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Proof step 53 from proof infinite-append-prefixnextnli (previous step)

ii(0) = 0, lf(0) = [], ∃ m. ii(m) ≤ n1 ∧ n1 < ii(m + 1), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextnli(lf(n), n), ∀ x, n. nextnli(x, n) = x + nextnli0(x, n), ∀ n. lf(n) = nf to ii(n), ∀ n. ii(n + 1) = ii(n) + # nextnli0(lf(n), n) ∧ (nf from ii(n)) to # nextnli0(lf(n), n) = nextnli0(lf(n), n) ⊦ ∃ m. ii(m) ≤ n1 ∧ n1 < ii(m + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 54:
ii(0) = 0, lf(0) = [], ii(m) ≤ n1, n1 < ii(m + 1), ii(m) < ii(m + 1), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextnli(lf(n), n), ∀ x, n. nextnli(x, n) = x + nextnli0(x, n), ∀ n. lf(n) = nf to ii(n), ∀ n. ii(n + 1) = ii(n) + # nextnli0(lf(n), n) ∧ (nf from ii(n)) to # nextnli0(lf(n), n) = nextnli0(lf(n), n) ⊦ ii(m + 1) = 0, ∃ m. ii(m) ≤ n1 ∧ n1 < ii(m + 1)

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1