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Proof step 18 from proof fids-strpath (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 19:
fid ∈ sdt0.fids ↔ (∃ str, p. str ∈ sdt0 ∧ p ∈ sdt0[str] ∧ sdt0[str][p] = fnode(fid)), ∀ sdt1, fid. # sdt1 < # sdt0[str, tree] → true → (fid ∈ sdt1.fids ↔ (∃ str, p. str ∈ sdt1 ∧ p ∈ sdt1[str] ∧ sdt1[str][p] = fnode(fid))) ⊦ fid ∈ sdt0.fids ↔ (∃ str0, p. str0 ∈ sdt0[str, tree] ∧ p ∈ sdt0[str, tree][str0] ∧ sdt0[str, tree][str0][p] = fnode(fid)), str ∈ sdt0, fid ∈ tree.fids

Rule argument

subst: [sdt1, fid] → [sdt0, fid]

Additional information

Used simplifier rules: ⊦ ¬ str ∈ sdt → # sdt[str, tree] = # sdt + 1; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic[directory-tree-inst]%c: ⊦ m + n = n + m; from specification nat-basic[directory-tree-inst] Used terms: sdt0, fid