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Proof step 6 from proof fids-del (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ dnode(md, sdt).fids = sdt.fids
Proof step 8:
str ∈ sdt, p0 ∈ sdt[str], ∀ tree. p0 ∈ tree → (tree -- p0).fids = tree.fids \ tree[p0].fids ∨ p0 = ε ⊦ (dnode(md, sdt) -- (str ' + p0)).fids = sdt.fids \ sdt[str][p0].fids

Rule argument

Additional information

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