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Proof step 15 from proof btree-clean-alladrs (previous step)

# zbrar = BRANCH_SIZE, # asar = m + 1, # asar0 = m + 1, union(asar, 0, m + 1) = as -- adr, union(asar0, 0, m + 1) = alladrs -- adr, (( ( (λ n. ⌜zbrar[n].branch?⌝ ∧ btree(r0, zbrar[n].child, zbrar[n].adr, height0, false, key-range(zbrar, n, m + 1, ri), iast, asar[n], asar0[n], ok) )⊗_ (0, m + 1 ) * r0 => mkznode(rp, zbrar, false, false, m + 1)) * P) )(zh0), m < BRANCH_SIZE, disj(asar, 0, m + 1), disj(asar0, 0, m + 1), nsize(m + 1, nstat), sorted(zbrar, 0, m), unmodified(adr, mkznode(rp, zbrar, false, false, m + 1), iast), adr ∈ as, adr ∈ alladrs, adr ∈ iast, as ⊆ alladrs, as ⊆ dom iast, (r = null ⊃ P(zh);r0 = r ∧ zh0 = zh), ∀ rp, r, adr, ri, iast, as, alladrs, nstat, P, zh. ((btree(rp, r, adr, height0, false, ri, iast, as, alladrs, nstat) * P))(zh) → as = alladrs ⊦ # as = 0, # alladrs = 0, alladrs = ∅, as = ∅, as = alladrs

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: set-union, inst: node-store name: delete-shift seq: ⊦ adr ∈ as → (as -- adr = as0 ↔ as = as0 ++ adr ∧ ¬ adr ∈ as0) subst: [adr, as, as0] → [adr, as, union(asar, 0, m + 1)] flags: <all paths>

Additional information

Used terms: adr, as, union(asar, 0, m + 1)