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Proof step 44 from proof btree-adrs (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [n, m, n0]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, m, n0