abbreviated "skeleton tables". I cannot wrap my head around how you are meant to use repeated substitution (as he says) to arrive at a complete m-configuration table. The first example he gives is this:

I took this table and followed these instructions: "If we were to replace C throughout by q (say), B by r, and a by x, we should have a complete table for the m-configuration f (q, r, x). f is called an "m-configuration function" or "m-function"." Which had me end up with this:

Now, all I've done is replace the variables like he explicitly said but I am unsure how to use this table compared to the previous m-configuration tables. I doubt this is correct, as I feel I am missing something big. When I see f (q, r, x), what does this mean? My understanding is something like:

• We have function f, which takes three inputs, two being m-configurations and the third being a symbol.
• We pass the values originally passed to function f back into f again if ə is not found, or into function f1 if ə is found. Either way the head is moved once to the left.
• From f1 we return C if a is found under the head, if something other than a is found we pass the values back into f1 and move the head right, or if its something other than a or not a (how is this possible) then we pass the values into f2 and move the head right.
• ...

I feel as though I am wrong in several ways and I can't find much discussion about this paper on the internet (shocker) so any help would be much appreciated! Even just seeing an expanded table could be enlightening.

The expression $$\mathfrak{f}\,(\mathfrak{C},\mathfrak{B},a)$$ is an $$m$$-configuration (i.e., a state), where

• $$\mathfrak{C}$$ is an $$m$$-configuration (for example, $$\mathfrak{q}$$),
• $$\mathfrak{B}$$ is an $$m$$-configuration (for example, $$\mathfrak{r}$$), and
• $$a$$ is a tape symbol (for example, $$x$$).

You can read $$\mathfrak{f}\,(\mathfrak{q},\mathfrak{r},x)$$ as the instruction “set the machine head over the leftmost $$x$$ and enter $$m$$-configuration $$\mathfrak{q}$$, or if there is no $$x$$ on the tape, enter $$m$$-configuration $$\mathfrak{r}$$.”

1. Look up the $$m$$-configuration (column 1).
3. Execute the corresponding behavior (column 3).
4. Enter the corresponding $$m$$-configuration (column 4).

Example

Suppose the machine is in $$m$$-configuration $$\mathfrak{g}$$ and the machine head is at the index marked with a dot on the string below (using $$\varepsilon$$ because I don’t know how to type the schwa symbol): $$\varepsilon\varepsilon 0xx1x\underset{\bullet}{1}0$$ In one of the previous tables, we can define a rule like this:

$$m$$-config. symbol operations final $$m$$-config.
$$\mathfrak{g}$$ $$1$$ $$L$$ $$\mathfrak{f}(\mathfrak{q},\mathfrak{r},x)$$.

Then the machine will move left, enter the $$m$$-configuration $$\mathfrak{f}(\mathfrak{q},\mathfrak{r},x)$$ and proceed as follows:

$$\varepsilon\varepsilon 0xx1\underset{\bullet}x10\quad\mathfrak{f}(\mathfrak{q},\mathfrak{r},x)$$

$$\varepsilon\varepsilon 0xx\underset{\bullet}1x10\quad\mathfrak{f}(\mathfrak{q},\mathfrak{r},x)$$

$$\varepsilon\varepsilon 0x\underset{\bullet}x1x10\quad\mathfrak{f}(\mathfrak{q},\mathfrak{r},x)$$

$$\varepsilon\varepsilon 0\underset{\bullet}xx1x10\quad\mathfrak{f}(\mathfrak{q},\mathfrak{r},x)$$

$$\varepsilon\varepsilon \underset{\bullet}0xx1x10\quad\mathfrak{f}(\mathfrak{q},\mathfrak{r},x)$$

$$\varepsilon\underset{\bullet}\varepsilon 0xx1x10\quad\mathfrak{f}(\mathfrak{q},\mathfrak{r},x)$$

$$\underset{\bullet}\varepsilon\varepsilon 0xx1x10\quad\mathfrak{f}_1(\mathfrak{q},\mathfrak{r},x)$$

$$\varepsilon\underset{\bullet}\varepsilon 0xx1x10\quad\mathfrak{f}_1(\mathfrak{q},\mathfrak{r},x)$$

$$\varepsilon\varepsilon \underset{\bullet}0xx1x10\quad\mathfrak{f}_1(\mathfrak{q},\mathfrak{r},x)$$

$$\varepsilon\varepsilon 0\underset{\bullet}xx1x10\quad\mathfrak{f}_1(\mathfrak{q},\mathfrak{r},x)$$

$$\varepsilon\varepsilon 0\underset{\bullet}xx1x10\quad\mathfrak{q}$$

Of course, if no $$x$$ was on the tape, it would instead end up in $$m$$-configuration $$\mathfrak{r}$$.