Calculi of Lambda Conv: Question on Abstraction I'm reading Page 4 of the Calculi of Lambda Conversion by Church.

BI is 1, since it is the operation of composition with the identity transformation, and thus an iden- tity operation, but one applicable only to transformations. ... The reader may further verify that CK. is H, CT is 1, C1 is T, CI is T.

He has previously defined these identities before the above quote:
Bfgx is f(gx)
 Ix = x
 II = I
 Hx = I
 HH = I
 KII = I
 KHI = H
 KI = H
 1f = f
 I != 1
 Cfxy = (fyx) 
 Txf = (fx)

Appreciate anyone who can show how Church arrives at his conclusions in the above quote.  For the above quote, there are not even enough variables in  the identities which follow below the quote to substitute in for.  How does he arrive at these claims? 
Thanks
 A: I think you might have some confusion regarding what these functions are and how they work. Let's go through all of the definitions in slightly more modern language and hopefully get you on more stable footing. To prove the identities you're interested in, we need to know what $K, H, I$ and so on actually are.
On page 2, Church writes "The identity function $I$ is defined by the rule that $(I x)$ is $x$". Thus $I$ is a function with $I(x) = x$. In the lambda calculus it is traditional to not write parenthesis for function application, which is why you see it written as $Ix = x$.
Similarly, we see that $Hx = I$. That is, $H$ is a function which ignores its input and returns $I$. As an example:
$$Hxy = (Hx)y = Iy = y$$
Notice that we associate function application left. This is important. Instead of having functions of multiple arguments, we instead have functions of one argument which return new functions. This is often called Currying. As another example, we might have $+3$ is the function which adds 3 to a number. Then $(+3)5 = 8$ and $(+7)4 = 11$. In this language, $+$ is a function which takes $n$ and returns the function $+n$.
Next (on page 4), Church defines $K$ to be a function returning a constant function -- $Kxy = x$. Said another way, $K$ is a function which ignores its second input.
$K3$ is the function which sends $y$ to $3$, no matter what $y$ is. $Kx$ is the function which sends $y$ to $x$, whatever $x$ is. 
It is for this reason that $KII = I$, and $KHI = H$. In both cases we simply return the first input to $K$. We might just as well say
$$K \text{ red } \text{blue} = \text{red}$$ 
it is the same idea.
Now we can ask "Why does Church say that KI = H?". Well two functions are equal if and only if they do the same thing. That is, $f=g$ if and only if $fx = gx$ for every x. So let's compare $KI$ with $H$: $KIx$ ignores $x$ and returns $I$, but $Hx$ is defined to be the function which ignores $x$ and returns $I$! Since $KI$ and $H$ agree on all inputs, we can safely say that $KI = H$.
Next, Church defines $1fx = fx$. The function $1$ is like the identity function, in that $1f = f$ for every function $f$. This is again because $1fx = fx$ for each input, and so the functions are equal. Church clarifies that $I \neq 1$ because $I$ can take anything as input, whereas $1$ can only take functions as input.
Finally, we get to a big list:
$Txf = fx$, $Bfgx = f(gx)$, $Cfxy = fyx$, and so on.
Let's prove the claims which you mention:

$$CK = H$$
Slowly at first. Let's see what $CK$ and $H$ do to some inputs, $x$ and $y$.
$CKxy = Kyx$ by the definition of $C$.
Next, $Kyx = y$, by the definition of $K$.
This tells us $CKxy = y$.
Similarly, $Hxy = Iy$, by the definition of $H$.
Then $Iy = y$, by the definition of $I$.
Where did we end up? $Hxy = y$.
So for every $x$ and $y$, $CKxy = Hxy$, and thus $CK = H$.

$$CT = 1$$
A little bit faster now -- let's see what they do to a general input.
I'll use $f$ and $x$ instead of $x$ and $y$, because they give a clearer picture of what is happening. Obviously the names we choose for the variables does not matter, just as 
$f(x) = x^2$ and $g(y) = y^2$ are the same function.
$CTfx = Txf = fx$, by the definitions of $C$ and $T$.
$1fx = fx$, by the definition of $1$.
So $CTfx = 1fx$ for every $f$ and $x$, and $CT = 1$.

$$C1 = T$$
We're almost through! Can you guess how we'll prove this?
$C1xf = 1fx = fx$
$Txf = fx$
So $C1 = T$

$$CI = T$$
At this point you know the drill. But the completionist in me just had to keep going.
$CIxf = Ifx = fx$
$Txf = fx$
So $CI = T$.

There are many relations between these functions, and the combinatorics you can do with them can be lots of fun! As Church says, some of these relations are extremely complicated. This is the magic of lambda calculus -- even though the rules of the game are very simple (once you get used to substituting these functions into each other, it is extremely routine. Think about how each of the above proofs felt so similar!), the game itself can be extremely complicated. Indeed any computable function can be modeled by some (very long) string of these $K$s and $H$s and so on! The process of simplifying the terms as we were doing above is a kind of computation, after all. Have fun playing around with all this!

I hope this helps ^_^
