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Proof step 13 from proof btree-refed-subseteq-alladrs (previous step)

# zbrar = BRANCH_SIZE, # asar = # asar0, # asar0 = m + 1, adr ∈ as ↔ ¬ dirty0, union(asar, 0, # asar0) = as -- adr, union(asar0, 0, # asar0) = alladrs -- adr, (( ( (λ n. ⌜zbrar[n].branch?⌝ ∧ btree(r0, zbrar[n].child, zbrar[n].adr, height0, dirty0, key-range(zbrar, n, # asar0, fi), iast, asar[n], asar0[n], ok) )⊗_ (0, # asar0 ) * r0 => mkznode(rp, zbrar, false, dirty0, # asar0)) * P) )(zh0), disj(asar, 0, # asar0), disj(asar0, 0, # asar0), nsize(# asar0, nstat), sorted(zbrar, 0, m), unmodified(adr, mkznode(rp, zbrar, false, dirty0, # asar0), iast), adr ∈ alladrs, # asar0 ≤ BRANCH_SIZE, dirty0 → dirty, (r = null ⊃ adr ∈ as ∧ P(zh);r0 = r ∧ zh0 = zh), ∀ rp, r, adr, dirty, fi, iast, as, alladrs, nstat, P, zh. ((btree(rp, r, adr, height0, dirty, fi, iast, as, alladrs, nstat) * P))(zh) → as ⊆ alladrs ⊦ # alladrs = 0, # asar0 = 0, alladrs = ∅, as ⊆ alladrs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# zbrar = BRANCH_SIZE, # asar = m + 1, # asar0 = m + 1, adr ∈ as ↔ ¬ dirty0, union(asar, 0, m + 1) = as -- adr, union(asar0, 0, m + 1) = alladrs -- adr, (( ( (λ n. ⌜zbrar[n].branch?⌝ ∧ btree(r0, zbrar[n].child, zbrar[n].adr, height0, dirty0, key-range(zbrar, n, m + 1, fi), iast, asar[n], asar0[n], ok) )⊗_ (0, m + 1 ) * r0 => mkznode(rp, zbrar, false, dirty0, m + 1)) * P) )(zh0), m < BRANCH_SIZE, disj(asar, 0, m + 1), disj(asar0, 0, m + 1), nsize(m + 1, nstat), sorted(zbrar, 0, m), unmodified(adr, mkznode(rp, zbrar, false, dirty0, m + 1), iast), adr ∈ alladrs, dirty0 → dirty, (r = null ⊃ adr ∈ as ∧ P(zh);r0 = r ∧ zh0 = zh), ∀ rp, r, adr, dirty, fi, iast, as, alladrs, nstat, P, zh. ((btree(rp, r, adr, height0, dirty, fi, iast, as, alladrs, nstat) * P))(zh) → as ⊆ alladrs ⊦ # alladrs = 0, alladrs = ∅, as ⊆ alladrs

Rule argument

Additional information

Used simplifier rules: ⊦ BRANCH_SIZE ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)