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Proof step 3 from proof indices-where-in-h (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 4:
⊦ ar.indicesWhere(φ, 0) = []
Proof step 5:
m0 +1 ≤ # ar, ∀ ar, n, φ. m0 ≤ # ar → (n ∈ ar.indicesWhere(φ, m0) ↔ n < m0 ∧ φ(ar[n])) ⊦ n ∈ ar.indicesWhere(φ, m0 +1) ↔ n < m0 +1 ∧ φ(ar[n])

Rule argument

indvar: m

Additional information

Used simplifier rules: indiceswhere-base: ⊦ ar.indicesWhere(φ, 0) = [];%nat freely generated by 0, +1;;% ⊦ ¬ n ∈ []; (s)% ⊦ ¬ m < 0; (s)