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Proof step 6 from proof prefixn-toset-subseteq (previous step)

∀ adr. adr ∈ ax → adr ∈ as ⊦ prefixn(n, ax).toSet ⊆ as, prefixn(n, ax).toSet ⊆ ax.toSet

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ as ⊆ as0 ↔ (∀ adr. adr ∈ as → adr ∈ as0)
Proof step 8:
∀ adr. adr ∈ ax → adr ∈ as ⊦ prefixn(n, ax).toSet ⊆ ax.toSet, ∀ adr. adr ∈ prefixn(n, ax).toSet → adr ∈ as

Rule argument

spec: set-basic, inst: node-store name: Subset seq: ⊦ as ⊆ as0 ↔ (∀ adr. adr ∈ as → adr ∈ as0) subst: [as0, as] → [as, prefixn(n, ax).toSet] flags: <all paths>

Additional information

Used terms: as, prefixn(n, ax).toSet