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Proof step 2 from proof inter-union-diff-subseteq (previous step)

as1 ∩ as3 ⊆ as ⊦ (as0 ∪ (as1 \ as2)) ∩ as3 ⊆ as ↔ as0 ∩ as3 ⊆ as

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ as ⊆ as0 ↔ (∀ adr. adr ∈ as → adr ∈ as0)
Proof step 4:
as1 ∩ as3 ⊆ as ⊦ (∀ adr. adr ∈ (as0 ∪ (as1 \ as2)) ∩ as3 → adr ∈ as) ↔ as0 ∩ as3 ⊆ as

Rule argument

spec: set-basic, inst: node-store name: Subset seq: ⊦ as ⊆ as0 ↔ (∀ adr. adr ∈ as → adr ∈ as0) subst: [as0, as] → [as, (as0 ∪ (as1 \ as2)) ∩ as3] flags: <all paths>

Additional information

Used terms: as, (as0 ∪ (as1 \ as2)) ∩ as3