# Use of “without loss of generality”

Why do we use “without loss of generality” when writing proofs?
Is it necessary or convention? What “synonym” can be used?

I think this is great question, as the mathematical use of “without loss of generality” often varies from its literal meaning. The literal meaning is when you rephrase a general statement

$$P(x)P(x)$$ is true for all $$x∈Sx \in S$$,

using another set (which is easier to work with)

$$P(z)P(z)$$ is true for all $$z∈Tz \in T$$,

where $$PP$$ is some property of elements in $$SS$$ and $$TT$$, and it can be shown (or is known) that $$S=TS=T$$.

For example:

• We want to show that $$P(x)P(x)$$ is true for all $$x∈Zx \in \mathbb{Z}$$. Without loss of generality, we can assume that $$x=z+1x=z+1$$ for some $$z∈Zz \in \mathbb{Z}$$. [In this case, $$S=ZS=\mathbb{Z}$$ and $$T={z+1:z∈Z}T=\{z+1:z \in \mathbb{Z}\}$$.]

• We want to show that $$P(x)P(x)$$ is true for all $$x∈Zx \in \mathbb{Z}$$. Without loss of generality, we can assume that $$x=5q+rx=5q+r$$ where $$q,r∈Zq,r \in \mathbb{Z}$$ and $$0≤r. [In this case, $$S=ZS=\mathbb{Z}$$ and $$T={5q+r:q∈Z and r∈Z and 0≤r.]

In the above instances, indeed no generality has been lost, since in each case we can prove $$S=TS=T$$ (or, more likely, it would be assumed that the reader can deduce that $$S=TS=T$$). I.e., proving that $$P(z)P(z)$$ holds for $$z∈Tz \in T$$ is the same as proving that $$P(x)P(x)$$ holds for $$x∈Sx \in S$$.

The above cases are examples of clear-cut legitimate usage of "without loss of generality", but there is a widespread second use. Wikipedia writes:

The term is used before an assumption in a proof which narrows the premise to some special case; it is implied that the proof for that case can be easily applied to all others (or that all other cases are equivalent). Thus, given a proof of the conclusion in the special case, it is trivial to adapt it to prove the conclusion in all other cases.

[emphasis mine.]

So, paradoxically, "without loss of generality" is often used to highlight when the author has deliberately lost generality in order to simplify the proof. Thus, we are rephrasing a general statement:

$$P(x)P(x)$$ is true for all $$x∈Sx \in S$$,

as

$$P(z)P(z)$$ is true for all $$z∈Tz \in T$$, and

if $$x∈Sx \in S$$, then there exists $$z∈Tz \in T$$ for which $$P(x)P(x)$$ is true if $$P(z)P(z)$$ is true.

For example:

• Let $$SS$$ be a set of groups of order $$nn$$. We want to show $$P(G)P(G)$$ is true for all $$G∈SG \in S$$. Without loss of generality, assume the underlying set of $$GG$$ is $${0,1,…n−1}\{0,1,\ldots n-1\}$$ for some $$n≥1n \geq 1$$. [Here, $$TT$$ is a set of groups with underlying set $${0,1,…n−1}\{0,1,\ldots n-1\}$$ that are isomorphic to groups in $$SS$$, and the reader is assumed to be able to deduce that property $$PP$$ is preserved by isomorphism.]

My personal preference is to replace the second case with:

"It is sufficient to prove $$P(z)P(z)$$ for $$z∈Tz \in T$$, since [[for some reason]] it follows that $$P(x)P(x)$$ is true for all $$x∈Sx \in S$$."