CF 809C about proof of previous solution

(from https://math.stackexchange.com/questions/319551/nim-addition-binary-addition-without-carrying)

http://codeforces.com/contest/809/problem/C
（假设从0开始编号,要证明坐标(i,j)的数字为i XOR j）

…
So, to interpret your question, you have two different procedures for filling out a table of numbers which has a definite top and left edge, but continues indefinitely down and to the right. First is

Don’t fill in any cell before all the positions directly above and directly to the left of it are filled.
Write in each cell the smallest nonnegative integer that doesn’t appear in any cell directly above it or directly to the left of it
The second is

Write in the cell with (zero-based) coordinates (i,j)(i,j) the number i⊕ji\oplus j, where ⊕\oplus is the bitwise XOR operation (that is, as you say, binary addition with no carries).
And you want to prove that these two methods result in the same value in each cell of the table.

We can prove that by long induction on ii and jj. For the induction step we imagine that the cells above and to the left of (i,j)(i,j) have already been filled with the ⊕\oplus of their coordinates, and we want to prove that step (2) of the first procedure results in exactly i⊕ji\oplus j.

One half of this is easy: since ⊕\oplus is a group operation, i⊕ji\oplus j cannot equal i⊕bi\oplus b or a⊕ja\oplus j for any b≠jb\ne j or a≠ia\ne i – in particular not for any smaller aa or bb.

We then need to prove that i⊕ji\oplus j is the least number that hasn’t yet been used in the same row or column. In other words, every c<i⊕jc must already have been used somewhere.

Consider the most significant bit position where cc differs from i⊕ji\oplus j. Because cc is less than i⊕ji\oplus j, there must be a 0 in this position of cc and a 1 in the position of i⊕ji\oplus j. The latter 1 must come from either ii or jj. Assume it comes from ii (the other case is exactly similar). Then flipping that bit of ii produces a number a0a_0 such that a0⊕ja_0\oplus j agrees with cc up to and including the bit position of first difference. With appropriate adjustments to the less significant bits we can get an aa such that a⊕j=ca\oplus j=c. And this aa must be less than ii, because aa and ii first differ on a bit position that has 1 in ii and 0 in aa.

This completes the induction step, and thus the proof.

大致意思是先构造一个(i,j)(i,j)数字为i XOR j的矩阵,再证明小于i XOR j的数在同一行(?<i,j)(?、同一列(i,?<j)(i,? 上都出现过.
证明类似nim和：设i XOR j=d,对任一0 <=x < d,一定存在一个二进制位,x在这位为0,d在这位为1,那么可以通过减小i或j使新的i XOR j=x.

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