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Proof step 8 from proof inter-union-empty-one-empty (previous step)

n < m, n < # asar, adr ∈ as, adr ∈ asar[n], m ≤ # asar, ∀ adr. adr ∈ as ∩ union(asar, 0, m) ↔ adr ∈ ∅ ⊦ # as = 0, # asar[n] = 0, #(as ∩ asar[n]) = 0, asar[n] = ∅, as = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n < # asar, n < m, adr ∈ asar[n], adr ∈ as, m ≤ # asar, ∀ adr0. ¬ adr0 ∈ as ∩ union(asar, 0, m) ⊦ #(as ∩ asar[n]) = 0, # asar[n] = 0, # as = 0, asar[n] = ∅, as = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ adr ∈ ∅; (s)