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Proof step 2 from proof from-append-add (previous step)

⊦ (σ to n) + σ0 from n1 + n = σ0 from n1, # σ0 < ⌜n1⌝ , # σ < ⌜n⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])
Proof step 4:
⊦ #((σ to n) + σ0 from n1 + n) = #(σ0 from n1) ∧ (∀ n0. ⌜n0⌝ < #((σ to n) + σ0 from n1 + n) → ((σ to n) + σ0 from n1 + n)[n0] = (σ0 from n1)[n0]), # σ0 < ⌜n1⌝ , # σ < ⌜n⌝

Rule argument

spec: inst: name: extensionality seq: ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n]) subst: [σ1, σ0] → [σ0 from n1, (σ to n) + σ0 from n1 + n] flags: rewrite, current

Additional information

Used terms: σ0 from n1, (σ to n) + σ0 from n1 + n