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Proof step 4 from proof swap-from-to (previous step)

⊦ #((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⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Proof step 4 from proof swap-from-to (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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