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Proof step 32 from proof btree-adrs (previous step)

# zbrar = BRANCH_SIZE, # asar = # asar0, # asar0 = (n0 + n) + 1, zbrar[n] = mkzbranch(key, adr, r), union(asar0, 0, # asar0) = alladrs -- adr0, (( ( ( btree(rp0, r, adr, height0, true, key-range(zbrar, n, # asar0, fi), iast, asar[n], asar0[n], ok) * (λ n1. ⌜zbrar[n1].branch?⌝ ∧ btree(rp0, zbrar[n1].child, zbrar[n1].adr, height0, true, key-range(zbrar, n1, # asar0, fi), iast, asar[n1], asar0[n1], ok) )⊗_ (0, n )) * (λ n1. ⌜zbrar[n1].branch?⌝ ∧ btree(rp0, zbrar[n1].child, zbrar[n1].adr, height0, true, key-range(zbrar, n1, # asar0, fi), iast, asar[n1], asar0[n1], ok) )⊗_ (1 + n, n0 )) * P) )(zh1), zbrar[n].branch?, n < BRANCH_SIZE, n0 + n < BRANCH_SIZE, ((rp0 => mkznode(rp, zbrar, false, true, # asar0)))(zh2), disj(asar, 0, # asar0), disj(asar0, 0, # asar0), nsize(# asar0, nstat), sorted(zbrar, 0, n0 + n), adr0 ∈ alladrs, asar[n] ⊆ asar0[n], zh2 ⊥ zh1, ∀ n1. ¬ n0 + n < n1 → asar[n1] ⊆ 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 = ∅, rp0 = null, adr0 ∈ union(asar, 0, # asar0), asar[n] ⊆ dom iast, alladrs ⊆ dom iast, asar0[n] ⊆ dom iast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
# zbrar = BRANCH_SIZE, # asar = # asar0, # asar0 = (n0 + n) + 1, zbrar[n] = mkzbranch(key, adr, r), union(asar0, 0, # asar0) = alladrs -- adr0, (( ( ( btree(rp0, r, adr, height0, true, key-range(zbrar, n, # asar0, fi), iast, asar[n], asar0[n], ok) * (λ n1. ⌜zbrar[n1].branch?⌝ ∧ btree(rp0, zbrar[n1].child, zbrar[n1].adr, height0, true, key-range(zbrar, n1, # asar0, fi), iast, asar[n1], asar0[n1], ok) )⊗_ (0, n )) * (λ n1. ⌜zbrar[n1].branch?⌝ ∧ btree(rp0, zbrar[n1].child, zbrar[n1].adr, height0, true, key-range(zbrar, n1, # asar0, fi), iast, asar[n1], asar0[n1], ok) )⊗_ (1 + n, n0 )) * P) )(zh1), n < BRANCH_SIZE, n0 + n < BRANCH_SIZE, ((rp0 => mkznode(rp, zbrar, false, true, # asar0)))(zh2), disj(asar, 0, # asar0), disj(asar0, 0, # asar0), nsize(# asar0, nstat), sorted(zbrar, 0, n0 + n), adr0 ∈ alladrs, asar[n] ⊆ asar0[n], zh2 ⊥ zh1, ∀ n1. ¬ n0 + n < n1 → asar[n1] ⊆ 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 = ∅, rp0 = null, adr0 ∈ union(asar, 0, # asar0), asar[n] ⊆ dom iast, alladrs ⊆ dom iast, asar0[n] ⊆ dom iast

Rule argument

Additional information

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