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Proof step 51 from proof btree-clean-alladrs (previous step)

# zbrar = BRANCH_SIZE, # asar = (n0 + n) + 1, # asar0 = # asar, zbrar[n] = mkzbranch(key, adr1, r1), ((btree(r0, r1, adr1, height0, false, key-range(zbrar, n, # asar, ri), iast, asar[n], asar0[n], ok) * P0))(zh0), 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 ⊦ # 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)

User applied rule all left

And closed the goal!

Rule argument

pos: left 18 terms: r0, r1, adr1, key-range(zbrar, n, # asar, ri), iast, asar[n], asar0[n], ok, P0, zh0 abort: F discard: T computed: T

Additional information

Used simplifier rules: btree-refed-subseteq-alladrs: ⊦ (( btree(rp, r, adr, height, dirty, fi, iast, as, alladrs, nstat) * P) )(zh) → as ⊆ alladrs;%btree-adrs: ⊦ (( btree (rp, rp0, adr0, height, dirty, fi, iast, as, alladrs, nstat) * P) )(zh) → as ⊆ dom iast;%sep-ass: ⊦ (P * P3) * P4 = P * P3 * P4; from specification heap-sep[zheap]%sep-com: ⊦ P * P2 = P2 * P; from specification heap-sep[zheap]%sub-in: ⊦ adr ∈ as ∧ as ⊆ as0 → adr ∈ as0; (forward) from specification set-basic[node-store] Used terms: r0, r1, adr1, key-range(zbrar, n, # asar, ri), iast, asar[n], asar0[n], ok, P0, zh0