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Proof step 33 from proof enc-invert (previous step)

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: discard

Additional information

Used simplifier rules: ⊦ m * n ≤ m1 * n + m2 * n ↔ m ≤ m1 + m2 ∨ n = 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → (n ≤ m ↔ false); (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic