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Proof step 49 from proof btree-clean-alladrs (previous step)

# zbrar = BRANCH_SIZE, # asar = (n0 + n) + 1, # asar0 = # asar, zbrar[n] = mkzbranch(key, adr1, r1), n < BRANCH_SIZE, n0 + n < BRANCH_SIZE, disj(asar, 0, # asar), disj(asar0, 0, # asar), nsize(# asar, nstat), sorted(zbrar, 0, n0 + n), unmodified(adr, mkznode(rp, zbrar, false, false, # asar), iast), adr ∈ iast, adr0 ∈ asar0[n], asar[n] ⊆ asar0[n], asar[n] ⊆ 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, ∀ n1. ¬ n0 + n < n1 → asar[n1] ⊆ union(asar0, 0, # asar) ++ adr, ∀ n1. ¬ n0 + n < n1 → asar[n1] ⊆ dom iast, ∃ P0. P0 ≡ P * r0 => mkznode(rp, zbrar, false, false, # asar) * (λ n1. ⌜zbrar[n1].branch?⌝ ∧ btree(r0, zbrar[n1].child, zbrar[n1].adr, height0, false, key-range(zbrar, n1, # asar, ri), iast, asar[n1], asar0[n1], ok) )⊗_ (1 + n, n0 ) * (λ n1. ⌜zbrar[n1].branch?⌝ ∧ btree(r0, zbrar[n1].child, zbrar[n1].adr, height0, false, key-range(zbrar, n1, # asar, ri), iast, asar[n1], asar0[n1], ok) )⊗_ (0, n ) ∧ ((btree(r0, r1, adr1, height0, false, key-range(zbrar, n, # asar, ri), iast, asar[n], asar0[n], ok) * P0))(zh0) ⊦ # union(asar0, 0, # asar) = 0, # asar0[n] = 0, asar0[n] = ∅, union(asar0, 0, # asar) = ∅, adr0 ∈ asar[n], adr0 ∈ union(asar, 0, # asar), adr ∈ union(asar0, 0, # asar), adr ∈ union(asar, 0, # asar)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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