Why do later passes not mess up earlier sorts?

Prove that after pass p, sorted for digit p+1 to last digit

Prove by induction over p

- True for p=1. Apply Stable sort to digit 1, sorted after pass
- Assume true for p=i, prove true for p=i+1
- After pass i, compare two numbers x and y
- If x appears before y then one of following conditions must be true
- 1.
*x*<*y*for digit*i*, then*x*belongs before*y*in sorted order- 2.
*x*=*y*for digit*i*, then*x*<*y*for rest of number, placed in that order in earlier pass, not swapped because use stable sort