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Proof step 18 from proof eq1 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
n0 % m + 1 = m, m * (m + n0) / m + m ≤ n0 + 1, ∀ m. (n0 + m) / m * m ≤ n0 → m = 0 ⊦ m = 0

Rule argument

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m * (n + 1) = m * n + m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%n ≠ 0 ⊦ (m + n) % n = m % n; (s)