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Proof step 10 from proof indices-where-in-h (previous step)

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])

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 2

Additional information

Used simplifier rules: ⊦ ¬ n ∈ []; (s)% ⊦ n ∈ n0 ' ↔ n = n0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ ¬ n < n; (s)