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Proof step 21 from proof fids-strpath (previous step)

sdt0[str0][p] = fnode(fid), fid ∈ sdt0.fids, str0 ∈ sdt0, p ∈ sdt0[str0], ∀ sdt1, fid. # sdt1 < # sdt0[str, tree] → true → (fid ∈ sdt1.fids ↔ (∃ str, p. str ∈ sdt1 ∧ p ∈ sdt1[str] ∧ sdt1[str][p] = fnode(fid))) ⊦ str ∈ sdt0, fid ∈ tree.fids, ∃ str0, p. str0 ∈ sdt0[str, tree] ∧ p ∈ sdt0[str, tree][str0] ∧ sdt0[str, tree][str0][p] = fnode(fid)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: str0, p abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ str0 ∈ sdt ∧ ¬ str ∈ sdt → sdt[str, tree][str0] = sdt[str0]; (s)% ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws) Used terms: str0, p