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Proof step 8 from proof afs-cons-in-next (previous step)

dirs[ino] = mkdir(# entries, md, # links(ino, dirs), # subdirs(ino, dirs), entries), ino ∈ dirs, str ∈ entries, ROOT_INO ∈ dirs, # links(ino, dirs) ≤ 1, ∀ name. name ∈ entries → entries[name] ∈ dirs ∨ entries[name] ∈ files, ∀ ino0. (ino0 ∈ dirs ∧ ino0 ≠ ROOT_INO ∧ links(ino0, dirs) = ∅ → dirs[ino0].entries = ∅) ∧ ( ino0 ∈ dirs → ¬ ino0 ∈ files ∧ dirs[ino0].nlink = # links(ino0, dirs) ∧ dirs[ino0].nlink ≤ 1 ∧ dirs[ino0].size = # dirs[ino0].entries ∧ dirs[ino0].nsubdirs = # subdirs(ino0, dirs)) ∧ (ino0 ∈ files → ¬ ino0 ∈ dirs ∧ files[ino0].nlink = # links(ino0, dirs)) ∧ (∀ name. ino0 ∈ dirs ∧ name ∈ dirs[ino0].entries → dirs[ino0].entries[name] ∈ dirs ∨ dirs[ino0].entries[name] ∈ files) ∧ (∀ n. ino0 ∈ files ∧ n ∈ files[ino0].content → # files[ino0].content[n] = VFS_PAGE_SIZE) ⊦ entries = ∅, entries[str] ∈ dirs, entries[str] ∈ files, ino ∈ files, ino ≠ ROOT_INO ∧ links(ino, dirs) = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: str

Additional information

Used terms: str