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Proof step 12 from proof emptystream-cancel (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# 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: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)