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Proof step 8 from proof emptystream-cancel (previous step)

#(σ + σ⊥) = # σ⊥, # σ = ⌜n⌝ , ∀ n. ⌜n⌝ < #(σ + σ⊥) → (σ + σ⊥)[n] = (σ⊥)[n] ⊦ σ = σ⊥ to n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ (σ⊥)[n] = zero
Proof step 10:
# σ = ⌜n⌝ , ∀ n0. ⌜n0⌝ < ⌜n⌝ + # σ⊥ → (σ + σ⊥)[n0] = zero, # σ⊥ = ∞ ∨ σ = [] ⊦ σ = σ⊥ to n

Rule argument

Additional information

Used simplifier rules: at-emptystream: ⊦ (σ⊥)[n] = zero;%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∞1 = n∞1 + n∞2 ↔ n∞1 = ∞ ∨ n∞2 = ⌜0⌝ ; (s)%append-assoc: ⊦ (σ0 + σ3) + σ4 = σ0 + σ3 + σ4; from specification stream[buffer-inf]% ⊦ #(σ0 + σ2) = # σ0 + # σ2; (s)