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Proof step 2 from proof swap-from-to (previous step)

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

Rule argument

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

Additional information

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