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Proof step 20 from proof map-indices-at (previous step)

n < # ar, nx0 < # ar, m < # ar, n ∈ nx0, ∀ ar0, n0, f0. nx0 < # ar0 ∧ n0 < # ar0 → ar0.mapIndices(nx0, f0)[n0] = (n0 ∈ nx0 ⊃ f0(ar0[n0]);ar0[n0]) ⊦ ar[m, f(ar[m])].mapIndices(nx0, f)[n] = f(ar[n]), dups(nx0), m ∈ nx0

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: ar[m, f(ar[m])], n, f abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ # ar[n, a] = # ar; (s)%m ≠ n ⊦ ar[m, a][n] = ar[n]; (s) Used terms: ar[m, f(ar[m])], n, f