Laplace and First uniqueness theorem proof: Why potential difference?

On Griffith's "Introduction to Electrodynamics" page 120 the author states that when proving the First Uniqueness theorem: The solution to Laplace's equation in some volume V is uniquely determined if V is specified on the boundary surface S....
We suppose there are two solutions to Laplace equation: $$\nabla^2V_1=0\\\nabla^2V_2=0$$
the key to the proof is to look at potential difference: $$V_3 \equiv V_1 - V_2$$.
Why do we choose the potential difference. I understand the rest of the proof, how you get that $$V_2=V_1$$ and prove that it's valid... but why the difference? When you know the rest of the derivation, it's obvious, but what if I didn't know the rest of the proof.

If you want to prove that V1=V2, and you know that V1 does equal V2 at the boundary, a nice way to do it would be to prove that their difference V3=V1-V2 is equal to 0 everywhere. Proving that V3=0 is evidently much easier than proving uniqueness by itself, because you can exploit the only known fact from the problem - V1 equaling V2 at the boundary, and get a strong statement that V3 is 0 at the boundary, which you can use in conjunction with tha fact V3 also satisfies the Laplace's equation! On the other hand, using summ or product of V1 and V2 doesn't exploit the boundary condition in any way, so it's not in any way useful.

You can find so called 'happy ideas' all the time when reading the proof of any mathematical statement. And even if that key idea was somehow motivated by some sort of intuition, in general authors tend to omit mentioning that when writing the proof. A famous quote attributed$$^1$$ to Gauss is "A good building should not show its scaffolding when completed".

$$^1$$From W. Sartorius Von Waltershausen, Carl Friedrich Gauss: A memorial, see it translated here.

• Thanks for nothing. Commented Jul 18, 2020 at 0:59
• Sorry, it was not my intention to offend you in any way. I posted my answer because your question (v1) doesn't seem to be about the proof of the theorem itself; I understand the rest of the proof, you say. But rather, as you also say: When you know the rest of the derivation, it's obvious, but what if I didn't know the rest of the proof. From this last sentence, it seems that you are not used to mathematical proofs, hence my answer.
– Urb
Commented Jul 18, 2020 at 11:06
• No problem. I only wish to know why we looked at the potential difference and not potential summation or product or anything like that. "The key is to look at the potential difference" is the quote from Griffiths and my question is "Why?" Commented Jul 18, 2020 at 11:19

To make it short (the type of argumentation already well explained by "vucko" ) I will give a formal summary of vuckos post:

If $$V_1$$ and $$V_2$$ fulfill the same boundary conditions, therefore $$V_3 = V_2 - V_1$$ has zero value as boundary conditions. Furthermore

$$\Delta V_3 = \Delta (V_2 -V_1)=\Delta V_2 -\Delta V_1 =0$$

Therefore $$V_3$$ fulfills the Laplace equation with zero boundary conditions. Therefore $$V_3=0$$. Consequently $$V_2=V_1$$. Q.E.D.