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Proof step 11 from proof align-down-add-ls (previous step)

n2 + n1 = m1 * m + n0, (m1 + m0) * m < m0 * m + n2, n2 < m, n0 < m, m ≤ n2 + n1 ⊦ m = 0, n0 = 0, n1 = 0, n2 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n2 + n1 = m1 * m + n0, n0 < m, n2 < m, (m1 + m0) * m < m0 * m + n2, n0 < n2 + n1, m ≤ n2 + n1 ⊦ m1 = 0, n0 = 0, n1 = 0, n2 = 0

Rule argument

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; (ws)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)