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Proof step 2 from proof ls-01 (previous step)

m / n < m0 / n0, n ≤ n0, m0 ≤ m ⊦ n = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n
Proof step 4:
m = m1 * n + n1 ∧ n1 < n, m1 < m0 / n0, n ≤ n0, m0 ≤ m ⊦ n = 0

Rule argument

spec: nat-div inst: name: div-elim seq: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n subst: [m, n, m0, n0] → [m, n, m1, n1] flags: rewrite, current

Additional information

Used terms: m, n, m1, n1