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Proof step 8 from proof max-snd-nls (previous step)

maxenc(m) + (maxenc(m) * maxenc(m) + maxenc(m)) / 2 < m, (maxenc(m) * maxenc(m) + maxenc(m)) / 2 ≤ m, m ≤ ((maxenc(m) + 2) * maxenc(m) + maxenc(m)) / 2, maxenc(m) + (maxenc(m) * maxenc(m) + maxenc(m)) / 2 < ((maxenc(m) + 2) * maxenc(m) + maxenc(m)) / 2, (maxenc(m) * maxenc(m) + maxenc(m)) / 2 ≤ ((maxenc(m) + 2) * maxenc(m) + maxenc(m)) / 2 ⊦ (maxenc(m) + 2) * maxenc(m) + maxenc(m) = 0, maxenc(m) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat-div inst: name: two-le-02 seq: ⊦ m % 2 = 0 ∧ n % 2 = 0 → (m / 2 ≤ n / 2 ↔ m ≤ n) subst: [n, m] → [(maxenc(m) + 2) * maxenc(m) + maxenc(m), maxenc(m) * maxenc(m) + maxenc(m)] flags: rewrite

Additional information

Used simplifier rules: square-even-02: ⊦ (m * m + m) % 2 = 0; from specification nat-div%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%square-even-03: ⊦ ((n + 2) * n + n) % 2 = 0; from specification nat-div Used terms: (maxenc(m) + 2) * maxenc(m) + maxenc(m), maxenc(m) * maxenc(m) + maxenc(m)