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Proof step 18 from proof fids-put1 (previous step)

tree[parent(p0)].dir?, p0 ∈ tree, ∀ tree0, tree2. tree0[parent(p0)].dir? ∧ p0 ∈ tree0 → tree0[p0, tree2].fids = (tree0.fids ∪ tree2.fids) \ tree0[p0].fids ⊦ 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 all left

And obtained the following new goals

Proof step 19:
tree[parent(p0)].dir?, p0 ∈ tree, tree[parent(p0)].dir? ∧ p0 ∈ tree → tree[p0, tree1].fids = (tree.fids ∪ tree1.fids) \ tree[p0].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

pos: left 3 terms: tree, tree1 abort: F discard: T computed: F

Additional information

Used terms: tree, tree1