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Proof step 9 from proof btree-inv-alloc (previous step)

((btree(null, rp, adr, height, dirty, ri, nodes(istore), as, alladrs, ok) * true))(zh0), ((btree(null, null, adr, height0, false, fi, nodes(istore0), alladrs0, alladrs0, ok) * true))(zh0), inv(istore), refsize-cons(istore, as), refsize-cons(istore0, alladrs0), refsize-cons(istore[n, indexblock(indx, lax, 0)], as), adr ∈ istore, alladrs0 ⊆ dom nodes(istore0), alladrs ⊆ dom nodes(istore), alladrs0 ⊆ dom nodes(istore), as ⊆ dom nodes(istore), as ⊆ alladrs ⊦ n ∈ istore, (adr0 ∈ alladrs → adr0 ∈ istore[n, indexblock(indx, lax, 0)]) ∧ ((btree(null, rp, adr, height, dirty, ri, nodes(istore[n, indexblock(indx, lax, 0)]), as, alladrs, ok) * true))(zh0)

User applied rule case distinction

And obtained the following new goals

Proof step 10:
((btree(null, null, adr, height0, false, fi, nodes(istore0), alladrs0, alladrs0, ok) * true))(zh0), ((btree(null, rp, adr, height, dirty, ri, nodes(istore), as, alladrs, ok) * true))(zh0), inv(istore), refsize-cons(istore[n, indexblock(indx, lax, 0)], as), refsize-cons(istore0, alladrs0), refsize-cons(istore, as), adr ∈ istore, as ⊆ alladrs, as ⊆ dom nodes(istore), alladrs0 ⊆ dom nodes(istore), alladrs ⊆ dom nodes(istore), alladrs0 ⊆ dom nodes(istore0) ⊦ ((btree(null, rp, adr, height, dirty, ri, nodes(istore[n, indexblock(indx, lax, 0)]), as, alladrs, ok) * true))(zh0), n ∈ istore

Rule argument

pos: right 2

Additional information

Used simplifier rules: sub-in: ⊦ adr ∈ as ∧ as ⊆ as0 → adr ∈ as0; (forward) from specification set-basic[node-store]%ne-01: ⊦ adr ∈ as → as ≠ ∅; (forward) from specification set-basic[node-store]% ⊦ adr ∈ dom iast ↔ adr ∈ iast; (s)% ⊦ adr ∈ nodes(istore) ↔ adr ∈ istore; (s)% ⊦ ¬ n ∈ istore ∧ adr ∈ istore → adr ∈ istore[n, iblock]; (ss)