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Proof step 12 from proof diff-subseteq-subseteq (previous step)

adr ∈ as0, as1 ⊆ as2, ∀ adr. adr ∈ as0 → adr ∈ as2 ⊦ # as0 = 0, as0 = ∅, adr ∈ as1, adr ∈ as2

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: adr

Additional information

Used terms: adr