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Proof step 18 from proof inter-union-diff-subseteq (previous step)

adr ∈ as1, adr ∈ as3, as1 ∩ as3 ⊆ as, ∀ adr. adr ∈ as0 ∩ as3 → adr ∈ as ⊦ #(as0 ∪ (as1 \ as2)) = 0, # as3 = 0, # as1 = 0, as1 = ∅, as3 = ∅, as0 ∪ (as1 \ as2) = ∅, adr ∈ as2, adr ∈ as, adr ∈ as0

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 19:
adr ∈ as3, adr ∈ as1, as1 ∩ as3 ⊆ as, adr ∈ as0 ∩ as3 → adr ∈ as, ∀ adr. adr ∈ as0 ∩ as3 → adr ∈ as ⊦ # as1 = 0, # as3 = 0, #(as0 ∪ (as1 \ as2)) = 0, as1 = ∅, as3 = ∅, as0 ∪ (as1 \ as2) = ∅, adr ∈ as0, adr ∈ as, adr ∈ as2

Rule argument

pos: left 4 terms: adr

Additional information

Used terms: adr