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

((btree(rp1, r1, adr, height1, false, ri1, iast, as1, alladrs1, nstat) * P1))(zh1), ((btree(rp0, r0, adr, 0, false, ri0, iast, as0, alladrs0, nstat) * P0))(zh0) ⊦ 0 = height1 ∧ as0 = as1 ∧ ri0 = ri1 ∧ alladrs0 = alladrs1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
((btree(rp0, r0, adr, 0, false, ri0, iast, ∅ ++ adr, ∅ ++ adr, nstat) * P0))(zh0), ((btree(rp1, r1, adr, height1, false, ri1, iast, alladrs1, alladrs1, nstat) * P1))(zh1), adr ∈ iast, alladrs1 ⊆ dom iast ⊦ 0 = height1 ∧ ri0 = ri1 ∧ ∅ ++ adr = alladrs1

Rule argument

Additional information

Used simplifier rules: btree-clean-alladrs: ⊦ (( btree(rp, r, adr, height, false, ri, iast, as, alladrs, nstat) * P) )(zh) → as = alladrs;%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;%btree-leaf-adr: ⊦ ((btree(rp, r, adr, 0, dirty, ri, iast, as, alladrs, nstat) * P) )(zh) → alladrs = {adr} ;% ⊦ adr ∈ dom iast ↔ adr ∈ iast; (s)% ⊦ adr ∈ ∅ ++ adr0 ↔ adr = adr0; (s)% ⊦ ∅ ++ adr ⊆ as ↔ adr ∈ as; (s)% ⊦ {adr} = ∅ ++ adr; (s)% ⊦ as ⊆ as; (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]