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Proof step 13 from proof fids-strpath (previous step)

fid ∈ tree.fids, ∀ sdt1, fid. # sdt1 < # sdt0[str, tree] → true → (fid ∈ sdt1.fids ↔ (∃ str, p. str ∈ sdt1 ∧ p ∈ sdt1[str] ∧ sdt1[str][p] = fnode(fid))) ⊦ str ∈ sdt0, ∃ str0, p. str0 ∈ sdt0[str, tree] ∧ p ∈ sdt0[str, tree][str0] ∧ sdt0[str, tree][str0][p] = fnode(fid)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ fid ∈ tree.fids ↔ (∃ p. p ∈ tree ∧ tree[p] = fnode(fid))
Proof step 15:
∃ p. p ∈ tree ∧ tree[p] = fnode(fid), ∀ sdt1, fid. # sdt1 < # sdt0[str, tree] → true → (fid ∈ sdt1.fids ↔ (∃ str, p. str ∈ sdt1 ∧ p ∈ sdt1[str] ∧ sdt1[str][p] = fnode(fid))) ⊦ str ∈ sdt0, ∃ str0, p. str0 ∈ sdt0[str, tree] ∧ p ∈ sdt0[str, tree][str0] ∧ sdt0[str, tree][str0][p] = fnode(fid)

Rule argument

spec: <toplevel> name: fids-in seq: ⊦ fid ∈ tree.fids ↔ (∃ p. p ∈ tree ∧ tree[p] = fnode(fid)) subst: [tree, fid] → [tree, fid] flags: <all paths>

Additional information

Used terms: tree, fid