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Proof step 3 from proof ls-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n
Proof step 5:
m0 = m1 * n0 + n1, m < m1 * n, m0 < m1 * n, n1 < n0, m0 ≤ m, n ≤ n0 ⊦ m1 = 0, n = 0, n0 = 0

Rule argument

spec: <toplevel> name: div-elim seq: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n subst: [m, n, m0, n0] → [m0, n0, m1, n1] flags: <rotated> <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ 0 ↔ m = 0; (s) Used terms: m0, n0, m1, n1