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Proof step 5 from proof restrict-further (previous step)

adr ∈ iast0, adr ∈ as1, adr ∈ as0, iast0 | as0 ⊆ iast1, as1 ⊆ as0 ⊦ # as1 = 0, # as0 = 0, iast0[adr] = iast1[adr], as0 = ∅, as1 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

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 | as0] flags: <all paths>

Additional information

Used terms: iast1, iast0 | as0