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Proof step 34 from proof btree-adrs (previous step)

# asar0 = m + 1, # asar = m + 1, # zbrar = BRANCH_SIZE, adr0 ∈ as ↔ ¬ dirty0, as = union(asar, 0, m + 1) ++ adr0, union(asar0, 0, m + 1) = alladrs -- adr0, (( (λ n. ⌜zbrar[n].branch?⌝ ∧ btree(r, zbrar[n].child, zbrar[n].adr, height0, dirty0, key-range(zbrar, n, m + 1, fi), iast, asar[n], asar0[n], ok) )⊗_ (0, m + 1 ) * P) )(zh1), m < BRANCH_SIZE, ((r => mkznode(rp, zbrar, false, dirty0, m + 1)))(zh2), disj(asar0, 0, m + 1), disj(asar, 0, m + 1), nsize(m + 1, nstat), sorted(zbrar, 0, m), unmodified(adr0, mkznode(rp, zbrar, false, dirty0, m + 1), iast), adr0 ∈ alladrs, adr0 ∈ as, as ⊆ alladrs, zh2 ⊥ zh1, dirty0 → dirty, (rp0 = null ⊃ adr0 ∈ as ∧ P(zh);r = rp0 ∧ zh2 ∪ zh1 = zh), ∀ rp, rp0, adr0, dirty, fi, iast, as, alladrs, nstat, P, zh. ((btree(rp, rp0, adr0, height0, dirty, fi, iast, as, alladrs, nstat) * P))(zh) → as ⊆ dom iast ⊦ # alladrs = 0, alladrs = ∅, adr0 ∈ union(asar, 0, m + 1), as ⊆ dom iast, alladrs ⊆ dom iast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
# asar0 = m + 1, # asar = m + 1, # zbrar = BRANCH_SIZE, union(asar0, 0, m + 1) = alladrs -- adr0, (( (λ n. ⌜zbrar[n].branch?⌝ ∧ btree(r, zbrar[n].child, zbrar[n].adr, height0, false, key-range(zbrar, n, m + 1, fi), iast, asar[n], asar0[n], ok) )⊗_ (0, m + 1 ) * P) )(zh1), m < BRANCH_SIZE, ((r => mkznode(rp, zbrar, false, false, m + 1)))(zh2), disj(asar0, 0, m + 1), disj(asar, 0, m + 1), nsize(m + 1, nstat), sorted(zbrar, 0, m), unmodified(adr0, mkznode(rp, zbrar, false, false, m + 1), iast), adr0 ∈ alladrs, zh2 ⊥ zh1, (rp0 = null ⊃ P(zh);r = rp0 ∧ zh2 ∪ zh1 = zh), ∀ n. ¬ m < n → asar[n] ⊆ alladrs, ∀ rp, rp0, adr0, dirty, fi, iast, as, alladrs, nstat, P, zh. ((btree(rp, rp0, adr0, height0, dirty, fi, iast, as, alladrs, nstat) * P))(zh) → as ⊆ dom iast ⊦ # alladrs = 0, alladrs = ∅, adr0 ∈ union(asar, 0, m + 1), union(asar, 0, m + 1) ++ adr0 ⊆ dom iast, alladrs ⊆ dom iast

Rule argument

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ 0 ≤ m; (s)%⊦ union(asar, n, m) ⊆ as ↔ (∀ n0. n ≤ n0 ∧ n0 < n + m → asar[n0] ⊆ as); (s)% ⊦ adr ∈ as → (as0 ++ adr ⊆ as ↔ as0 ⊆ as); (s)% ⊦ ¬ adr ∈ as → (adr ∈ as ++ adr0 ↔ adr = adr0); (s)