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Proof step 5 from proof fids-sub (previous step)

tree = dnode(md, sdt), p0 ∈ tree[str '], str ' ∈ tree, ∀ tree. p0 ∈ tree → tree[p0].fids ⊆ tree.fids ⊦ tree[str ' + p0].fids ⊆ tree.fids

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ dnode(md, sdt).fids = sdt.fids
Proof step 7:
str ∈ sdt, p0 ∈ sdt[str], ∀ tree. p0 ∈ tree → tree[p0].fids ⊆ tree.fids ⊦ sdt[str][p0].fids ⊆ sdt.fids

Rule argument

Additional information

Used simplifier rules: fids-dnode: ⊦ dnode(md, sdt).fids = sdt.fids;%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str ∈ sdt → dnode(md, sdt)[str ' + p] = sdt[str][p]; (s)%str ∈ sdt ⊦ dnode(md, sdt)[str '] = sdt[str]; (s)% ⊦ str ' ∈ dnode(md, sdt) ↔ str ∈ sdt; (s)