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Proof step 9 from proof extensionality-to (previous step)

⌜n⌝ < # σ0, # σ0 = # σ1, (σ0 to n) + ⌜σ0[n] '⌝ = σ1 to n + 1, ⌜n⌝ < # σ1, ∀ n. ¬ # σ1 < ⌜n⌝ → σ0 to n = σ1 to n ⊦ σ0[n] = σ1[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information