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Proof step 6 from proof btree-clean-alladrs (previous step)

((btree(rp, r, adr, height0 +1, false, ri, iast, as, alladrs, nstat) * P))(zh), ∀ rp, r, adr, ri, iast, as, alladrs, nstat, P, zh. ((btree(rp, r, adr, height0, false, ri, iast, as, alladrs, nstat) * P))(zh) → as = alladrs ⊦ as = alladrs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
((btree(rp, r, adr, height0 + 1, false, ri, iast, as, alladrs, nstat) * P))(zh), as ⊆ dom iast, as ⊆ alladrs, ∀ rp, r, adr, ri, iast, as, alladrs, nstat, P, zh. ((btree(rp, r, adr, height0, false, ri, iast, as, alladrs, nstat) * P))(zh) → as = alladrs ⊦ as = 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]% ⊦ m +1 = m + 1; (s)