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Proof step 15 from proof fids-sub (previous step)

p0 ∈ tree, ∀ fid. count(fid, tree[p0].fids) ≤ count(fid, tree.fids) ⊦ str ∈ sdt0, tree[p0].fids ⊆ sdt0.fids ∪ tree.fids

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: bag-union, inst: fidbag name: count-subset seq: ⊦ fidb ⊆ fidb0 ↔ (∀ fid. count(fid, fidb) ≤ count(fid, fidb0)) subst: [fidb0, fidb] → [sdt0.fids ∪ tree.fids, tree[p0].fids] flags: <all paths>

Additional information

Used terms: sdt0.fids ∪ tree.fids, tree[p0].fids