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Proof step 2 from proof restrict-add-subseteq-in (previous step)

adr ∈ iast0, iast0 | (as ++ adr) ⊆ iast1 ⊦ adr ∈ iast1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ iast ⊆ iast0 ↔ (∀ adr. adr ∈ iast → adr ∈ iast0 ∧ iast[adr] = iast0[adr])
Proof step 4:
adr ∈ iast0, ∀ adr0. adr0 ∈ iast0 | (as ++ adr) → adr0 ∈ iast1 ∧ (iast0 | (as ++ adr))[adr0] = iast1[adr0] ⊦ adr ∈ iast1

Rule argument

spec: store, inst: index-store-address-store name: Subset seq: ⊦ iast ⊆ iast0 ↔ (∀ adr. adr ∈ iast → adr ∈ iast0 ∧ iast[adr] = iast0[adr]) subst: [iast0, iast] → [iast1, iast0 | (as ++ adr)] flags: <all paths>

Additional information

Used terms: iast1, iast0 | (as ++ adr)