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

m < n ⊦ (σ to n)[m] = σ[m], # σ < ⌜n⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < n ⊦ (σ to n)[m] = σ[m], # σ < ⌜n⌝

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic