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Proof step 16 from proof alignup-le (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 17:
m1 % n = 0, n * m0 / n + n < m1, m1 ≤ m0 ⊦ m0 % n = 0, n = 0

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m0 ≠ 0 → (m0 * m < m0 * n ↔ m < n); (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%ls-02: n ≠ 0 ⊦ m / n < m0 / n → m < m0; (forward) from specification nat-div%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)