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Proof step 6 from proof inter-union-diff-subseteq (previous step)

as1 ∩ as3 ⊆ as ⊦ (∀ adr. adr ∈ (as0 ∪ (as1 \ as2)) ∩ as3 → adr ∈ as) ↔ (∀ adr. adr ∈ as0 ∩ as3 → adr ∈ as)

User applied rule case distinction

And obtained the following new goals

Proof step 7:
as1 ∩ as3 ⊆ as, ∀ adr. adr ∈ (as0 ∪ (as1 \ as2)) ∩ as3 → adr ∈ as ⊦ ∀ adr. adr ∈ as0 ∩ as3 → adr ∈ as
Proof step 9:
as1 ∩ as3 ⊆ as, ∀ adr. adr ∈ as0 ∩ as3 → adr ∈ as ⊦ ∀ adr. adr ∈ (as0 ∪ (as1 \ as2)) ∩ as3 → adr ∈ as

Rule argument

pos: right 1

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