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Proof step 15 from proof indices-where-in-h (previous step)

n < # ar, m0 < # ar, φ(ar[n]), n ≤ m0, ∀ ar, n, φ. m0 ≤ # ar → (n ∈ ar.indicesWhere(φ, m0) ↔ n < m0 ∧ φ(ar[n])) ⊦ φ(ar[m0]), n ∈ ar.indicesWhere(φ, m0), n ∈ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
m0 < # ar, n < # ar, φ(ar[n]), n ≤ m0, ∀ ar, n, φ. m0 ≤ # ar → (n ∈ ar.indicesWhere(φ, m0) ↔ n < m0 ∧ φ(ar[n])) ⊦ φ(ar[m0]), n ∈ ar.indicesWhere(φ, m0)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n ∈ []; (s)