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Proof step 21 from proof btree-adrs (previous step)

# 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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
# zbrar = BRANCH_SIZE, # asar = # asar0, # asar0 = m + 1, adr0 ∈ as ↔ ¬ dirty0, union(asar, 0, # asar0) = as -- adr0, union(asar0, 0, # asar0) = alladrs -- 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(asar, 0, # asar0), disj(asar0, 0, # asar0), nsize(# asar0, nstat), sorted(zbrar, 0, m), 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 ⊦ # alladrs = 0, # asar0 = 0, alladrs = ∅, alladrs ⊆ dom iast, as ⊆ dom iast

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# asar0, m]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # asar0, m