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Proof step 2 from proof map-indices-at (previous step)

n < # ar, valid-indices(ar, nx) ⊦ ar.mapIndices(nx, f)[n] = (n ∈ nx ⊃ f(ar[n]);ar[n]), dups(nx)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ valid-indices(ar, nx) ↔ nx < # ar
Proof step 4:
nx < # ar, n < # ar ⊦ ar.mapIndices(nx, f)[n] = (n ∈ nx ⊃ f(ar[n]);ar[n]), dups(nx)

Rule argument

Additional information

Used simplifier rules: valid-indices: ⊦ valid-indices(ar, nx) ↔ nx < # ar;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic