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Proof step 3 from proof subdirs-add-entry (previous step)

ino ∈ dirs, ino0 ∈ dirs ⊦ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries[str, ino0])]) = subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]) ++ str

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ strings = strings0 ↔ (∀ str. str ∈ strings ↔ str ∈ strings0)
Proof step 5:
ino0 ∈ dirs, ino ∈ dirs ⊦ ∀ str0. str0 ∈ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries[str, ino0])]) ↔ str0 ∈ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]) ++ str

Rule argument

spec: set-basic, inst: dir-entries name: Extension seq: ⊦ strings = strings0 ↔ (∀ str. str ∈ strings ↔ str ∈ strings0) subst: [strings0, strings] → [ subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]) ++ str, subdirs (ino, dirs[ino, mkdir(n, md, n0, n1, entries[str, ino0])])] flags: <all paths>

Additional information

Used terms: subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries)]) ++ str, subdirs(ino, dirs[ino, mkdir(n, md, n0, n1, entries[str, ino0])])