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Proof step 6 from proof square-inject (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ m0 * (m + n) = m0 * m + m0 * n
Proof step 8:
((m + n) * (m + n + 1)) % 2 = 0 ∧ 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

Rule argument

spec: nat-mult inst: name: dist seq: ⊦ m0 * (m + n) = m0 * m + m0 * n subst: [m0, n, m] → [2, n0, ((m0 + n0) * (m0 + n0 + 1)) / 2] flags: rewrite

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: 2, n0, ((m0 + n0) * (m0 + n0 + 1)) / 2