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Proof step 16 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, ⌜n⌝ = #(σ⊥ to n) ∧ (n0 < n → σ[n0] = (σ⊥ to n)[n0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
⊦ (σ⊥)[n] = zero
Proof step 18:
# σ = ⌜n⌝ , ∀ n0. ⌜n0⌝ < ⌜n⌝ + # σ⊥ → (σ + σ⊥)[n0] = zero, # σ⊥ = ∞ ∨ σ = [] ⊦ ⌜n⌝ = #(σ⊥ to n) ∧ (n0 < n → σ[n0] = (σ⊥ to n)[n0])

Rule argument

Additional information

Used simplifier rules: at-emptystream: ⊦ (σ⊥)[n] = zero;% ⊦ m ≤ m; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)%append-assoc: ⊦ (σ0 + σ3) + σ4 = σ0 + σ3 + σ4; from specification stream[buffer-inf]% ⊦ ¬ ⌜m⌝ < # σ0 → (σ0 + σ2)[m] = σ2[m - (# σ0).n]; (s)%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf% ⊦ # σ0 = ⌜0⌝ ↔ σ0 = []; (s)% ⊦ ⌜n⌝ < n∞ + ⌜n⌝ ↔ n∞ ≠ ⌜0⌝ ; (s)