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Proof step 38 from proof from-append-add (previous step)

# x = n2 + n1, ⌜x0⌝ = (σ to n) + ⌜x⌝ , n0 < n2 ⊦ (x0 from n1 + n)[n0] = x[n1 + n0], x = [], # σ < ⌜n⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 39:
⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])
Proof step 40:
# ⌜x0⌝ = #((σ to n) + ⌜x⌝ ), # x = n2 + n1, n0 < n2, ∀ n0. ⌜n0⌝ < # ⌜x0⌝ → ⌜x0⌝ [n0] = ((σ to n) + ⌜x⌝ )[n0] ⊦ (x0 from n1 + n)[n0] = x[n1 + n0], x = [], # σ < ⌜n⌝

Rule argument

spec: <toplevel> name: extensionality seq: ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n]) subst: [σ1, σ0] → [(σ to n) + ⌜x⌝ , ⌜x0⌝ ] flags: <all paths>

Additional information

Used terms: (σ to n) + ⌜x⌝ , ⌜x0⌝