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Proof step 20 from proof fids-put3 (previous step)

tree[p0] = dnode(md0, sdt), p0 ∈ tree, ∀ tree, md0, sdt, md1. tree[p0] = dnode(md0, sdt) ∧ p0 ∈ tree → tree[p0, dnode(md1, sdt)].fids = tree.fids ⊦ tree[p0, dnode(md1, sdt)].fids = tree.fids, p0 = ε, str ∈ sdt1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: tree, md0, sdt, md1

Additional information

Used terms: tree, md0, sdt, md1