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Proof step 9 from proof union-eq-rem-subseteq (previous step)

n < # asar, adr0 ∈ asar[n], ∀ adr1. adr1 ∈ union(asar, 0, # asar) ↔ adr1 ≠ adr ∧ adr1 ∈ as ⊦ # asar[n] = 0, asar[n] = ∅, adr0 ∈ as

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: adr0

Additional information

Used simplifier rules: in-index-in-union: ⊦ adr ∈ asar[n] ∧ n < m → adr ∈ union(asar, 0, m); Used terms: adr0