Htykuut

I"m wondering whether it's valid to carry logical operations across false statements, same as as you would across true statements.

Compare:

$$begin{alignat*}{3}
& 1 < 2 \
Rightarrow & 1 + 1 < 2 + 1 \
Rightarrow & frac{1}{3} < frac{1}{2}\
end{alignat*}$$

With:

$$begin{alignat*}{3}
& 2 < 1 &&hspace{2cm} text{is false} \
Rightarrow & 2 + 1 < 1 + 1 &&hspace{2cm} text{is false} \
Rightarrow & frac{1}{2} < frac {1}{3} &&hspace{2cm} text{is false} \
end{alignat*}$$

$(2<1) → (3<2)$ is True, because the antecedent is False (see Material conditional).

Logical operations are carried on statements, either true or false.

But valid logical operations license the truth of the conclusion only when applied to true premises.

If I understand you right, you're asking, in essence,

If I have two claims $A$ and $B$ and I have proved $ARightarrow B$, would that also work for proving $neg A Rightarrow neg B$?

No, you can't do that. For an example where it goes wrong, you can reason
$$ x > 2 implies x^2 > 4 $$
but there are examples of $x$ where $x>2$ is false yet $x^2>4$ is true -- e.g. this is the case for $x=-3$.

The conclusion is your particular example is indeed true, but you need to justify it differently. One option is to simply write "$xge y$" instead of "$x < y$ is false"; it turns out that each of your rewritings is still true with a different inequality sign.

Another option is to note that each of the $Rightarrow$ in your original reasoning could actually be $Leftrightarrow$ -- and it does generally hold that if you have proved $ALeftrightarrow B$, then you'll also know $neg ALeftrightarrowneg B$.

Logic doesn’t care about what is actually true; it just concerns itself with what can be inferred from what. Therefore, logic can be applied just as well to true statements as false statements.

Thus, for example, if you have as an axiom that whenever $x <y$ then $x+1<y+1$, then you can use that to infer $3<2$ from $2<1$ just as much as you can infer that $2<3$ from $1<2$

No.

2 < 1 is false. Therefore, "2 < 1 implies that (2 + 1) < (1 + 1)" is a true statement, because F => T and F => F are both true in classical logic. However, by the same reasoning, we can't infer 2 < 1 as false, and "2 < 1 implies that (2 + 1) < (1 + 1)" as true, that (2 + 1) < (1 + 1) is false, since (2 + 1) < (1 + 1) could be true.