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Proof step 30 from proof btree-adrs (previous step)

# asar0 = m + 1, # asar = m + 1, # zbrar = BRANCH_SIZE, m = n0 + n, union(asar0, 0, m + 1) = alladrs -- adr0, (( ( ( btree(rp0, zbrar[n].child, zbrar[n].adr, height0, true, key-range(zbrar, n, m + 1, fi), iast, asar[n], asar0[n], ok) * (λ n. ⌜zbrar[n].branch?⌝ ∧ btree(rp0, zbrar[n].child, zbrar[n].adr, height0, true, key-range(zbrar, n, m + 1, fi), iast, asar[n], asar0[n], ok) )⊗_ (0, n )) * (λ n. ⌜zbrar[n].branch?⌝ ∧ btree(rp0, zbrar[n].child, zbrar[n].adr, height0, true, key-range(zbrar, n, m + 1, fi), iast, asar[n], asar0[n], ok) )⊗_ (1 + n, n0 )) * P) )(zh1), zbrar[n].branch?, m < BRANCH_SIZE, n < BRANCH_SIZE, ((rp0 => mkznode(rp, zbrar, false, true, m + 1)))(zh2), disj(asar0, 0, m + 1), disj(asar, 0, m + 1), nsize(m + 1, nstat), sorted(zbrar, 0, m), adr0 ∈ alladrs, n ≤ m, asar[n] ⊆ asar0[n], zh2 ⊥ zh1, ∀ 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 = ∅, rp0 = null, adr0 ∈ union(asar, 0, m + 1), asar0[n] ⊆ dom iast, alladrs ⊆ dom iast, asar[n] ⊆ dom iast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
# asar0 = (n0 + n) + 1, # asar = # asar0, # zbrar = BRANCH_SIZE, union(asar0, 0, # asar0) = alladrs -- adr0, (( ( ( btree(rp0, zbrar[n].child, zbrar[n].adr, height0, true, key-range(zbrar, n, # asar0, fi), iast, asar[n], asar0[n], ok) * (λ n1. ⌜zbrar[n1].branch?⌝ ∧ btree(rp0, zbrar[n1].child, zbrar[n1].adr, height0, true, key-range(zbrar, n1, # asar0, fi), iast, asar[n1], asar0[n1], ok) )⊗_ (0, n )) * (λ n1. ⌜zbrar[n1].branch?⌝ ∧ btree(rp0, zbrar[n1].child, zbrar[n1].adr, height0, true, key-range(zbrar, n1, # asar0, fi), iast, asar[n1], asar0[n1], ok) )⊗_ (1 + n, n0 )) * P) )(zh1), zbrar[n].branch?, n0 + n < BRANCH_SIZE, n < BRANCH_SIZE, ((rp0 => mkznode(rp, zbrar, false, true, # asar0)))(zh2), disj(asar0, 0, # asar0), disj(asar, 0, # asar0), nsize(# asar0, nstat), sorted(zbrar, 0, n0 + n), adr0 ∈ alladrs, asar[n] ⊆ asar0[n], zh2 ⊥ zh1, ∀ 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, ∀ n1. ¬ n0 + n < n1 → asar[n1] ⊆ alladrs ⊦ # alladrs = 0, alladrs = ∅, rp0 = null, adr0 ∈ union(asar, 0, # asar0), asar0[n] ⊆ dom iast, alladrs ⊆ dom iast, asar[n] ⊆ dom iast

Rule argument

Additional information

Used simplifier rules: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)