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Proof step 5 from proof fids-put3 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
tree[str ' + p0] = dnode(md0, sdt), str ' ∈ tree, p0 ∈ tree[str '], ∀ tree, md0, sdt, md1. tree[p0] = dnode(md0, sdt) ∧ p0 ∈ tree → tree[p0, dnode(md1, sdt)].fids = tree.fids ⊦ tree[str ' + p0, dnode(md1, sdt)].fids = tree.fids

Rule argument

Additional information

Used simplifier rules: a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p]; (s)% ⊦ str + p = str ' + p; (s)