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Proof step 9 from proof afs-cons-init (previous step)

⊦ 0 = # subdirs(ROOT_INO, (∅ ++ ROOT_INO)[ROOT_INO, mkdir(0, md, 0, 0, ∅)]), subdirs(ROOT_INO, (∅ ++ ROOT_INO)[ROOT_INO, mkdir(0, md, 0, 0, ∅)]) = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ strings = strings0 ↔ (∀ str. str ∈ strings ↔ str ∈ strings0)

Rule argument

spec: set-basic, inst: dir-entries name: Extension seq: ⊦ strings = strings0 ↔ (∀ str. str ∈ strings ↔ str ∈ strings0) subst: [strings0, strings] → [∅, subdirs (ROOT_INO, (∅ ++ ROOT_INO)[ROOT_INO, mkdir(0, md, 0, 0, ∅)])] flags: <all paths>

Additional information

Used simplifier rules: subdirs: ⊦ str ∈ subdirs(ino, dirs) ↔ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] ∈ dirs;% ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)% ⊦ ROOT_INO ≠ 0; (s)% ⊦ ¬ ino ∈ ∅; (s)%ino ≠ 0, ¬ ino ∈ dirs ⊦ ino0 ∈ dirs ++ ino ↔ ino0 = ino ∨ ino0 ∈ dirs; (s)% ⊦ dirs[ino, ad][ino0] = (ino0 = ino ∧ ino ∈ dirs ⊃ ad;dirs[ino0]); (s)% ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s)% ⊦ ¬ str ∈ ∅; (s)% ⊦ ¬ str ∈ ∅; (s) Used terms: ∅, subdirs(ROOT_INO, (∅ ++ ROOT_INO)[ROOT_INO, mkdir(0, md, 0, 0, ∅)])