Understanding La Disjonction de Cas Reasoning

[archaic]

I am reading a pdf where, under a "classic ways of reasoning" section, the author introduced a method called la disjonction de cas, which I think in English would be "case by case" reasoning. He enounced it as follows:
$$\text{Let }\mathrm A,\,\mathrm B\text{ and }\mathrm C\text{ be three propositions, then:}\\
\text{This implication is always true: } ((\mathrm A\Rightarrow\mathrm C)\wedge(\mathrm B\Rightarrow\mathrm C))\Rightarrow((\mathrm A \vee \mathrm B)\Rightarrow\mathrm C)$$
I am not sure I understand the point of this, here's how I am thinking about it:
If I can show that $((\mathrm A\Rightarrow\mathrm C)\wedge(\mathrm B\Rightarrow\mathrm C))$ is true, then I have proved that $((\mathrm A \vee \mathrm B)\Rightarrow\mathrm C)$ is true. I know nothing about the truth value of $\mathrm C$, so I should prove that both $\mathrm A$ and $\mathrm B$ are true to force the truthfulness of the proposition on the left side of the main implication and thus that on the right.
I feel that I am missing something, though, or that I am not seeing the main point. If you could evaluate my reasoning, and/or add something more, I'd be grateful.

EDIT: I should prove that the implications on both sides of the conjunction are true not $\mathrm A$ and $\mathrm B$, since showing the latter propositions to be doesn't imply that the right side of the main implication is true as $\mathrm C$ might not follow.

 

[Mark44]

I am reading a pdf where, under a "classic ways of reasoning" section, the author introduced a method called la disjonction de cas, which I think in English would be "case by case" reasoning. He enounced it as follows:
$$\text{Let }\mathrm A,\,\mathrm B\text{ and }\mathrm C\text{ be three propositions, then:}\\
\text{This implication is always true: } ((\mathrm A\Rightarrow\mathrm C)\wedge(\mathrm B\Rightarrow\mathrm C))\Rightarrow((\mathrm A \vee \mathrm B)\Rightarrow\mathrm C)$$
I am not sure I understand the point of this, here's how I am thinking about it:
If I can show that $((\mathrm A\Rightarrow\mathrm C)\wedge(\mathrm B\Rightarrow\mathrm C))$ is true, then I have proved that $((\mathrm A \vee \mathrm B)\Rightarrow\mathrm C)$ is true.

No, that's not how it works. If you are trying to prove the overall implication, you would need to also show that $(A \vee B) \Rightarrow C$ is true.

I know nothing about the truth value of $\mathrm C$, so I should prove that both $\mathrm A$ and $\mathrm B$ are true to force the truthfulness of the proposition on the left side of the main implication and thus that on the right.
I feel that I am missing something, though, or that I am not seeing the main point. If you could evaluate my reasoning, and/or add something more, I'd be grateful.

EDIT: I should prove that the implications on both sides of the conjunction are true not $\mathrm A$ and $\mathrm B$, since showing the latter propositions to be doesn't imply that the right side of the main implication is true as $\mathrm C$ might not follow.

One way to establish the overall implication is to use a truth table. From my work, it looks like the overall implication actually goes both ways.

What the implication is saying is that if A implies C and B implies C, then either A or B implies C.
Here's a simple example of the implication being used.

Define A as the statement $x = 1$.
Define B as the statement $x = -1$.
Define C as the statement $x^2 = 1$.
Clearly $A \Rightarrow C$, and $B \Rightarrow C$, so $(A \Rightarrow C) \wedge (B \Rightarrow C)$
Then, if either x = 1 or x = -1, then $x^2$ will be 1. In symbols, $(A \vee B) \Rightarrow C$.

 

[archaic]

No, that's not how it works. If you are trying to prove the overall implication, you would need to also show that (A∨B)⇒C(A∨B)⇒C(A \vee B) \Rightarrow C is true.

But the overall implication is always true! If $\mathrm P\Rightarrow\mathrm Q$ is true, then if I show that $\mathrm P$ is true, it follows that $\mathrm Q$ is true using the truth table of the implication (keeping in mind that implication is true).

 

[Mark44]

But the overall implication is always true! If $\mathrm P\Rightarrow\mathrm Q$ is true, then if I show that $\mathrm P$ is true, it follows that $\mathrm Q$ is true using the truth table of the implication (keeping in mind that implication is true).

It's not clear to me what you're trying to do. If you are merely using the implication, then yes, if P is true, Q must also be true. OTOH, if you are trying to prove the implication, you must show that when P is true, it necessarily follows that Q will be true. Note that for an implication $P \Rightarrow Q$, the only situation in which the implication is false is when P is true but Q is false. All other combinations of truth values result in a true implication.