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Proof step 7 from proof swap-to-from-app (previous step)

⊦ (σ to n0 + m) + σ0 from m = ((σ to m + n0) from m) + σ0, # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n])
Proof step 9:
⊦ #((σ to n0 + m) + σ0 from m) = #(((σ to m + n0) from m) + σ0) ∧ (∀ n. ⌜n⌝ < #((σ to n0 + m) + σ0 from m) → ((σ to n0 + m) + σ0 from m)[n] = (((σ to m + n0) from m) + σ0)[n]), # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Rule argument

spec: inst: name: extensionality seq: ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0[n] = σ1[n]) subst: [σ1, σ0] → [((σ to m + n0) from m) + σ0, (σ to n0 + m) + σ0 from m] flags: rewrite, current

Additional information

Used terms: ((σ to m + n0) from m) + σ0, (σ to n0 + m) + σ0 from m