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Proof step 2 from proof alignup-upper-bound (previous step)

is-aligned(m0, n), m0 ≥ m ⊦ n = 0, align↑(m, n) ≤ m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ is-aligned(n, m) ↔ n % m = 0
Proof step 4:
0 ≠ m ⊦ align↑(n, m) = (n % m = 0 ⊃ n;(n / m)+1 * m)
Proof step 5:
m0 % n = 0, m < (m % n = 0 ⊃ m;n * m / n + n), m0 < (m % n = 0 ⊃ m;n * m / n + n), m ≤ m0 ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: is-aligned-def: ⊦ is-aligned(n, m) ↔ n % m = 0;%alignup-def: 0 ≠ m ⊦ align↑(n, m) = (n % m = 0 ⊃ n;(n / m)+1 * m);% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m * (n + 1) = m * n + m; (s)% ⊦ m +1 = m + 1; (s)% ⊦ n ≥ m ↔ m ≤ n; (s)