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Proof step 8 from proof max-def (previous step)

¬ ((m0 + 1) * (m0 + 2)) / 2 ≤ m, ¬ ((m0 + 1) * (m0 + 2)) / 2 ≤ m, true, ∀ m2, m3, n. m2 - m3 < m - m0 → true → (n * (n + 1)) / 2 ≤ m2 ∧ m2 < ((n + 1) * (n + 2)) / 2 ∧ (m3 * (m3 + 1)) / 2 ≤ m2 → maxenc(m3, m2) = n ⊦ (n * (n + 1)) / 2 ≤ (m0 * (m0 + 1)) / 2 + 0 ∧ (m0 * (m0 + 1)) / 2 + 0 < ((n + 1) * (n + 2)) / 2 ∧ 0 ≤ m0, (n * (n + 1)) / 2 ≤ m ∧ m < ((n + 1) * (n + 2)) / 2 ∧ (m0 * (m0 + 1)) / 2 ≤ m → m0 = n

User applied rule case distinction

And obtained the following new goals

Proof step 9:
¬ ((m0 + 1) * (m0 + 2)) / 2 ≤ m, ¬ ((m0 + 1) * (m0 + 2)) / 2 ≤ m, true, ∀ m2, m3, n. m2 - m3 < m - m0 → true → (n * (n + 1)) / 2 ≤ m2 ∧ m2 < ((n + 1) * (n + 2)) / 2 ∧ (m3 * (m3 + 1)) / 2 ≤ m2 → maxenc(m3, m2) = n ⊦ (n * (n + 1)) / 2 ≤ (m0 * (m0 + 1)) / 2 + 0, (n * (n + 1)) / 2 ≤ m ∧ m < ((n + 1) * (n + 2)) / 2 ∧ (m0 * (m0 + 1)) / 2 ≤ m → m0 = n

Rule argument

pos: right 1

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%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lez: ⊦ 0 ≤ m; from specification nat%z: ⊦ 0 + m = m; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%ls-02: n ≠ 0 ⊦ m / n < m0 / n → m < m0; (forward) from specification nat-div%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%two-zero-01-01: ⊦ m / 2 = 0 ↔ ¬ ¬(m = 0 ∨ m = 1); from specification nat-div%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%square-one: ⊦ m * m + m ≠ 1; from specification nat-mult%trans: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%ref-1: ⊦ ¬ n < n; from specification nat-basic1