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Proof step 2 from proof aligndown-is-smaller (previous step)

n0 < align↓(n, m) ⊦ m = 0, n0 < n

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:
n < (n % m = 0 ⊃ n;n / m * m), n0 < (n % m = 0 ⊃ n;n / m * m), n ≤ n0 ⊦ m = 0

Rule argument

Additional information

Used simplifier rules: aligndown-def: 0 ≠ m ⊦ align↓(n, m) = (n % m = 0 ⊃ n;n / m * m);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)