Error in Intuitive Understanding of First-Order Logic w/ 'And' and 'If-Then

[julypraise]

One of the ordered field, F, property is the following (i):

(i) for every x, y, and z, if both x, y, z in F and y < z, then x + y < x + z.

Now please think about (i'):

(i') for every x, y, and z, if x, y, z in F implies y < z, then x + y < x + z.

I cannot prove that (i) and (ii) are equivalent simply using some simple logical rules. But when I inspect (i') in a intuitive sense, it seems (i') does not have a difference in terms of meaning with (i). Moreover, (i') is equivalent to (i'') which is the following:

(i'') for every x, y, z in F, if y < z, then x + y < x + z.

Especially when I think about this (i''), it seems really similar in terms of meaning to (i''). Yet, I cannot make (i) in the same form as (i'').

Thus, my question are these:

(1) Is (i) equivalent to (i')?
(2) If they are not, why does my intuitive understanding of those two sentences' meaning makes error? Any opinion?


My plausible guess for the question (2) is that it is because (i') implies (i). But anyway, I'm not sure whether it is a right anwer.



Further question:

What is wrong with the following proof, or is the following proof correct:

(->). Suppose (X and Y) implies Z. Suppose X. Suppose Y. Since X and Y, Z is true. Thus, Y implies Z. Thus X implies (Y -> Z).
(<-). Suppose X implies (Y implies Z). Suppose X and Y. Since X is true, Y implies Z. Since Y is true, Z is true too. Thus (X and Y) implies Z.
In conclusion, we proved that that (X and Y) implies Z is equivalnet to that X implies (Y -> Z).

Remark.
When proving the above in Fitch form style, I didn't find any contradiction. Please try that too and comment whatever you think.

 

[Jhero]

The proof is incorrect. The problem lies in the first step, where you assume that "(X and Y) implies Z". This does not necessarily have to be true, so you cannot use it as a starting point for your proof. Instead, you should start by assuming that X implies (Y -> Z), and then try to prove that (X and Y) implies Z.