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Proof step 3 from proof mult (previous step)

∀ n, n0. n ≠ 0 → (n0 + m0 * n) / n = n0 / n + m0 ⊦ n = 0, (n0 + n * m0) / n = n0 / n + m0

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
n ≠ 0 ⊦ (n0 + n) / n = n0 / n + 1
Proof step 5:
((m0 * n + n0) + n) / n = (m0 * n + n0) / n + 1, ∀ n, n0. n ≠ 0 → (n0 + m0 * n) / n = n0 / n + m0 ⊦ n = 0, (n0 + n * m0) / n = n0 / n + m0

Rule argument

spec: nat-div inst: name: add seq: n ≠ 0 ⊦ (n0 + n) / n = n0 / n + 1 subst: [n, n0] → [n, m0 * n + n0] flags: current, []

Additional information

Used terms: n, m0 * n + n0