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Proof step 5 from proof map-ifn-at (previous step)

n < # ar, m ≤ # ar ⊦ ar.mapIndices(ar.indicesWhere(φ, m), f)[n] = (n < m ∧ φ(ar[n]) ⊃ f(ar[n]);ar[n])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ valid-indices(ar, nx) ∧ ¬ dups(nx) ∧ n < # ar → ar.mapIndices(nx, f)[n] = (n ∈ nx ⊃ f(ar[n]);ar[n])
Proof step 7:
n < # ar, m ≤ # ar ⊦ ar.mapIndices(ar.indicesWhere(φ, m), f)[n] = (n < m ∧ φ(ar[n]) ⊃ f(ar[n]);ar[n]), valid-indices(ar, ar.indicesWhere(φ, m)) ∧ ¬ dups(ar.indicesWhere(φ, m)) ∧ n < # ar
Proof step 10:
n < # ar, valid-indices(ar, ar.indicesWhere(φ, m)), m ≤ # ar ⊦ (n ∈ ar.indicesWhere(φ, m) ⊃ f(ar[n]);ar[n]) = (n < m ∧ φ(ar[n]) ⊃ f(ar[n]);ar[n]), dups(ar.indicesWhere(φ, m))

Rule argument

spec: <toplevel> name: map-indices-at seq: ⊦ valid-indices(ar, nx) ∧ ¬ dups(nx) ∧ n < # ar → ar.mapIndices(nx, f)[n] = (n ∈ nx ⊃ f(ar[n]);ar[n]) subst: [n, f, nx, ar] → [n, f, ar.indicesWhere(φ, m), ar] flags: <all paths>

Additional information

Used terms: n, f, ar.indicesWhere(φ, m), ar