Claim
(a, b) < (c, d) with a,b,c,d ∈ [0,M) ⇔ a · M + b < c · M + d
(a, b) = (c, d) with a,b,c,d ∈ [0,M) ⇔ a · M + b = c · M + d
(a, b) > (c, d) with a,b,c,d ∈ [0,M) ⇔ a · M + b > c · M + d
Proof by case distinction
| relation to prove | constraint | evaluation | conclusion |
|---|---|---|---|
| (a, b) < (c, d) | a < c | a · M + M – 1 < c · M | ✓ |
| (a, b) < (c, d) | a = c ∧ b < d | a · M + b < c · M + d | 0 + b < 0 + d ✓ |
| (a, b) = (c, d) | a = c ∧ b = d | a · M + b = c · M + d | 0 + 0 = 0 + 0 ✓ |
| (a, b) > (c, d) | a = c ∧ b > d | a · M + b > c · M + d | 0 + b > 0 + d ✓ |
| (a, b) > (c, d) | a > c | a · M > c · M + M – 1 | ✓ |
Rationale
This just exploits the nature of a lexicographical order itself for bounded domains. Consider two binary numbers and they get compared. This integer comparison equals the lexicographical tuple comparison of its digit expansion. I just wanted to make this explicit as my head was screwing around with it.