def isqrt_le

bend-collections-laws-crypto@1.0.0.0 · proofs/math/natural/sqrtn.bend:142

import Base
import bend-collections-laws-crypto@1.0.0.0/proofs/lib/nat.bend as N
import bend-collections-laws-crypto@1.0.0.0/proofs/lib/lemmas/proofs/nat_algebra.bend as A
import bend-collections-laws-crypto@1.0.0.0/src/math/natural.bend as M
import bend-collections-laws-crypto@1.0.0.0/src/math/pow2.bend as P2

142# isqrt(n)^2 <= n   (Mathlib Nat.sqrt_le')
143def isqrt_le(+n: Nat) -> {Nat.is_le(Nat.pow(M.isqrt(n), 2n), n) == True{} : Bool}:
144  +g = P2.pow2t(1n+Nat.div(M.bit_length(n), 2n))
145  %Equal.sym(Nat, Nat.pow(M.isqrt(n), 2n), Nat.mul(M.isqrt(n), M.isqrt(n)), Equal.cong(Nat, Nat, z => Nat.mul(M.isqrt(n), z), Nat.mul(M.isqrt(n), 1n), M.isqrt(n), A.mul_one(M.isqrt(n)))) : {Nat.is_le(_, n) == True{} : Bool}
146  iter_sq(g, n, g, M.sqrt_next(n, g), Nat.is_lt(M.sqrt_next(n, g), g), {==}, {==}, N.le_refl(g))