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Proof step 5 from proof is-aligned-of-mod (previous step)

m0 = m2 * m1 + n0, n0 = 0, n0 < m1 ⊦ (n % m0) % m1 = 0 ↔ n % m1 = 0, m0 = 0, m1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ (n % (m2 * m1)) % m1 = 0 ↔ n % m1 = 0, m1 = 0, m2 = 0

Rule argument

Additional information

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