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Proof step 4 from proof fids-put1 (previous step)

tree0[parent(p)].dir?, p ∈ tree0 ⊦ tree0[p, tree1].fids = (tree0.fids ∪ tree1.fids) \ tree0[p].fids, p = ε

User applied rule structural induction

And obtained the following new goals

Proof step 5:
tree0[parent(str + p0)].dir?, str + p0 ∈ tree0, ∀ tree0, tree1. tree0[parent(p0)].dir? ∧ p0 ∈ tree0 → tree0[p0, tree1].fids = (tree0.fids ∪ tree1.fids) \ tree0[p0].fids ∨ p0 = ε ⊦ str + p0 = ε, tree0[str + p0, tree1].fids = (tree0.fids ∪ tree1.fids) \ tree0[str + p0].fids

Rule argument

indvar: p

Additional information

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