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Proof step 9 from proof fids-put3 (previous step)

sdt0[str][p0] = dnode(md0, sdt), str ∈ sdt0, p0 ∈ sdt0[str], ∀ tree, md0, sdt, md1. tree[p0] = dnode(md0, sdt) ∧ p0 ∈ tree → tree[p0, dnode(md1, sdt)].fids = tree.fids ⊦ dnode(md, sdt0)[str ' + p0, dnode(md1, sdt)].fids = sdt0.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:
sdt0[str][p0] = dnode(md0, sdt), p0 ∈ sdt0[str], ∀ tree, md0, sdt, md1. tree[p0] = dnode(md0, sdt) ∧ p0 ∈ tree → tree[p0, dnode(md1, sdt)].fids = tree.fids, ∃ tree, sdt1. sdt0 = sdt1[str, tree] ∧ ¬ str ∈ sdt1 ⊦ dnode(md, sdt0)[str ' + p0, dnode(md1, sdt)].fids = sdt0.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] → [sdt0, str] flags: <all paths>

Additional information

Used terms: sdt0, str