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Proof step 3 from proof max-def (previous step)

m < ((n + 1) * (n + 2)) / 2, (n * (n + 1)) / 2 ≤ m, (m0 * (m0 + 1)) / 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, m) = n

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ((m + 1) * (m + 2)) / 2 ≤ n → maxenc(m, n) = maxenc(m + 1, n)
Proof step 5:
m < ((n + 1) * (n + 2)) / 2, (m0 * (m0 + 1)) / 2 ≤ m, (n * (n + 1)) / 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, m) = n, ((m0 + 1) * (m0 + 2)) / 2 ≤ m
Proof step 27:
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

Rule argument

spec: <toplevel> name: maxi-rec seq: ⊦ ((m + 1) * (m + 2)) / 2 ≤ n → maxenc(m, n) = maxenc(m + 1, n) subst: [n, m] → [m, m0] flags: <all paths>

Additional information

Used terms: m, m0