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Proof step 2 from proof btree-repl-dirty-address (previous step)

((btree(rp, r, adr0, height, true, ri, iast, as, alladrs, nstat) * P))(zh), zh[r].dirty, r ∈ zh ⊦ ((btree(rp, r, adr1, height, true, ri, iast, as, (alladrs -- adr0) ++ adr1, nstat) * P))(zh), adr1 ∈ alladrs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
((btree(rp, r, adr0, height, true, ri, iast, as, alladrs, nstat) * P))(zh), zh[r].dirty, r ∈ zh, as ⊆ dom iast, as ⊆ alladrs ⊦ ((btree(rp, r, adr1, height, true, ri, iast, as, (alladrs -- adr0) ++ adr1, nstat) * P))(zh), adr1 ∈ alladrs

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]