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Proof step 6 from proof disj-in-both (previous step)

n1 < m, n0 < m, adr ∈ asar[n1], adr ∈ asar[n0], ∀ n, n2. n < m ∧ n2 < m ∧ n ≠ n2 → asar[n] ∩ asar[n2] = ∅ ⊦ # asar[n1] = 0, # asar[n0] = 0, asar[n1] = ∅, asar[n0] = ∅, n0 = n1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n0, n1

Additional information

Used simplifier rules: int-a: ⊦ (as ∩ as0) ∩ as1 = as ∩ as0 ∩ as1; from specification set-union[address-list-set]%int-c: ⊦ as ∩ as0 = as0 ∩ as; from specification set-union[address-list-set]% ⊦ as ∩ as0 = ∅ ∧ adr ∈ as → (adr ∈ as0 ↔ false); (as) Used terms: n0, n1