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Proof step 20 from proof fids-put1 (previous step)

tree[p0, tree1].fids = (tree.fids ∪ tree1.fids) \ tree[p0].fids, tree[parent(p0)].dir?, p0 ∈ tree ⊦ dnode(md, sdt0[str, tree])[str ' + p0, tree1].fids = ((sdt0.fids ∪ tree.fids) ∪ tree1.fids) \ tree[p0].fids, p0 = ε, str ∈ sdt0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
fidb ⊆ fidb0 ⊦ (fidb0 \ fidb) ∪ fidb1 = (fidb0 ∪ fidb1) \ fidb
Proof step 22:
p ∈ tree ⊦ tree[p].fids ⊆ tree.fids
Proof step 23:
tree[p0, tree1].fids = (tree.fids \ tree[p0].fids) ∪ tree1.fids, tree[parent(p0)].dir?, p0 ∈ tree, tree[p0].fids ⊆ tree.fids ⊦ dnode(md, sdt0[str, tree])[str ' + p0, tree1].fids = ((sdt0.fids ∪ tree.fids) ∪ tree1.fids) \ tree[p0].fids, p0 = ε, str ∈ sdt0

Rule argument

spec: bag-union, inst: fidbag name: union-diff-distr seq: fidb ⊆ fidb0 ⊦ (fidb0 \ fidb) ∪ fidb1 = (fidb0 ∪ fidb1) \ fidb subst: [fidb, fidb1, fidb0] → [tree[p0].fids, tree1.fids, tree.fids] flags: <rotated> <all paths>

Additional information

Used simplifier rules: fids-sub: p ∈ tree ⊦ tree[p].fids ⊆ tree.fids; Used terms: tree[p0].fids, tree1.fids, tree.fids