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Proof step 38 from proof infinite-append-prefixnextli (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
lf(0) = [], ii(0) = 0, nextli0(lf(n0)) = [], ii(n0 + 1) = ii(n0), ii(m) ≤ ii(n0) + 1, ii(n0) + 1 < ii(m + 1), ii(m) < ii(m + 1), ∀ m. n0 < m → ii(m) = ii(n0) ∧ lf(m) = lf(n0), ∀ x. nextli(x) = x + nextli0(x), ∀ n. lf(n + 1) = nextli(lf(n)), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n) = nf to ii(n), ∀ n. ii(n + 1) = ii(n) + # nextli0(lf(n)) ∧ (nf from ii(n)) to # nextli0(lf(n)) = nextli0(lf(n)) ⊦ ii(m + 1) = 0

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