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Proof step 8 from proof square-inject (previous step)

((m + n) * (m + n + 1)) % 2 = 0, (m + n) * (m + n + 1) + 2 * n = 2 * ((m0 + n0) * (m0 + n0 + 1)) / 2 + 2 * n0 ⊦ ((m + n) * (m + n + 1)) / 2 + n = ((m0 + n0) * (m0 + n0 + 1)) / 2 + n0 ↔ m = m0 ∧ n = n0, 2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ n % m = 0 ∧ m ≠ 0 → m * n / m = n
Proof step 10:
((m + n) * (m + n + 1)) % 2 = 0, ((m0 + n0) * (m0 + n0 + 1)) % 2 = 0, (m + n) * (m + n + 1) + 2 * n = (m0 + n0) * (m0 + n0 + 1) + 2 * n0 ⊦ ((m + n) * (m + n + 1)) / 2 + n = ((m0 + n0) * (m0 + n0 + 1)) / 2 + n0 ↔ m = m0 ∧ n = n0, 2 = 0

Rule argument

spec: <toplevel> name: mult-div seq: ⊦ n % m = 0 ∧ m ≠ 0 → m * n / m = n subst: [n, m] → [(m0 + n0) * (m0 + n0 + 1), 2] flags: <all paths>

Additional information

Used simplifier rules: square-even-02: ⊦ (m * m + m) % 2 = 0;% ⊦ m * (n + 1) = m * n + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ (* m0) * m + m = (m0 + 1) * m; (s) Used terms: (m0 + n0) * (m0 + n0 + 1), 2