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Proof step 26 from proof btree-istore-reduce-refed (previous step)

adr ∈ as ↔ ¬ dirty0, ((r => mkznode(rp, zbrar, false, dirty0, nused) * bnode(r, zbrar, 0, nused, nused, height0 + 1, dirty0, ri, iast0, as -- adr, alladrs -- adr) * P) )(zh0), nsize(nused, nstat), unmodified(adr, mkznode(rp, zbrar, false, dirty0, nused), iast0), adr ∈ alladrs, iast0 | as ⊆ iast1, as ⊆ dom iast0, as ⊆ alladrs, dirty0 → dirty, (rp0 = null ⊃ adr ∈ as ∧ P(zh);r = rp0 ∧ zh0 = zh), ∀ rp, rp0, adr, dirty, ri, iast0, as, alladrs, nstat, P, zh, iast1. ((btree(rp, rp0, adr, height0, dirty, ri, iast0, as, alladrs, nstat) * P))(zh) ∧ iast0 | as ⊆ iast1 ∧ as ⊆ dom iast0 ∧ as ⊆ alladrs → ((btree(rp, rp0, adr, height0, dirty, ri, iast1, as, alladrs, nstat) * P))(zh) ⊦ # alladrs = 0, alladrs = ∅, unmodified(adr, mkznode(rp, zbrar, false, dirty0, nused), iast1) ∧ (( r => mkznode(rp, zbrar, false, dirty0, nused) * bnode(r, zbrar, 0, nused, nused, height0 + 1, dirty0, ri, iast1, as -- adr, alladrs -- adr) * P) )(zh0)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 3

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