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

align↑(m0, n) < align↑(m1, n) ⊦ n = 0, m0 < m1

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:
(m0 % n = 0 ⊃ m0;n * m0 / n + n) < (m1 % n = 0 ⊃ m1;n * m1 / n + n), m1 ≤ m0 ⊦ n = 0

Rule argument

Additional information

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