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Proof step 11 from proof indices-where-in-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
m0 < # ar, n < # ar, φ(ar[n]), n ≤ m0, ∀ ar, n, φ. m0 ≤ # ar → (n ∈ ar.indicesWhere(φ, m0) ↔ n < m0 ∧ φ(ar[n])) ⊦ n ∈ (φ(ar[m0]) ⊃ m0 ';[]), n ∈ ar.indicesWhere(φ, m0)

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)