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Proof step 2 from proof aligndown-is-aligned-add (previous step)

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