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

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

User applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%two-mdiv2-02: ⊦ (2 * m) / 2 = m; from specification nat-div%le-num: 0 ≠ m ⊦ (* m) * n ≤ m * n0 ↔ n ≤ n0; from specification nat-mult%add-num-03: n < n0 ⊦ (* n) ≠ m0 + (* n0); from specification nat%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-basic1%le-div-to-mult: m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m; from specification nat-div%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%two-rec: ⊦ (m + 2) / 2 = m / 2 + 1; from specification nat-div%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-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult