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Proof step 29 from proof btree-refed-subseteq-alladrs (previous step)

# asar0 = # asar, # asar = (n1 + n) + 1, # zbrar = BRANCH_SIZE, zbrar[n] = mkzbranch(key, adr0, r0), (( ( ( ( btree(r, r0, adr0, height0, true, key-range(zbrar, n, # asar, fi), iast, asar[n], asar0[n], ok) * (λ n0. ⌜zbrar[n0].branch?⌝ ∧ btree(r, zbrar[n0].child, zbrar[n0].adr, height0, true, key-range(zbrar, n0, # asar, fi), iast, asar[n0], asar0[n0], ok) )⊗_ (0, n )) * (λ n0. ⌜zbrar[n0].branch?⌝ ∧ btree(r, zbrar[n0].child, zbrar[n0].adr, height0, true, key-range(zbrar, n0, # asar, fi), iast, asar[n0], asar0[n0], ok) )⊗_ (1 + n, n1 )) * r => mkznode(rp, zbrar, false, true, # asar)) * P) )(zh0), zbrar[n].branch?, n1 + n < BRANCH_SIZE, n < BRANCH_SIZE, n0 < BRANCH_SIZE, disj(asar0, 0, # asar), disj(asar, 0, # asar), nsize(# asar, nstat), sorted(zbrar, 0, n1 + n), n0 ≤ n1 + n, ∀ rp, r, adr, dirty, fi, iast, as, alladrs, nstat, P, zh. ((btree(rp, r, adr, height0, dirty, fi, iast, as, alladrs, nstat) * P))(zh) → as ⊆ alladrs ⊦ adr ∈ union(asar0, 0, # asar), adr ∈ union(asar, 0, # asar), asar[n] ⊆ union(asar0, 0, # asar), asar[n0] ⊆ union(asar0, 0, # asar)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
# asar0 = # asar, # asar = (n1 + n) + 1, # zbrar = BRANCH_SIZE, zbrar[n] = mkzbranch(key, adr0, r0), (( ( ( ( btree(r, r0, adr0, height0, true, key-range(zbrar, n, # asar, fi), iast, asar[n], asar0[n], ok) * (λ n0. ⌜zbrar[n0].branch?⌝ ∧ btree(r, zbrar[n0].child, zbrar[n0].adr, height0, true, key-range(zbrar, n0, # asar, fi), iast, asar[n0], asar0[n0], ok) )⊗_ (0, n )) * (λ n0. ⌜zbrar[n0].branch?⌝ ∧ btree(r, zbrar[n0].child, zbrar[n0].adr, height0, true, key-range(zbrar, n0, # asar, fi), iast, asar[n0], asar0[n0], ok) )⊗_ (1 + n, n1 )) * r => mkznode(rp, zbrar, false, true, # asar)) * P) )(zh0), n1 + n < BRANCH_SIZE, n < BRANCH_SIZE, n0 < BRANCH_SIZE, disj(asar0, 0, # asar), disj(asar, 0, # asar), nsize(# asar, nstat), sorted(zbrar, 0, n1 + n), n0 ≤ n1 + n, ∀ rp, r, adr, dirty, fi, iast, as, alladrs, nstat, P, zh. ((btree(rp, r, adr, height0, dirty, fi, iast, as, alladrs, nstat) * P))(zh) → as ⊆ alladrs ⊦ adr ∈ union(asar0, 0, # asar), adr ∈ union(asar, 0, # asar), asar[n] ⊆ union(asar0, 0, # asar), asar[n0] ⊆ union(asar0, 0, # asar)

Rule argument

Additional information

Used simplifier rules: ⊦ mkzbranch(key, adr, r).branch?; (s)