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Proof step 2 from proof alignup-add-le (previous step)

⊦ align↑(n0, m) + n1 ≤ n0 ↔ n1 = 0 ∧ is-aligned(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)+1 * m)
Proof step 4:
⊦ is-aligned(n, m) ↔ n % m = 0
Proof step 5:
(n0 % m = 0 ⊃ n0;m * n0 / m + m) + n1 ≤ n0 ⊦ m = 0, n1 = 0 ∧ n0 % m = 0

Rule argument

Additional information

Used simplifier rules: alignup-def: 0 ≠ m ⊦ align↑(n, m) = (n % m = 0 ⊃ n;(n / m)+1 * m);%is-aligned-def: ⊦ is-aligned(n, m) ↔ n % m = 0;% ⊦ m ≤ m; (s)% ⊦ n + 0 = n; (s)%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)