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Proof step 26 from proof enc-invert (previous step)

maxenc(m) * (maxenc(m) + 1) = 2 * n0, m < ((maxenc(m) + 1) * (maxenc(m) + 2)) / 2, 2 * m < (maxenc(m) + 1) * (maxenc(m) + 2), (maxenc(m) * (maxenc(m) + 1)) / 2 ≤ m, maxenc(m) * (maxenc(m) + 1) ≤ 2 * m, n0 ≤ m ⊦ m < n0, m ≤ maxenc(m) + n0

User applied rule simplifier

And obtained the following new goals

Proof step 27:
maxenc(m) * maxenc(m) + maxenc(m) = 2 * n0, maxenc(m) * maxenc(m) < (maxenc(m) + 2) * maxenc(m) + 2, 2 * n0 < (maxenc(m) + 2) * maxenc(m) + maxenc(m) + 2, maxenc(m) + n0 < m, 2 * m < (maxenc(m) + 2) * maxenc(m) + maxenc(m) + 2, 2 * n0 ≤ (maxenc(m) + 2) * maxenc(m) + maxenc(m), m * 2 ≤ (maxenc(m) + 2) * maxenc(m) + maxenc(m), n0 ≤ m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m + n < m + n0 ↔ n < n0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%0 ≠ m ⊦ (* m) * n ≤ m * n0 ↔ n ≤ n0; (s)% ⊦ (2 * m) / 2 = m; (s)%n < n0 ⊦ (* n) ≠ m0 + (* n0); (s)% ⊦ m < n → n ≠ 0; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ (m + 2) / 2 = m / 2 + 1; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m * (n + 1) = m * n + m; (s)