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Proof step 20 from proof btree-adrs (previous step)

adr0 ∈ as ↔ ¬ dirty0, (( ( ⌜0 + nused ≤ nused ∧ nused ≤ # zbrar ∧ # zbrar = BRANCH_SIZE⌝ ∧ ⌜nused ≠ 0 ∧ sorted(zbrar, 0, nused - 1)⌝ ∧ _∃ (λ asar. _∃ (λ asar0. ⌜ # asar = nused ∧ disj(asar, 0, nused) ∧ union(asar, 0, nused) = as -- adr0 ∧ # asar0 = nused ∧ disj(asar0, 0, nused) ∧ union(asar0, 0, nused) = alladrs -- adr0 ⌝ ∧ (λ n. ⌜zbrar[n].branch?⌝ ∧ btree(r, zbrar[n].child, zbrar[n].adr, height0, dirty0, key-range(zbrar, n, nused, fi), iast, asar[n], asar0[n], ok) )⊗_ (0, nused )))) * P) )(zh1), ((r => mkznode(rp, zbrar, false, dirty0, nused)))(zh2), nsize(nused, nstat), unmodified(adr0, mkznode(rp, zbrar, false, dirty0, nused), iast), adr0 ∈ alladrs, as ⊆ alladrs, zh2 ⊥ zh1, dirty0 → dirty, (rp0 = null ⊃ adr0 ∈ as ∧ P(zh);r = rp0 ∧ zh2 ∪ zh1 = zh), ∀ 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 = ∅, as ⊆ dom iast, alladrs ⊆ dom iast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
# asar = # asar0, # zbrar = BRANCH_SIZE, adr0 ∈ as ↔ ¬ dirty0, union(asar0, 0, # asar0) = alladrs -- adr0, union(asar, 0, # asar0) = as -- adr0, (( (λ n. ⌜zbrar[n].branch?⌝ ∧ btree(r, zbrar[n].child, zbrar[n].adr, height0, dirty0, key-range(zbrar, n, # asar0, fi), iast, asar[n], asar0[n], ok) )⊗_ (0, # asar0 ) * P) )(zh1), ((r => mkznode(rp, zbrar, false, dirty0, # asar0)))(zh2), disj(asar0, 0, # asar0), disj(asar, 0, # asar0), nsize(# asar0, nstat), sorted(zbrar, 0, # asar0 - 1), unmodified(adr0, mkznode(rp, zbrar, false, dirty0, # asar0), iast), adr0 ∈ alladrs, # asar0 ≤ BRANCH_SIZE, as ⊆ alladrs, zh2 ⊥ zh1, dirty0 → dirty, (rp0 = null ⊃ adr0 ∈ as ∧ P(zh);r = rp0 ∧ zh2 ∪ zh1 = zh), ∀ 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 ⊦ # asar0 = 0, # alladrs = 0, alladrs = ∅, as ⊆ dom iast, alladrs ⊆ dom iast

Rule argument

Additional information

Used simplifier rules: ⊦ ((_∃ Pa0 * P0))(zh) ↔ (∃ asar. ((Pa0(asar) * P0))(zh)); (s)% ⊦ ⌜bv⌝ (zh) ↔ bv; (s)% ⊦ ((P ∧ P2))(zh) ↔ P(zh) ∧ P2(zh); (s)%sep-com: ⊦ P * P2 = P2 * P; from specification heap-sep[zheap]%sep-ass: ⊦ (P * P3) * P4 = P * P3 * P4; from specification heap-sep[zheap]%and-com: ⊦ (P ∧ P2) = (P2 ∧ P); from specification heap-sep[zheap]% ⊦ (⌜bv⌝ ∧ P) * P2 = (⌜bv⌝ ∧ P * P2); (s)%and-ass: ⊦ ((P ∧ P3) ∧ P4) = (P ∧ P3 ∧ P4); from specification heap-sep[zheap]% ⊦ m ≤ m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)