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Proof step 65 from proof btree-clean-fwd (previous step)

# zbrar0 = BRANCH_SIZE, # asar = # asar0, # asar2 = # asar0, # asar1 = # asar0, # zbrar = BRANCH_SIZE, # asar0 = (n0 + 1) + n, zbrar0[n] = mkzbranch(key, adr0, r4), zbrar[n] = mkzbranch(key, adr0, r3), asar1[n] = asar2[n], asar[n] = asar0[n], union(asar2, 0, # asar0) = alladrs1 -- adr, union(asar1, 0, # asar0) = alladrs1 -- adr, union(asar0, 0, # asar0) = alladrs0 -- adr, union(asar, 0, # asar0) = alladrs0 -- adr, ((btree(r, r3, adr0, height0, false, key-range(zbrar, n, # asar0, ri1), iast, asar2[n], asar2[n], ok) * P2))(zh3), ((btree(r2, r4, adr0, height, false, key-range(zbrar0, n, # asar0, ri0), iast, asar0[n], asar0[n], ok) * P))(zh), n < BRANCH_SIZE, n0 + n < BRANCH_SIZE, ((r => mkznode(rp1, zbrar, false, false, # asar0)))(zh4), ((r2 => mkznode(rp0, zbrar0, false, false, # asar0)))(zh5), disj(asar2, 0, # asar0), disj(asar1, 0, # asar0), disj(asar0, 0, # asar0), disj(asar, 0, # asar0), nsize(# asar0, nstat), sorted(zbrar0, 0, n0 + n), sorted(zbrar, 0, n0 + n), unmodified(adr, mkznode(rp0, zbrar0, false, false, # asar0), iast), unmodified(adr, mkznode(rp1, zbrar, false, false, # asar0), iast), adr ∈ alladrs0, adr ∈ alladrs1, adr ∈ iast, asar0[n] ⊆ dom iast, asar2[n] ⊆ dom iast, alladrs0 ⊆ dom iast, alladrs1 ⊆ dom iast, zh4 ⊥ zh3, zh5 ⊥ zh, ((btree(r2, r4, adr0, height, false, key-range(zbrar0, n, # asar0, ri0), iast, asar0[n], asar0[n], ok) * P))(zh) ∧ ((btree(r, r3, adr0, height0, false, key-range(zbrar, n, # asar0, ri1), iast, asar2[n], asar2[n], ok) * P2))(zh3) → height = height0 ∧ asar0[n] = asar2[n] ∧ key-range(zbrar0, n, # asar0, ri0) = key-range(zbrar, n, # asar0, ri1) ∧ asar0[n] = asar2[n], (r1 = null ⊃ P1(zh1);r = r1 ∧ zh4 ∪ zh3 = zh1), (r0 = null ⊃ P0(zh0);r2 = r0 ∧ zh5 ∪ zh = zh0) ⊦ # alladrs0 = 0, # alladrs1 = 0, alladrs0 = ∅, alladrs1 = ∅, key-range(zbrar0, n, # asar0, ri0) = key-range(zbrar0, n, # asar0, ri1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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