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Proof step 2 from proof aligndown-add-same0 (previous step)

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

Rule argument

Additional information

Used simplifier rules: aligndown-def: 0 ≠ m ⊦ align↓(n, m) = (n % m = 0 ⊃ n;n / m * m);%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)