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Proof step 142 from proof infinite-append-butlast (previous step)

(σ from m0 + 1) to # y = ⌜y⌝ , ii(2 + n1) = 2 + m0 + # y, ii(0) = 0, ii(n1 + 1) = m0 + 1, lf(2 + n1) = (σ from m0 + 1)[# y] ' + x1, ⌜# y⌝ + ⌜m0⌝ + ⌜1⌝ < # σ, (σ from m0 + 1)[# y] ∈ lf(2 + n1), (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).rest) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. n ≠ 0 → ii(n) ≠ 0 ⊦ σ[m0 + # y + 1] = (σ from m0 + 1)[# y], # σ < ⌜2⌝ + ⌜m0⌝ + ⌜# y⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: stream inst: name: get-from seq: ⊦ ⌜n⌝ + ⌜m⌝ < # σ → (σ from n)[m] = σ[n + m] subst: [m, n, σ] → [# y, m0 + 1, σ] flags: rewrite

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf Used terms: # y, m0 + 1, σ