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Proof step 4 from proof map-indices-at (previous step)

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

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 5:
⊦ ar.mapIndices([], f) = ar
Proof step 6:
n < # ar, m + nx0 < # ar, ∀ ar, n, f. nx0 < # ar ∧ n < # ar → ar.mapIndices(nx0, f)[n] = (n ∈ nx0 ⊃ f(ar[n]);ar[n]) ∨ dups(nx0) ⊦ ar.mapIndices(m + nx0, f)[n] = (n ∈ m + nx0 ⊃ f(ar[n]);ar[n]), dups(m + nx0)

Rule argument

term: nx

Additional information

Used simplifier rules: mapindices-base: ⊦ ar.mapIndices([], f) = ar;%nat-list freely generated by [], +;;% ⊦ ¬ n ∈ []; (s)