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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ fid ∈ fidb ∪ fidb0 ↔ fid ∈ fidb ∨ fid ∈ fidb0
Proof step 11:
∀ sdt1, fid. # sdt1 < # sdt0[str, tree] → true → (fid ∈ sdt1.fids ↔ (∃ str, p. str ∈ sdt1 ∧ p ∈ sdt1[str] ∧ sdt1[str][p] = fnode(fid))) ⊦ fid ∈ sdt0.fids ∨ fid ∈ tree.fids ↔ (∃ str0, p. str0 ∈ sdt0[str, tree] ∧ p ∈ sdt0[str, tree][str0] ∧ sdt0[str, tree][str0][p] = fnode(fid)), str ∈ sdt0

Rule argument

spec: bag-union, inst: fidbag name: in-union seq: ⊦ fid ∈ fidb ∪ fidb0 ↔ fid ∈ fidb ∨ fid ∈ fidb0 subst: [fidb0, fidb, fid] → [tree.fids, sdt0.fids, fid] flags: <all paths>

Additional information

Used terms: tree.fids, sdt0.fids, fid