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

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

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) + ((σ from n) to n1)) = #(σ to n + n1) ∧ (∀ n0. ⌜n0⌝ < #((σ to n) + ((σ from n) to n1)) → ((σ to n) + ((σ from n) to n1))[n0] = (σ to n + n1)[n0]), # σ < ⌜n⌝ + ⌜n1⌝

Rule argument

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

Additional information

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