Back to Theorem 4 again.

I understand Part A, I also understand theorem 5 which proves that addition is associative, however, when I get to theorem 6, proving that addition is commutative I am referred back to Theorem 4, part B.

In part B the first equation is:

x + 1 = x’, fine, I take it that I could also write y + 1 = y’ since x and y are both any elements of N.

I do not take it that I can write 1 + x = x’ , nor that I can write 1 + y = y’ since addition has not yet been shown to be commutative, i.e. that the order of the operands doesn't matter.

In Part B, section I of the proof I have trouble with the line that states:

x + y’ = (y’)’ = (x + y)’

This part of the proof has x = 1 so we get:

1 + y’ = (y’)’ = (1 + y)’

So how do you go from 1 + y’ to (y’)’ , i.e. 1 + y’ = (y’)’ ?

I could see y’ + 1 = (y’)’ following from x + 1 = x', or y + 1 = y' but that is not the case.

In Part B section II, I have a similar problem with:

x’ + y' = (x + y')’, again, I could see

y' + x’ = (y' + x)’ , but, again that is not what we have and you would still have the operands reversed on the right side.

I have been stumped for a couple of weeks now. Probably I am missing something simple, time to get humble.

Of course all of the statements are true, they just seem to have left a theorem out that establishes that they are true. Frankly it seems that the fundamental properties should be written:

x + 1 = 1 + x = x’, and

x + y’ = y’ + x = (x +y)’ …. the definition of Addition, Rev B.

Thanks.