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Proof step 7 from proof fids-sub (previous step)

str ∈ sdt, p0 ∈ sdt[str], ∀ tree. p0 ∈ tree → tree[p0].fids ⊆ tree.fids ⊦ sdt[str][p0].fids ⊆ sdt.fids

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ str ∈ sdt ↔ (∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0)
Proof step 9:
p0 ∈ sdt[str], ∀ tree. p0 ∈ tree → tree[p0].fids ⊆ tree.fids, ∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0 ⊦ sdt[str][p0].fids ⊆ sdt.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