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Proof step 5 from proof eq1 (previous step)

((n0 + 1) + m) / m * m ≤ n0 + 1, ∀ m. (n0 + m) / m * m ≤ n0 → m = 0 ⊦ m = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
n ≠ 0 ⊦ m % n + 1 = n → (m + 1) / n = m / n + 1
Proof step 7:
((n0 + 1) + m) / m * m ≤ n0 + 1, ∀ m. (n0 + m) / m * m ≤ n0 → m = 0 ⊦ m = 0, m ≠ 0 ∧ (m + n0) % m + 1 = m
Proof step 18:
(m + n0) % m + 1 = m, ((m + n0) / m + 1) * m ≤ n0 + 1, ∀ m. (n0 + m) / m * m ≤ n0 → m = 0 ⊦ m = 0

Rule argument

spec: nat-div name: div-succ seq: n ≠ 0 ⊦ m % n + 1 = n → (m + 1) / n = m / n + 1 subst: [n, m] → [m, m + n0] flags: <all paths>

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m, m + n0