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Proof step 17 from proof btree-adrs (previous step)

adr0 ∈ as ↔ ¬ dirty0, nsize(nused, nstat), unmodified(adr0, mkznode(rp, zbrar, false, dirty0, nused), iast), adr0 ∈ alladrs, as ⊆ alladrs, dirty0 → dirty, (rp0 = null ⊃ adr0 ∈ as ∧ P(zh);r = rp0 ∧ zh0 = 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, ∃ zh2, zh1. zh2 ⊥ zh1 ∧ zh0 = zh2 ∪ zh1 ∧ ((r => mkznode(rp, zbrar, false, dirty0, nused)))(zh2) ∧ ((bnode(r, zbrar, 0, nused, nused, height0 + 1, dirty0, fi, iast, as -- adr0, alladrs -- adr0) * P))(zh1) ⊦ # alladrs = 0, alladrs = ∅, as ⊆ dom iast, alladrs ⊆ dom iast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
adr0 ∈ as ↔ ¬ dirty0, ((bnode(r, zbrar, 0, nused, nused, height0 + 1, dirty0, fi, iast, as -- adr0, alladrs -- adr0) * 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 = ∅, alladrs ⊆ dom iast, as ⊆ dom iast

Proof step 17 from proof btree-adrs (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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