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Proof step 10 from proof valid-indices-nodups-h (previous step)

n0 < # ar, φ(ar[n0]), ∀ ar0, φ0. ¬ dups(ar0.indicesWhere(φ0, n0)) ∨ ¬ n0 ≤ # ar0, dups(ar.indicesWhere(φ, n0)) ∨ dups(n0 ') ∨ (∃ n. n ∈ ar.indicesWhere(φ, n0) ∧ n ∈ n0 ') ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n0 < # ar, φ(ar[n0]), ∀ ar0, φ0. ¬ dups(ar0.indicesWhere(φ0, n0)) ∨ ¬ n0 ≤ # ar0, dups(ar.indicesWhere(φ, n0)) ∨ n0 ∈ ar.indicesWhere(φ, n0) ∧ n = n0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ n0 ' ↔ n = n0; (s)% ⊦ ¬ dups(n '); (s)