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Proof step 12 from proof fids-del (previous step)

p0 ∈ tree, ∀ tree. p0 ∈ tree → (tree -- p0).fids = tree.fids \ tree[p0].fids ∨ p0 = ε ⊦ (dnode(md, sdt0[str, tree]) -- (str ' + p0)).fids = (sdt0.fids ∪ tree.fids) \ tree[p0].fids, str ∈ sdt0

User applied rule all left

And obtained the following new goals

Proof step 13:
p0 ∈ tree, p0 ∈ tree → (tree -- p0).fids = tree.fids \ tree[p0].fids ∨ p0 = ε ⊦ (dnode(md, sdt0[str, tree]) -- (str ' + p0)).fids = (sdt0.fids ∪ tree.fids) \ tree[p0].fids, str ∈ sdt0

Rule argument

pos: left 2 terms: tree abort: F discard: T computed: T

Additional information

Used terms: tree