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Proof step 5 from proof btree-refed-subseteq-alladrs (previous step)

((btree(rp, r, adr, height0 +1, dirty, fi, iast, as, alladrs, nstat) * P))(zh), ∀ rp, r, adr, dirty, fi, iast, as, alladrs, nstat, P, zh. ((btree(rp, r, adr, height0, dirty, fi, iast, as, alladrs, nstat) * P))(zh) → as ⊆ alladrs ⊦ as ⊆ alladrs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
((btree(rp, r, adr, height0 + 1, dirty, fi, iast, as, alladrs, nstat) * P))(zh), ∀ rp, r, adr, dirty, fi, iast, as, alladrs, nstat, P, zh. ((btree(rp, r, adr, height0, dirty, fi, iast, as, alladrs, nstat) * P))(zh) → as ⊆ alladrs ⊦ as ⊆ alladrs

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)