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Proof step 99 from proof infinite-append-prefixnextli (previous step)

n0 ≤ m, ii(n) = ii(n0), lf(n) = lf(n0), nextli0(lf(n0)) = [], nextli(lf(n0)) = lf(n0), # lf(n0) = ii(n0), lf(0) = [], ii(0) = 0, n0 < n, ∀ m. n ≤ m → nextli0(lf(m)) = [] ∧ ii(m) = ii(n0), ∀ x. nextli(x) = x + nextli0(x), ∀ n. lf(n + 1) = nextli(lf(n)), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. nextli0(lf(n)) = [] → (∀ m. n < m → lf(m) = lf(n) ∧ nextli0(lf(m)) = [] ∧ ii(m) = ii(n)), ∀ m. nextli0(lf(m)) = [] ↔ n0 ≤ m, ∀ n0. ii(n0 + 1) = ii(n0) + # nextli0(lf(n0)) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (lf(n) from ii(n0)) to # nextli0(lf(n0)) = nextli0(lf(n0)), ∀ n0. lf(n0) = lf(n) to ii(n0), ∀ m. n0 < m → lf(m) = lf(n0) ∧ nextli0(lf(m)) = [] ∧ ii(m) = ii(n0) ⊦ ii(m) = ii(n0), n = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 18 terms: m

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat Used terms: m