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Proof step 5 from proof max-snd-nle (previous step)

2 * m ≤ maxenc(m) + (maxenc(m) * maxenc(m) + maxenc(m)) / 2 ⊦ m = 1, m = 0

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: nat-pair inst: name: max-size-01 seq: ⊦ maxenc(m) * (maxenc(m) + 1) ≤ 2 * m ∧ 2 * m < (maxenc(m) + 1) * (maxenc(m) + 2) subst: [m] → [m] flags: current, []

Additional information

Used simplifier rules: max-zero-01: ⊦ maxenc(m) = 1 ↔ m = 1 ∨ m = 2;%max-zero: ⊦ maxenc(m) = 0 ↔ m = 0;%max-two: ⊦ maxenc(2) = 1;%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%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%add-num-03: n < n0 ⊦ (* n) ≠ m0 + (* n0); from specification nat%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%le-div-to-mult: m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m; from specification nat-div%coef-20: ⊦ m * n ≤ m1 * n + n ↔ ¬ ¬(m ≤ m1 + 1 ∨ n = 0); from specification nat-mult%numeral-26: m2 ≠ 0 ⊦ (* m2) * m ≤ m + n0 ↔ (m2 - 1) * m ≤ n0; from specification nat-mult%one: ⊦ m0 * 1 = m0; from specification nat-mult%one-02: ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); from specification nat Used terms: m