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Proof step 3 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, (∃ height1, dirty0, as1, alladrs1. ((btree(null, rp, adr, height1, dirty0, ri, nodes(istore[n, indexblock(indx, lax, 0)]), as1, alladrs1, ok) * true))(zh0) ∧ refsize-cons(istore[n, indexblock(indx, lax, 0)], as1) ∧ alladrs1 ⊆ dom nodes(istore[n, indexblock(indx, lax, 0)])) ∧ (∃ height, as, alladrs. ((btree(null, null, adr, height, false, fi, nodes(istore0), as, alladrs, ok) * true))(zh0) ∧ refsize-cons(istore0, as) ∧ alladrs ⊆ dom nodes(istore[n, indexblock(indx, lax, 0)]))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

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