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Proof step 54 from proof btree-ri-unique (previous step)

# asar1 = (n0 + 1) + n, # asar0 = # asar1, # zbrar0 = BRANCH_SIZE, # asar = # asar1, # asar2 = # asar1, adr ∈ as0 ↔ adr ∈ as1, adr ∈ as1 ↔ ¬ dirty1, zbrar0[n] = mkzbranch(key, adr0, r), union(asar2, 0, # asar1) = alladrs0 -- adr, union(asar1, 0, # asar1) = as0 -- adr, union(asar, 0, # asar1) = as1 -- adr, union(asar0, 0, # asar1) = alladrs1 -- adr, ((btree(rp0, r, adr0, height0, dirty1, key-range(zbrar0, n, # asar1, ri1), iast, asar[n], asar0[n], ok) * P2))(zh), ((btree(rp0, r, adr0, height0, dirty1, key-range(zbrar0, n, # asar1, ri0), iast, asar1[n], asar2[n], ok) * P))(zh), n < BRANCH_SIZE, n0 + n < BRANCH_SIZE, n + 1 < BRANCH_SIZE, # asar1 < BRANCH_SIZE, disj(asar2, 0, # asar1), disj(asar1, 0, # asar1), disj(asar, 0, # asar1), disj(asar0, 0, # asar1), sorted(zbrar0, 0, n0 + n), unmodified(adr, mkznode(rp, zbrar0, false, dirty1, # asar1), iast), adr ∈ alladrs1, adr ∈ alladrs0, asar[n] ⊆ asar0[n], asar[n] ⊆ dom iast, as1 ⊆ dom iast, as1 ⊆ alladrs1, as0 ⊆ dom iast, as0 ⊆ alladrs0, asar1[n] ⊆ asar2[n], asar1[n] ⊆ dom iast, dirty1 → dirty, ((btree(rp0, r, adr0, height0, dirty1, key-range(zbrar0, n, # asar1, ri1), iast, asar[n], asar0[n], ok) * P2))(zh) ∧ ((btree(rp0, r, adr0, height0, dirty1, key-range(zbrar0, n, # asar1, ri0), iast, asar1[n], asar2[n], ok) * P))(zh) ∧ asar[n] ⊆ asar0[n] ∧ asar[n] ⊆ dom iast ∧ asar1[n] ⊆ asar2[n] ∧ asar1[n] ⊆ dom iast → key-range(zbrar0, n, # asar1, ri1) = key-range(zbrar0, n, # asar1, ri0) ∨ r = null ⊦ # alladrs0 = 0, # alladrs1 = 0, alladrs0 = ∅, alladrs1 = ∅, key-range(zbrar0, n, # asar1, ri0) = key-range(zbrar0, n, # asar1, ri1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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