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Proof step 13 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 ⊦ dnode(md, sdt1[str, tree])[str ' + p0, dnode(md1, sdt)].fids = sdt1.fids ∪ tree.fids, str ∈ sdt1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: directory-tree name: put-rec seq: p ≠ ε ⊦ str ∈ sdt → dnode(md, sdt)[str ' + p, tree0] = dnode(md, sdt[str, sdt[str][p, tree0]]) subst: [tree0, p, str, sdt, md] → [dnode(md1, sdt), p0, str, sdt1[str, tree], md] flags: <all paths>

Additional information

Used simplifier rules: fids-dnode: ⊦ dnode(md, sdt).fids = sdt.fids;%fids-ins: ⊦ ¬ str ∈ sdt → sdt[str, tree].fids = sdt.fids ∪ tree.fids;% ⊦ str ∈ sdt[str, tree]; (s)% ⊦ tree[ε] = tree; (s)% ⊦ p + ε = p; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ dnode(md, sdt)[str ', tree0] = dnode(md, sdt[str, tree0]); (s)% ⊦ sdt[str, tree0][str, tree] = sdt[str, tree]; (s)% ⊦ fidb ∪ fidb0 = fidb ∪ fidb1 ↔ fidb0 = fidb1; (s)%union-ass: ⊦ fidb ∪ fidb0 ∪ fidb1 = (fidb ∪ fidb0) ∪ fidb1; from specification bag-union[fidbag]%union-com: ⊦ fidb ∪ fidb0 = fidb0 ∪ fidb; from specification bag-union[fidbag]% ⊦ ε ∈ tree; (s)% ⊦ tree[ε, tree0] = tree0; (s) Used terms: dnode(md1, sdt), p0, str, sdt1[str, tree], md