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Proof step 8 from proof indices-where-in-h (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
m0 < # ar, n ∈ ar.indicesWhere(φ, m0), ∀ ar, n, φ. m0 ≤ # ar → (n ∈ ar.indicesWhere(φ, m0) ↔ n < m0 ∧ φ(ar[n])) ⊦ ¬ m0 < n ∧ φ(ar[n])
Proof step 10:
m0 < # ar, n ∈ (φ(ar[m0]) ⊃ m0 ';[]), ∀ ar, n, φ. m0 ≤ # ar → (n ∈ ar.indicesWhere(φ, m0) ↔ n < m0 ∧ φ(ar[n])) ⊦ ¬ m0 < n ∧ φ(ar[n])

Rule argument

pos: left 3

Proof step 8 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