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Proof step 10 from proof emptystream-cancel (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ σ = σ2 ↔ # σ = # σ2 ∧ (∀ n. ⌜n⌝ < # σ → σ[n] = σ2[n])
Proof step 12:
# 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])

Rule argument

spec: stream, inst: buffer-inf name: extensionality seq: ⊦ σ = σ2 ↔ # σ = # σ2 ∧ (∀ n. ⌜n⌝ < # σ → σ[n] = σ2[n]) subst: [σ2, σ] → [σ⊥ to n, σ] flags: <all paths>

Additional information

Used terms: σ⊥ to n, σ