Is this logical reasoning correct?

[frb]

Suppose I want to show that two functions f and g are equal. A way to prove this could be to prove the statement:
$$f(x) = n \Leftrightarrow g(x) = n$$

Is it enough to show one side of the implication?
Prove the following statement:
$$f(x) = n \Rightarrow g(x) = n$$

and reason as follows, suppose $$f(x) \neq n$$,
$$\Rightarrow \exists m\neq n: f(x)=m$$
$$\Rightarrow g(x)=m\neq n$$
Which would mean that I have shown the converse implication, and thus I have equivalence.

 

[tiny-tim]

Prove the following statement:
$$f(x) = n \Rightarrow g(x) = n$$

and reason as follows, suppose $$f(x) \neq n$$,
$$\Rightarrow \exists m\neq n: f(x)=m$$
$$\Rightarrow g(x)=m\neq n$$
Which would mean that I have shown the converse implication, and thus I have equivalence.

Hi frb!

Sorry, but I don't understand your proof, so I'm going to say no, it's not correct.

 

[frb]

Suppose I have proven the statement f(x) = n implies g(x)=n.

To obtain equivalence I have to prove f(x) != n implies g(x) != n.

So i reason as follows, suppose f(x) != n,
f(x) surely has another value, let this value be m, so f(x) = m, and m != n.
f(x) = m implies g(x) = m, and that implies that g(x) != n.

 

[tiny-tim]

Suppose I have proven the statement f(x) = n implies g(x)=n.

To obtain equivalence I have to prove f(x) != n implies g(x) != n.

So i reason as follows, suppose f(x) != n,
f(x) surely has another value, let this value be m, so f(x) = m, and m != n.
f(x) = m implies g(x) = m, and that implies that g(x) != n.

ah … that seems fine …

but if you want to prove f = g, which is the same as, for all x, f(x) = g(x),

then if you know for all x, f(x) = n implies g(x)=n,

then for all x, f(x) = g(x) … you don't need any more logic than that.

 

[frb]

of course. My brain is fried due to too much studying. I tend to make things more difficult then. I thought that there should be a flaw or something. Thanks though!

 

[n_bourbaki]

This proof assumes that f and g have the same domain.

It fails if they have different domains. But then by definition the functions are not equal, so there was no need to go any further.