Without loss of generality

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Without loss of generality (abbreviated to WLOG or WOLOG and less commonly stated as without any loss of generality) is a frequently used expression in mathematics. The term is used before an assumption in a proof which narrows the scope of analysis to some subset of possible instantiations of the premise; it is implied that the structure of the proof on this subset can be generalized to all others. Thus, through a subproof depending on a subset of the premise, we show that conclusions are proven to follow from the full premise.

This often requires the presence of symmetry. For example, if two numbers are called x, y, and it is known that x < y, then any relationship proved based on this assumption will hold for the complementary relation, y < x, because the roles of x and y are interchanged, but the proof is symmetric in the two variables. In other words, if we know that P(x, y) is true if and only if P(y, x) is true, then without loss of generality it is enough to show P(x, y) is true (since P(y, x) then immediately follows, by symmetry). (In this context, we call P symmetric.)

Next to WLOG there must be present an assumption. To check that there is no loss of generality, write out the entire proof (without making the simplifying assumption) and then see if the proof which you wrote out follows out of a proof of just a part of the premise.

[edit] Example

Consider the following theorem (the simplest case of Ramsey's theorem and also an example of Dirichlet's pigeonhole principle):

Three objects are each painted either red or blue; there must be two objects of the same color.

The proof:

Assume without loss of generality that the first object is red. If either of the other two objects is red, we are finished; if not, the other two objects must both be blue and we are still finished.

We begin the full proof by listing all the permutations, separating those with R first from those with B first:

RRR
RRB
RBR
RBB
BRR (inverse of #4)
BRB (inverse of #3)
BBR (inverse of #2)
BBB (inverse of #1)

of which there are eight, as we expect (2 × 2 × 2). We now see that the separated lists are equivalent under our assumptions (the first half can be converted to the second half by converting all Rs to Bs and vice versa), so we can apply our analysis to the simpler half-list of permutations beginning with R.

We scan the shorter list (permutations 1-4) and see that there are two objects of the same color in every case. In permutations 1-3, at least one object beyond the first is red, and in permutation 4 both of the non-first objects are blue.

We reduced the premise as we did by noticing, first, that the order of the objects doesn't matter; second, that we are interested not in the kind of color but in its count. Both assumptions are consistent with our conclusion, which required that two objects be of the same color -- a loose requirement.