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Proof step 14 from proof fids-put2 (previous step)

sdt[str][parent(p0)].dir?, str ∈ sdt, parent(p0) ∈ sdt[str], ∀ tree, tree0. tree[parent(p0)].dir? ∧ parent(p0) ∈ tree → tree[p0, tree0].fids = tree.fids ∪ tree0.fids ∨ p0 ∈ tree ⊦ sdt[str, sdt[str][p0, tree1]].fids = sdt.fids ∪ tree1.fids, p0 = ε, p0 ∈ sdt[str]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ str ∈ sdt ↔ (∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0)
Proof step 16:
sdt[str][parent(p0)].dir?, parent(p0) ∈ sdt[str], ∀ tree, tree0. tree[parent(p0)].dir? ∧ parent(p0) ∈ tree → tree[p0, tree0].fids = tree.fids ∪ tree0.fids ∨ p0 ∈ tree, ∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0 ⊦ sdt[str, sdt[str][p0, tree1]].fids = sdt.fids ∪ tree1.fids, p0 = ε, p0 ∈ sdt[str]

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