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Proof step 4 from proof from-append-add (previous step)

⊦ #((σ to n) + σ0 from n1 + n) = #(σ0 from n1) ∧ (∀ n0. ⌜n0⌝ < #((σ to n) + σ0 from n1 + n) → ((σ to n) + σ0 from n1 + n)[n0] = (σ0 from n1)[n0]), # σ0 < ⌜n1⌝ , # σ < ⌜n⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ # σ0 < ⌜n1⌝ , # σ < ⌜n⌝ , #((σ to n) + σ0 from n1 + n) = #(σ0 from n1) ∧ (⌜n0⌝ < #((σ to n) + σ0 from n1 + n) → ((σ to n) + σ0 from n1 + n)[n0] = (σ0 from n1)[n0])

Proof step 4 from proof from-append-add (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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