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Proof step 8 from proof btree-restrict-union-proper (previous step)

((btree(rp, r, adr, height, dirty, ri, iast, asar[n], alladrs, nstat) * P))(zh), n < # asar, asar[n] ⊆ dom iast, asar[n] ⊆ alladrs ⊦ ((btree(rp, r, adr, height, dirty, ri, iast | union(asar, 0, # asar), asar[n], alladrs, nstat) * P))(zh), iast | asar[n] ⊆ iast | union(asar, 0, # asar)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ iast ⊆ iast0 ↔ (∀ adr. adr ∈ iast → adr ∈ iast0 ∧ iast[adr] = iast0[adr])
Proof step 10:
((btree(rp, r, adr, height, dirty, ri, iast, asar[n], alladrs, nstat) * P))(zh), n < # asar, asar[n] ⊆ alladrs, asar[n] ⊆ dom iast ⊦ ((btree(rp, r, adr, height, dirty, ri, iast | union(asar, 0, # asar), asar[n], alladrs, nstat) * P))(zh), ∀ adr. adr ∈ iast | asar[n] → adr ∈ iast | union(asar, 0, # asar) ∧ (iast | asar[n])[adr] = (iast | union(asar, 0, # asar))[adr]

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] → [iast | union(asar, 0, # asar), iast | asar[n]] flags: <all paths>

Additional information

Used terms: iast | union(asar, 0, # asar), iast | asar[n]