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Proof step 9 from proof valid-indices-indiceswhere-h-strong (previous step)

n0 < # ar, n0 < m, m ∈ (φ(ar[n0]) ⊃ n0 ';[]), ∀ m0, ar0, φ0. ¬ m0 ∈ ar0.indicesWhere(φ0, n0) ∨ ¬ n0 ≤ # ar0 ∨ ¬ n0 ≤ m0 ⊦

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 3

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%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s)