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Proof step 10 from proof max-rev (previous step)

((m0 + 1) * (m0 + 2)) / 2 ≤ m, true, ∀ m2, m3, n. m2 - m3 < m - m0 → true → maxenc(m3, m2) = n ∧ (m3 * (m3 + 1)) / 2 ≤ m2 → (n * (n + 1)) / 2 ≤ m2 ∧ m2 < ((n + 1) * (n + 2)) / 2 ⊦ maxenc(m0 + 1, m) = n ∧ (m0 * (m0 + 1)) / 2 ≤ m → (n * (n + 1)) / 2 ≤ m ∧ m < ((n + 1) * (n + 2)) / 2

User applied rule apply induction

And obtained the following new goals

Proof step 11:
((m0 + 1) * (m0 + 2)) / 2 ≤ m, true, ∀ m2, m3, n. m2 - m3 < m - m0 → true → maxenc(m3, m2) = n ∧ (m3 * (m3 + 1)) / 2 ≤ m2 → (n * (n + 1)) / 2 ≤ m2 ∧ m2 < ((n + 1) * (n + 2)) / 2 ⊦ m - m0 + 1 < m - m0, maxenc(m0 + 1, m) = n ∧ (m0 * (m0 + 1)) / 2 ≤ m → (n * (n + 1)) / 2 ≤ m ∧ m < ((n + 1) * (n + 2)) / 2

Rule argument

subst: [m2, m3, n] → [m, m0 + 1, n]

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%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat Used terms: m, m0 + 1, n