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Proof step 32 from proof btree-clean-alladrs (previous step)

# asar0 = m + 1, # asar = m + 1, # zbrar = BRANCH_SIZE, (( ( (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, n < BRANCH_SIZE, disj(asar0, 0, m + 1), disj(asar, 0, m + 1), nsize(m + 1, nstat), sorted(zbrar, 0, m), unmodified(adr, mkznode(rp, zbrar, false, false, m + 1), iast), adr0 ∈ asar[n], adr ∈ iast, n ≤ m, (r = null ⊃ P(zh);r0 = r ∧ zh0 = zh), ∀ n. ¬ m < n → asar[n] ⊆ dom iast, ∀ n. ¬ m < n → asar[n] ⊆ union(asar0, 0, m + 1) ++ adr, ∀ 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 ⊦ # asar[n] = 0, # union(asar, 0, m + 1) = 0, asar[n] = ∅, union(asar, 0, m + 1) = ∅, adr ∈ union(asar, 0, m + 1), adr ∈ union(asar0, 0, m + 1), adr0 ∈ union(asar0, 0, m + 1), adr ∈ asar[n]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 17 terms: n

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ ¬ adr ∈ as → (as ⊆ as0 ++ adr ↔ as ⊆ as0); (s)%sub-in: ⊦ adr ∈ as ∧ as ⊆ as0 → adr ∈ as0; (forward) from specification set-basic[node-store] Used terms: n