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Proof step 24 from proof enc-invert (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0
Proof step 26:
n0 ≤ m, (maxenc(m) * (maxenc(m) + 1)) / 2 ≤ m ∧ m < ((maxenc(m) + 1) * (maxenc(m) + 2)) / 2, maxenc(m) * (maxenc(m) + 1) ≤ 2 * m ∧ 2 * m < (maxenc(m) + 1) * (maxenc(m) + 2), ¬ m ≤ maxenc(m) + n0, maxenc(m) * (maxenc(m) + 1) = 2 * n0 ⊦ m < n0

Rule argument

spec: nat inst: name: ls-03 seq: n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0 subst: [m, n0, n] → [maxenc(m), n0, m] flags: rewrite

Additional information

Used simplifier rules: rsucc-02: ⊦ m * (n + 1) = m * 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%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%two-rec: ⊦ (m + 2) / 2 = m / 2 + 1; from specification nat-div%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat Used terms: maxenc(m), n0, m