Jayne's Equation 1.13 Derivation Dear Stack Exchange Members,
I'm reading 'Probability Theory - The Logic of of Science" by ET Jaynes, and I'm on pg. 11.
Jayne's says: *"...For example, we shall presently have  use for a rather elementary  theorem
if B'= AD then AB'=B' and BA'=A'*
..."
I'm not sure what he means by this.  What is D? What is this equation saying? How do I derive it?
I tried it out in excel and can't get the last column to be all trues.

A   B       D   B'=AD   AB'=B'  BA'=A'  E and F E = F
0   0       0   FALSE   FALSE   FALSE   FALSE   TRUE
1   0       0   FALSE   TRUE    FALSE   FALSE   TRUE
0   1       0   TRUE    TRUE    FALSE   FALSE   FALSE
1   1       0   TRUE    TRUE    TRUE    TRUE    TRUE
0   0       1   FALSE   FALSE   FALSE   FALSE   TRUE
1   0       1   TRUE    TRUE    FALSE   FALSE   FALSE
0   1       1   TRUE    TRUE    FALSE   FALSE   FALSE
1   1       1   FALSE   TRUE    TRUE    TRUE    FALSE

The code for the columns is as follows:
B'=AD - > =NOT(B2)=AND(A2,D2)
AB'=B' -> AND(A2,NOT(B2))=NOT(B2)
BA'=A' -> BA'=A'
E and F -> FALSE
=H2=E2  -> H2=E2
NB: I put column C is just blank to be more clear with my notation.
I would be grateful for any insight someone can provide.
EDIT: new table:

A   B       D   B'=AD   AB'=B'  BA'=A'  E and F E=F
0   0       0   FALSE   FALSE   FALSE   FALSE   TRUE
1   0       0   FALSE   TRUE    TRUE    TRUE    FALSE
0   1       0   TRUE    TRUE    TRUE    TRUE    TRUE
1   1       0   TRUE    TRUE    TRUE    TRUE    TRUE
0   0       1   FALSE   FALSE   FALSE   FALSE   TRUE
1   0       1   TRUE    TRUE    TRUE    TRUE    TRUE
0   1       1   TRUE    TRUE    TRUE    TRUE    TRUE
1   1       1   FALSE   TRUE    TRUE    TRUE    FALSE

Notice the rows where B' =AD is true?
See how AB'=B' and BA'=A' are also true in those rows?
I think what Jayne means is 
if B'= AD IS TRUE then THIS IMPLIES AB'=B' IS TRUE and BA'=A' IS TRUE*
That is, material implication.  Let's try it out.

A   B       D   B'=AD   AB'=B'  BA'=A'  E and F E=F      E->F
0   0       0   FALSE   FALSE   FALSE   FALSE   TRUE    TRUE
1   0       0   FALSE   TRUE    TRUE    TRUE    FALSE   TRUE
0   1       0   TRUE    TRUE    TRUE    TRUE    TRUE    TRUE
1   1       0   TRUE    TRUE    TRUE    TRUE    TRUE    TRUE
0   0       1   FALSE   FALSE   FALSE   FALSE   TRUE    TRUE
1   0       1   TRUE    TRUE    TRUE    TRUE    TRUE    TRUE
0   1       1   TRUE    TRUE    TRUE    TRUE    TRUE    TRUE
1   1       1   FALSE   TRUE    TRUE    TRUE    FALSE   TRUE

Yay, it works!  so, I assume Jayne means implies.
 A: Presumably $A$, $B$, $D$ are logical variables, multiplication is "and", and prime is "not".  So $B' = AD$ means "not-B is equivalent to (A and D)".
And then $AB' = A(AD) = AD = B'$.  Since $B'$ implies $A$, $A'$ implies $B$ which
means that $BA' = A'$.
EDIT: In your truth table, $BA'=A'$ should be true in all cases except
$A=0, B=0$, just like $AB'=B'$.
A: For a little more intuition/background on
$$\overline{B} = AD \implies A\overline{B} = \overline{B} \text{ and } B\overline{A}=\overline{A} $$
The if part of the statement is stating a certain relationship between $A$ and $B$, which includes an unrelated third statement $D$. Later, Jaynes uses this result while deriving the sum rule.
My intuition is that the $D$ is there to provide just enough of a relationship between $A$ and $B$, so a proof is possible, while being loose enough to allow the result to be general.
Instead of constructing truth tables, you could also verify these with plain boolean logic:
$$A\overline{B} = A \land AD = AD = \overline{B}$$
$$B\overline{A}= \overline{AD}\land\overline{A} = (\overline{A}+\overline{D})\overline{A} = \overline{A} + \overline{A}\land\overline{D} = \overline{A}(1+\overline{D}) = \overline{A}$$
