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Proof step 3 from proof fids-del (previous step)

p ∈ tree ⊦ (tree -- p).fids = tree.fids \ tree[p].fids, p = ε

User applied rule structural induction

And obtained the following new goals

Proof step 4:
str + p0 ∈ tree, ∀ tree. p0 ∈ tree → (tree -- p0).fids = tree.fids \ tree[p0].fids ∨ p0 = ε ⊦ str + p0 = ε, (tree -- (str + p0)).fids = tree.fids \ tree[str + p0].fids

Rule argument

indvar: p

Additional information

Used simplifier rules: path freely generated by ε, +;;