# The 'formal' way to prove equivalence?

In school we were always taught to prove equivalence by splitting an equation into $LHS$ and $RHS$ and working with each side individually until $LHS = RHS$.

For example, prove:

$$2^{k + 1} - 2 = 2(2^k - 1)$$

Which could be done as follows:

$$RHS = 2(2^k - 1)$$

$$= 2^1 \cdot 2^k - 2$$

$$= 2^{k + 1} - 2$$

$$= LHS$$

However, Wikipedia says (although it does not cite any sources):

This abbreviation is seldom if ever used in print; it is very informal.

Is Wikipedia correct, and if this is "very informal", what is the proper, 'formal' way to prove equivalence (without the $LHS$ and $RHS$ abbreviations)?

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I'm sure I've seen that in print... – apnorton Feb 6 '13 at 12:12
I would at least agree that it is seldom used. Usually you just write the actual expression for the left/right hand side. So in this example, replace LHS by $2^{k+1}-2$ and RHS by $2(2^k-1)$. It makes it clearer anyway, because you don't have to look back to see what the left and right hand sides actually were. – Matthew Pressland Feb 6 '13 at 12:13
As a referee, I would flag any uses of "LHS" or "RHS" in a paper as needing to be corrected before the paper is accepted. – Carl Mummert Feb 6 '13 at 12:46

Prop. $2^{k + 1} - 2 = 2(2^k - 1)$
Proof: $2^{k+1} - 2 = 2^1\cdot 2^k -2=2\cdot (2^k-1)$. -QED-