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Proof step 3 from proof two-loup-04 (previous step)

m + 1 < (m + n) / 2 - 1, m + 1 < n, m < n, n < (m + n) / 2 - 1, m < (m + n) / 2 - 1, 2 + m < (m + n) / 2 - 1 ⊦ m + 1 = n

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:
m + n = m0 * 2 + n0, 2 + m < m0 - 1, m < m0 - 1, n < m0 - 1, m < n, m + 1 < n, m + 1 < m0 - 1, n0 < 2 ⊦ m + 1 = n, 2 = 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] → [m + n, 2, m0, n0] flags: <rotated> <all paths>

Additional information

Used terms: m + n, 2, m0, n0