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Proof step 4 from proof btree-restrict-union-proper (previous step)

((btree(rp, r, adr, height, dirty, ri, iast, asar[n], alladrs, nstat) * P))(zh), n < # asar ⊦ ((btree(rp, r, adr, height, dirty, ri, iast | union(asar, 0, # asar), asar[n], alladrs, nstat) * P))(zh) ↔ true

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
((btree(rp, r, adr, height, dirty, ri, iast, asar[n], alladrs, nstat) * P))(zh), n < # asar, asar[n] ⊆ dom iast, asar[n] ⊆ alladrs ⊦ ((btree(rp, r, adr, height, dirty, ri, iast | union(asar, 0, # asar), asar[n], alladrs, nstat) * P))(zh)

Rule argument

Additional information

Used simplifier rules: btree-refed-subseteq-alladrs: ⊦ (( btree(rp, r, adr, height, dirty, fi, iast, as, alladrs, nstat) * P) )(zh) → as ⊆ alladrs;%btree-adrs: ⊦ (( btree (rp, rp0, adr0, height, dirty, fi, iast, as, alladrs, nstat) * P) )(zh) → as ⊆ dom iast;%sep-com: ⊦ P * P2 = P2 * P; from specification heap-sep[zheap]%sep-ass: ⊦ (P * P3) * P4 = P * P3 * P4; from specification heap-sep[zheap]