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Proof step 5 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), adr ∈ istore, alladrs0 ⊆ dom nodes(istore0), alladrs ⊆ dom nodes(istore), alladrs0 ⊆ dom nodes(istore), as ⊆ dom nodes(istore), as ⊆ alladrs ⊦ n ∈ istore, ((btree(null, rp, adr, height, dirty, ri, nodes(istore[n, indexblock(indx, lax, 0)]), as, alladrs, ok) * true))(zh0) ∧ refsize-cons(istore[n, indexblock(indx, lax, 0)], as) ∧ alladrs ⊆ dom nodes(istore[n, indexblock(indx, lax, 0)])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
((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, alladrs ⊆ dom nodes(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)

Rule argument

Additional information

Used simplifier rules: ⊦ ∅.size = 0; (s)% ⊦ as ⊆ dom nodes(istore) ∧ ¬ n ∈ istore → as | n = ∅; (s)% ⊦ indexblock(indx, lax, m).ref-size = m; (s)%⊦ refsize-cons(istore, as) → ( refsize-cons(istore[n, iblock], as) ↔ ¬ n ∈ istore ∨ istore[n].ref-size = (as | n).size → iblock.ref-size = (as | n).size); (ss)