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Proof step 40 from proof infinite-append-prefixnextli (previous step)

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 + 1 ≤ n0 → ii(m + 1) ≤ ii(n0), ∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
lf(0) = [], ii(0) = 0, nextli0(lf(n0)) = [], ii(n0 + 1) = ii(n0), n0 ≤ m, ii(n0) < ii(m + 1), 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: lels-03: ⊦ m < n → m ≤ n; from specification nat%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat