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It's easier to explain the question on an example. Let's consider this pretty simple problem:

We have to prove that $(-1)a=-a$ where $a \in R $.

My question is, can I prove this in the following way?

$$(-1)a\stackrel{?}{=}-a \Rightarrow (-1)a+a\stackrel{?}{=}0 \Rightarrow \\ a(-1+1)\stackrel{?}{=}0 \Rightarrow a*0\stackrel{?}{=}0 \Rightarrow 0=0 $$

Is this a correct proof?

Since we arrived to an obvious truth from uncertainty, can we conclude the problem proven?

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  • $\begingroup$ @khajveh No. You cannot assume what you want to prove. $\endgroup$ Commented Feb 29, 2016 at 22:26

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It's logically OK if you do not write $\Rightarrow$. Instead write $\Leftrightarrow$ everywhere. If you write $\Rightarrow$, it means you're proving the opposite direction (of what you want to prove). You need to prove the backward direction.

Another way to prove it is to write the same sequence of equations but in the reverse direction (starting from $0=0$). Then you can everywhere write $\Rightarrow$ and it's perfectly fine too.

Write both of these and think what you logically say by them.

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$$ 5 \stackrel{?}{=} 3 \Rightarrow 5*0 \stackrel{?}{=} 3*0 \Rightarrow 0 = 0 $$ shows that it is not allowed to prove a statement by showing that its assumption implies another true statement. However, your 'proof' does suggest a correct way to prove what you want to show: can you reverse the arrows?

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