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

⊦ align↓(align↓(n0, n) + n1, n) = align↓(n0, n) + 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:
⊦ (((n0 % n = 0 ⊃ n0;n0 / n * n) + n1) % n = 0 ⊃ (n0 % n = 0 ⊃ n0;n0 / n * n) + n1;((n0 % n = 0 ⊃ n0;n0 / n * n) + n1) / n * n) = (n0 % n = 0 ⊃ n0;n0 / n * n) + (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);