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Proof step 13 from proof indices-where-in-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
n < m0, n < # ar, m0 < # ar, n + 1 < # ar, φ(ar[n]), φ(ar[m0]), ∀ ar, n, φ. m0 ≤ # ar → (n ∈ ar.indicesWhere(φ, m0) ↔ n < m0 ∧ φ(ar[n])) ⊦ n = m0, n ∈ ar.indicesWhere(φ, m0)

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ n ∈ n0 ' ↔ n = n0; (s)