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Proof step 9 from proof aligndown-add-not-smaller (previous step)

(n0 + n1) / m * m < (n0 % m = 0 ⊃ n0;n0 / m * m) ⊦ (n0 + n1) % m = 0, m = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ 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%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 < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic