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Proof step 4 from proof max-snd-nle (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 * m ≤ maxenc(m) + (maxenc(m) * maxenc(m) + maxenc(m)) / 2 ⊦ m = 1, m = 0

Rule argument

Additional information

Used simplifier rules: max-nls: ⊦ ¬ m < (maxenc(m) * maxenc(m) + maxenc(m)) / 2;%two-02: ⊦ m + m = 2 * m; from specification nat-mult%le: n ≤ m ⊦ (m - n) + m0 ≤ n0 ↔ m + m0 ≤ n0 + n; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%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