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Proof step 39 from proof fids-put1 (previous step)

tree[parent(p0)].dir?, count(fid, tree.fids ∪ tree1.fids) < count(fid, tree[p0].fids), p0 ∈ tree, ∀ fid. count(fid, tree[p0].fids) ≤ count(fid, tree.fids) ⊦ p0 = ε, sdt0.fids ∪ ((tree.fids ∪ tree1.fids) \ tree[p0].fids) = ((sdt0.fids ∪ tree.fids) ∪ tree1.fids) \ tree[p0].fids, str ∈ sdt0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 40:
⊦ count(fid, fidb ∪ fidb0) = count(fid, fidb) + count(fid, fidb0)
Proof step 41:
tree[parent(p0)].dir?, count(fid, tree.fids) + count(fid, tree1.fids) < count(fid, tree[p0].fids), p0 ∈ tree, ∀ fid. count(fid, tree[p0].fids) ≤ count(fid, tree.fids) ⊦ p0 = ε, sdt0.fids ∪ ((tree.fids ∪ tree1.fids) \ tree[p0].fids) = ((sdt0.fids ∪ tree.fids) ∪ tree1.fids) \ tree[p0].fids, str ∈ sdt0

Rule argument

spec: bag-union, inst: fidbag name: union seq: ⊦ count(fid, fidb ∪ fidb0) = count(fid, fidb) + count(fid, fidb0) subst: [fidb0, fidb, fid] → [tree1.fids, tree.fids, fid] flags: <all paths>

Additional information

Used terms: tree1.fids, tree.fids, fid