Is Adding Zero Valid in Inequality Proofs?

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Hi I'm doing the first chapter of Spivak's Calculus and just a little concerned about a
particular thing he does in the chapter.

He is talking about the trichotomy axiom and that if a > b then a - b, this can be
understood as expressing (a - b) > 0 and then the axiom can be interpreted as
(a - b) > 0, (a - b) = 0 or (a - b) < 0 (which means that - (a - b) > 0).

My concern is whether it's okay to write the proof that if a < b, and c < d then
a + c < b + d as follows:

If a < b then 0 < (b - a)
If c < d then 0 < (d - c)

Spivak writes that (b - a) + (d - c) = (b + d) - (a + c) and it follows that a + c < b + d
but it makes more sense to me to fill in the details as follows:

If a < b then 0 < (b - a)
If c < d then 0 < (d - c)
0 + 0 < (b - a) + (d - c)
0 < (b + d) - (a + c)
a + c < b + d

My concern is that it's a bit weird to add 0 + 0 on the left, I mean if you add something
to an inequality you should add it to both sides right?
0 < (b - a)
0 + (d - c) < (b - a) + (d - c)
d - c < (b - a) + (d - c)

But this is just circular, you've still got the same thing, that 0 < (b - a).
I ask this because in every inequality proof I've seen so far the method used is to add
to both sides:
I refer you to page 4 of the first chapter of Thomas Calculus as an example:

2x - 1 < x + 3
2x - 1 + 1 < x + 3 + 1
2x + 0 < x + 4
x < 2

How do I make sense out of this, I can see the sense in adding like:
0 + 0 < (b - a) + (d - c)
but am insecure as to whether this is valid or not.

It is the 5th problem of the first chapter of Spivak if you've done it yourself but it
really relates to the description in the text tbh

 

[╔(σ_σ)╝]

Your method is completely valid and the inequality is correct as long as each step is true :-).

 

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Great, I did most of the proofs that way & then it occurred to me that it might all be wrong, thanks