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Proof step 8 from proof fids-del (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ str ∈ sdt ↔ (∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0)
Proof step 10:
p0 ∈ sdt[str], ∀ tree. p0 ∈ tree → (tree -- p0).fids = tree.fids \ tree[p0].fids ∨ p0 = ε, ∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0 ⊦ (dnode(md, sdt) -- (str ' + p0)).fids = sdt.fids \ sdt[str][p0].fids

Rule argument

spec: store, inst: directory-tree-inst name: exin seq: ⊦ str ∈ sdt ↔ (∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0) subst: [sdt, str] → [sdt, str] flags: <all paths>

Additional information

Used terms: sdt, str