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Proof step 27 from proof fids-put1 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
⊦ dnode(md, sdt).fids = sdt.fids
Proof step 29:
⊦ ¬ str ∈ sdt → sdt[str, tree].fids = sdt.fids ∪ tree.fids
Proof step 30:
tree[p0, tree1].fids = (tree.fids ∪ tree1.fids) \ tree[p0].fids, tree[parent(p0)].dir?, p0 ∈ tree, tree[p0].fids ⊆ tree.fids ⊦ p0 = ε, sdt0.fids ∪ tree[p0, tree1].fids = ((sdt0.fids ∪ tree.fids) ∪ tree1.fids) \ tree[p0].fids, str ∈ sdt0

Rule argument

Additional information

Used simplifier rules: fids-dnode: ⊦ dnode(md, sdt).fids = sdt.fids;%fids-ins: ⊦ ¬ str ∈ sdt → sdt[str, tree].fids = sdt.fids ∪ tree.fids;% ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws)% ⊦ sdt[str, tree0][str, tree] = sdt[str, tree]; (s)% ⊦ sdt[str, tree][str] = tree; (s)