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Proof step 11 from proof subseteq-union-diff-one (previous step)

n0 < # asar, disj(asar, 0, # asar), adr ∈ asar[n0], ∀ adr. adr ∈ asar[n0] → adr ∈ as ⊦ # asar[n0] = 0, asar[n0] = ∅, adr ∈ as

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: adr

Additional information

Used simplifier rules: disj-unique: n0 < # asar, n1 < # asar ⊦ disj(asar, 0, # asar) ∧ adr ∈ asar[n0] → (adr ∈ asar[n1] ↔ n0 = n1); Used terms: adr