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Proof step 6 from proof aligndown-add-same0 (previous step)

n0 + n1 = m0 * m + n2, n2 < m, n0 % m + n1 < m ⊦ (n2 = 0 ⊃ n0 + n1;m0 * m) = (n0 % m = 0 ⊃ n0;n0 / m * m)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
n0 + n1 = m0 * m + n2, n0 = m1 * m + n3, n3 + n1 < m, n2 < m, n3 < m ⊦ (n2 = 0 ⊃ n0 + n1;m0 * m) = (n3 = 0 ⊃ n0;m1 * m), m = 0

Rule argument

spec: nat-div name: div-elim seq: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n subst: [m, n, m0, n0] → [n0, m, m1, n]

Additional information

Used simplifier rules: ⊦ 0 * n = 0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ¬ m < 0; (s)%n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: n0, m, m1, n