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Proof step 12 from proof indices-where-in-h (previous step)

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)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 13:
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 '
Proof step 15:
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 ∈ []

Rule argument

pos: right 1

Proof step 12 from proof indices-where-in-h (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information