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Proof step 27 from proof max-def (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

subst: [m2, m3, n] → [m, m0 + 1, n]

Additional information

Used simplifier rules: ⊦ m * (n + 1) = m * n + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ (m + 2) / 2 = m / 2 + 1; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ n < n; (s)% ⊦ (m + n) / 2 < n ↔ m < n; (s)% ⊦ m0 * m < m ↔ m0 = 0 ∧ m ≠ 0; (ws)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ ¬ n + (m - n) < m; (s) Used terms: m, m0 + 1, n