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Proof step 30 from proof fids-in (previous step)

tree[p] = fnode(fid), fid ∈ tree.fids, p ∈ tree, ∀ fid0. fid0 ∈ tree.fids → (∃ p. p ∈ tree ∧ tree[p] = fnode(fid0)), ∀ sdt1, fid, md. # sdt1 < # sdt0[str, tree] → fid ∈ sdt1.fids ∧ (∀ str. str ∈ sdt1 → (∀ fid. fid ∈ sdt1[str].fids → (∃ p. p ∈ sdt1[str] ∧ sdt1[str][p] = fnode(fid)))) → (∃ p. p ∈ dnode(md, sdt1) ∧ dnode(md, sdt1)[p] = fnode(fid)) ⊦ str ∈ sdt0, ∃ p. p ∈ dnode(md, sdt0[str, tree]) ∧ dnode(md, sdt0[str, tree])[p] = fnode(fid)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: str + p abort: F discard: F computed: F

Additional information

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