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Proof step 2 from proof align-down-add-eq (previous step)

⊦ align↓(n0 + n1, m) = n0 ↔ n1 < m ∧ is-aligned(n0, m), m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 ≠ m ⊦ align↓(n, m) = (n % m = 0 ⊃ n;n / m * m)
Proof step 4:
⊦ is-aligned(n, m) ↔ n % m = 0
Proof step 5:
((n0 + n1) % m = 0 ⊃ n0 + n1;(n0 + n1) / m * m) = n0 ⊦ m = 0, n1 < m ∧ n0 % m = 0

Rule argument

Additional information

Used simplifier rules: aligndown-def: 0 ≠ m ⊦ align↓(n, m) = (n % m = 0 ⊃ n;n / m * m);%is-aligned-def: ⊦ is-aligned(n, m) ↔ n % m = 0;%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ n % m = 0 ∧ m ≠ 0 → m * n / m = n; (s)% ⊦ m < n → m / n = 0; (s)% ⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) / n = m / n + m0 / n; (s)% ⊦ n + 0 = n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n → m % n = m; (s)% ⊦ m0 % n = 0 ∧ 0 ≠ n → (m + m0) % n = m % n; (s)