Are all "without loss of generality"-questions logic-based?

Question

This Wikipedia article and AoPS page have the same proof that is an example of WLOG. These and other WLOG proofs seem to be connected with logical statements. Am I wrong? However, a question I asked a long time ago had an answer that involved WLOG. It, on the other hand, did not involve logic, as far as I know. Here is a snippet of the answer:

Applying the affine substitution $u = \frac{x - a}{b - a}$, $dx = \frac{du}{b - a}$ transforms the integral to $$(-1)^k (\beta - \alpha)^{j + k + 1} \int_0^1 u^j (1 - u)^k \,du$$ so, there's no loss of generality if we consider $\alpha = 0, \beta = 1$ and consider just the integral above, which also eliminates the pesky minus sign.

What does he mean when he references WLOG? I do not see logic or contradictions. Comparing this is the example on Wikipedia, the example addresses all possible cases, but this seems to not follow the pattern.

Question: Is there a checklist to follow? For example, do you need to describe all cases? And how are the uses of WLOG similar in the wikipedia article and the case I described above?

Edit: This is not similar to other questions as I ask about WLOG in relation to substituting for integrals, specifically.

WLOG- Without Loss of Generality. It's very useful when applicable. It lets you reduce a general problem to only needing to consider a specific case. "WLOG x is even and y is odd. then x=2p, y=2q+1, x+y=2(p+q)+1. It follows that the sum of any even number any odd number is odd." Otherwise you'd have to repeat the proof almost identically for the case when y is even. — TurlocTheRed, Sep 11 '23 at 18:39
The OP is asking whether WLOG is restricted to logical equivalences. That is not covered by the question cited as a duplicate. Hence my vote to reopen. — Rob Arthan, Sep 11 '23 at 23:17

score 4 · Answer 1 · answered Sep 10 '23 at 18:23

4

WLOG means "without loss of generality". Here this means that it is sufficient that the method works for integrals with the bounds $0$ and $1$ since every other integral can be transformed into such an integral. Therefore we can assume that the bounds are $0$ and $1$ without losing anything.

answered Sep 10 '23 at 18:23

Peter

84,454

Is this true for other integrals? For example, if I had a question: $\int^b_a (x^a)\cdot(x^b)dx$, could I simply assume $a=0,b=1$? – sreysus Sep 10 '23 at 18:29
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No, you can't "simply assume" anything. If you wish, you could transform to an integral with bounds $0$ and $1$. But that won't transform away the other $a$ and $b$ in the integrand. – Robert Israel Sep 11 '23 at 21:21
I am sure you know how I meant it , namely that it is sufficient that the method works for those bounds to work always. And I think the vast majority of the readers as well. – Peter Sep 12 '23 at 11:08

score 2 · Accepted Answer · answered Sep 12 '23 at 23:20

2

"Without loss of generality, let us prove $B$" at the beginnning a proof of $A$ is really just a short-hand for saying: "it is easy to see that $B$ implies $A$, so let's prove $B$ instead". A typical use would be to prove a result about the order relations between points on the real line by assuming WLOG that one of the points is $0$. This depends on mathematical facts and not just logical facts.

In most uses of the phrase "without loss of generality", $A$ and $B$ are equivalent. If they aren't, then it is clearer to say "it suffices to prove $B$" or "let us prove the stronger statement $B$".

answered Sep 12 '23 at 23:20

Rob Arthan

48,577

So, is WLOG informal as you don't prove A and only work your proof by an assumption? If not, why and would you see WLOG in formal papers? – sreysus Sep 13 '23 at 00:49
@sreysus, It is considered to be formal enough when any educated reader can easily adapt the proof to the original, general case. When properly used, it is a good practice as the reader can focus on the key idea of the proof itself while not being distracted by all the other nitty-gritty details needed for the general case. You can indeed find its usage throughout actual, formal math papers. – Sangchul Lee Sep 13 '23 at 01:29

score 1 · Answer 3 · answered Sep 12 '23 at 23:50

To add on to the answer by Rob Arthan, I would say that "without loss of generality, we may assume $\varphi$" often means that $A$ is in the form of some statement with hypotheses, and $B$ is in the form of $A$ but with the extra hypothesis $\varphi$ added in.

So, for example, let us consider the following simple example of a WLOG argument:

For any $x, y \in \mathbb{R}$, we have $\min(x,y) + \max(x,y) = x+y$.

Proof: By symmetry, without loss of generality, we may assume that $x \le y$. But in this case, $\min(x,y) = x$ and $\max(x,y) = y$, and the desired result follows immediately.

In this example, the symmetry argument which is being hinted at is to prove:

$$(\forall x, y \in \mathbb{R}, x \le y \rightarrow \min(x,y) + \max(x,y) = x + y) \\ \implies \\ (\forall x, y \in \mathbb{R}, \min(x,y) + \max(x,y) = x + y).$$ And in more detail, the way to prove this would be: suppose we name the assumption with added hypotheses $B$, and we have real numbers $x,y$. Then by trichotomy, we have either $x<y$, $x=y$, or $x>y$. In the first two cases, then $x \le y$, and the desired conclusion follows directly from $B$. In the case $x>y$, on the other hand, then we have $y \le x$, so applying $B$ with $x$ and $y$ reversed gives us $\min(y,x) + \max(y,x) = y+x$. Now, substituting $\min(y,x) = \min(x,y)$, $\max(y,x) = \max(x,y)$, and $y+x = x+y$ gives the desired conclusion in this case as well.

Are all "without loss of generality"-questions logic-based?

3 Answers3