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Proof step 2 from proof two-loup-08 (previous step)

m < (m + n) / 2, m + 1 < n0, n < (m + n) / 2, m < n, n < n0, n0 ≤ (m + n) / 2, m < n0, m + 1 < (m + n) / 2 ⊦ m + n = 0, m = 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 + n = m0 * 2 + n1 ∧ n1 < 2 ∧ 2 ≠ 0, m < m0, m + 1 < n0, n < m0, m < n, n < n0, n0 ≤ m0, m < n0, m + 1 < m0 ⊦ m + n = 0, m = 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, 2, m0, n1] flags: rewrite, current

Additional information

Used terms: m + n, 2, m0, n1