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Proof step 4 from proof valid-indices-indiceswhere-h (previous step)

n ≤ # ar ⊦ ∀ n0. n0 ∈ ar.indicesWhere(φ, n) → n0 < # ar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 ∈ ar.indicesWhere(φ, n), n ≤ # ar, # ar ≤ n0, n ≤ n0 ⊦

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)