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Proof step 15 from proof fids-del (previous step)

(tree -- p0).fids = tree.fids \ tree[p0].fids, p0 ∈ tree ⊦ (dnode(md, sdt0[str, tree]) -- (str ' + p0)).fids = (sdt0.fids ∪ tree.fids) \ tree[p0].fids, p0 = ε, str ∈ sdt0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
p ≠ ε ⊦ str ∈ sdt → dnode(md, sdt) -- (str ' + p) = dnode(md, sdt[str, sdt[str] -- p])
Proof step 17:
(tree -- p0).fids = tree.fids \ tree[p0].fids, p0 ∈ tree, str ∈ sdt0[str, tree] ⊦ dnode(md, sdt0[str, tree][str, sdt0[str, tree][str] -- p0]).fids = (sdt0.fids ∪ tree.fids) \ tree[p0].fids, p0 = ε, str ∈ sdt0

Rule argument

spec: directory-tree name: del-recin seq: p ≠ ε ⊦ str ∈ sdt → dnode(md, sdt) -- (str ' + p) = dnode(md, sdt[str, sdt[str] -- p]) subst: [p, str, sdt, md] → [p0, str, sdt0[str, tree], md] flags: <all paths>

Additional information

Used simplifier rules: ⊦ str ∈ sdt[str, tree]; (s) Used terms: p0, str, sdt0[str, tree], md