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Proof step 14 from proof emptystream-cancel (previous step)

# σ = ⌜n⌝ , ⌜n⌝ < ⌜n⌝ + # σ⊥ → (σ + σ⊥)[n] = zero, ∀ n0. ⌜n0⌝ < ⌜n⌝ + # σ⊥ → (σ + σ⊥)[n0] = zero, # σ⊥ = ∞ ∨ σ = [] ⊦ ⌜n⌝ = #(σ⊥ to n) ∧ (n0 < n → σ[n0] = (σ⊥ to n)[n0])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ ⌜m⌝ < # σ0 → (σ0 + σ2)[m] = σ0[m]
Proof step 16:
# σ = ⌜n⌝ , ⌜n⌝ < ⌜n⌝ + # σ⊥ → (σ + σ⊥)[n] = zero, ∀ n0. ⌜n0⌝ < ⌜n⌝ + # σ⊥ → (σ + σ⊥)[n0] = zero, # σ⊥ = ∞ ∨ σ = [] ⊦ ⌜n⌝ < # σ, ⌜n⌝ = #(σ⊥ to n) ∧ (n0 < n → σ[n0] = (σ⊥ to n)[n0])

Rule argument

spec: stream, inst: buffer-inf name: get-append-ls seq: ⊦ ⌜m⌝ < # σ0 → (σ0 + σ2)[m] = σ0[m] subst: [m, σ2, σ0] → [n, σ⊥, σ] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ⌜n⌝ < n∞ + ⌜n⌝ ↔ n∞ ≠ ⌜0⌝ ; (s)% ⊦ # σ0 = ⌜0⌝ ↔ σ0 = []; (s)%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% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ ¬ n < n; (s) Used terms: n, σ⊥, σ