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Proof step 9 from proof fids-put1 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ str ∈ sdt ↔ (∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0)
Proof step 11:
dnode(md, sdt)[parent(str ' + p0)].dir?, p0 ∈ sdt[str], ∀ tree0, tree1. tree0[parent(p0)].dir? ∧ p0 ∈ tree0 → tree0[p0, tree1].fids = (tree0.fids ∪ tree1.fids) \ tree0[p0].fids ∨ p0 = ε, ∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0 ⊦ dnode(md, sdt)[str ' + p0, tree1].fids = (sdt.fids ∪ tree1.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