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Proof step 6 from proof split-prefix-append-to (previous step)

# σ0 = ∞, (σ0 to n) + σ1 = σ0 ⊦ # σ1 = ∞ ∧ (⌜n0⌝ < # σ1 → σ1[n0] = (σ0 from n)[n0])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ σ = σ2 ↔ # σ = # σ2 ∧ (∀ n. ⌜n⌝ < # σ → σ[n] = σ2[n])
Proof step 8:
#((σ0 to n) + σ1) = # σ0, # σ0 = ∞, ∀ n0. ⌜n0⌝ < #((σ0 to n) + σ1) → ((σ0 to n) + σ1)[n0] = σ0[n0] ⊦ # σ1 = ∞ ∧ (⌜n0⌝ < # σ1 → σ1[n0] = (σ0 from 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] → [σ0, (σ0 to n) + σ1] flags: <all paths>

Additional information

Used terms: σ0, (σ0 to n) + σ1