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Proof step 3 from proof subdirs-put (previous step)

dirs[p_ino] = mkdir(m, md, n0, n1, entries), p_ino ∈ dirs ⊦ (ino ∈ dirs ⊃ subdirs(p_ino, dirs) ++ str;(str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs))) = (subdirs(p_ino, dirs) \ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ ∅ ++ str;∅)) ∪ (ino ∈ dirs ⊃ ∅ ++ 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:
dirs[p_ino] = mkdir(m, md, n0, n1, entries), p_ino ∈ dirs ⊦ ∀ str0. str0 ∈ (ino ∈ dirs ⊃ subdirs(p_ino, dirs) ++ str;(str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str;subdirs(p_ino, dirs))) ↔ str0 ∈ (subdirs(p_ino, dirs) \ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ ∅ ++ str;∅)) ∪ (ino ∈ dirs ⊃ ∅ ++ str;∅)

Rule argument

spec: set-basic, inst: dir-entries name: Extension seq: ⊦ strings = strings0 ↔ (∀ str. str ∈ strings ↔ str ∈ strings0) subst: [strings0, strings] → [ ( subdirs(p_ino, dirs) \ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ ∅ ++ str;∅)) ∪ (ino ∈ dirs ⊃ ∅ ++ str;∅), (ino ∈ dirs ⊃ subdirs(p_ino, dirs) ++ str ;(str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str ;subdirs(p_ino, dirs) ) )] flags: <all paths>

Additional information

Used terms: ( subdirs(p_ino, dirs) \ (str ∈ entries ∧ entries[str] ∈ dirs ⊃ ∅ ++ str;∅)) ∪ (ino ∈ dirs ⊃ ∅ ++ str;∅), (ino ∈ dirs ⊃ subdirs(p_ino, dirs) ++ str ;(str ∈ entries ∧ entries[str] ∈ dirs ⊃ subdirs(p_ino, dirs) -- str ;subdirs(p_ino, dirs) ) )