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Proof step 3 from proof btree-clean-alladrs (previous step)

((btree(rp, r, adr, 0, false, ri, iast, as, alladrs, nstat) * P))(zh) ⊦ as = alladrs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
((btree(rp, r, adr, 0, false, ri, iast, ∅, ∅ ++ adr, nstat) * P))(zh) ⊦

Rule argument

Additional information

Used simplifier rules: btree-leaf-adr: ⊦ ((btree(rp, r, adr, 0, dirty, ri, iast, as, alladrs, nstat) * P) )(zh) → alladrs = {adr} ;%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;% ⊦ ∅ ⊆ as; (s)% ⊦ ∅ ≠ as ++ adr; (s)% ⊦ as ≠ ∅ ++ adr → (as ⊆ ∅ ++ adr ↔ as = ∅); (s)% ⊦ {adr} = ∅ ++ adr; (s)%sep-com: ⊦ P * P2 = P2 * P; from specification heap-sep[zheap]%sep-ass: ⊦ (P * P3) * P4 = P * P3 * P4; from specification heap-sep[zheap]