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Proof step 6 from proof afs-cons-abs-dir-subdirs (previous step)

fs[ri[inodekey(ino)]].nsubdirs = 0, fs[ri[inodekey(ino)]].directory, fs[ri[inodekey(ROOT_INO)]].directory, inodekey(ino) ∈ ri, inodekey(ROOT_INO) ∈ ri, ∀ ino0. (inodekey(ino0) ∈ ri ∧ fs[ri[inodekey(ino0)]].directory ∧ ino0 ≠ ROOT_INO ∧ links(ino0, abs-dirs(fs, ri)) = ∅ → abs-children(ino0, fs, ri) = ∅) ∧ ( inodekey(ino0) ∈ ri ∧ fs[ri[inodekey(ino0)]].directory → fs[ri[inodekey(ino0)]].nlink = # links(ino0, abs-dirs(fs, ri)) ∧ fs[ri[inodekey(ino0)]].nlink ≤ 1 ∧ fs[ri[inodekey(ino0)]].size = # abs-children(ino0, fs, ri) ∧ fs[ri[inodekey(ino0)]].nsubdirs = # subdirs(ino0, abs-dirs(fs, ri))) ∧ (inodekey(ino0) ∈ ri ∧ ¬ fs[ri[inodekey(ino0)]].directory → fs[ri[inodekey(ino0)]].nlink = # links(ino0, abs-dirs(fs, ri))) ∧ (∀ name. inodekey(ino0) ∈ ri ∧ fs[ri[inodekey(ino0)]].directory ∧ dentrykey(ino0, name) ∈ ri → inodekey(fs[ri[dentrykey(ino0, name)]].ino) ∈ ri ∧ fs[ri[inodekey(fs[ri[dentrykey(ino0, name)]].ino)]].directory ∨ inodekey(fs[ri[dentrykey(ino0, name)]].ino) ∈ ri ∧ ¬ fs[ri[inodekey(fs[ri[dentrykey(ino0, name)]].ino)]].directory) ∧ (∀ n. inodekey(ino0) ∈ ri ∧ ¬ fs[ri[inodekey(ino0)]].directory ∧ datakey(ino0, n) ∈ ri → # fs[ri[datakey(ino0, n)]].data = VFS_PAGE_SIZE) ⊦ # subdirs(ino, abs-dirs(fs, ri)) = 0, subdirs(ino, abs-dirs(fs, ri)) = ∅

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: ino abort: F discard: F computed: T

Additional information

Used simplifier rules: size-zero-02: ⊦ # strings = 0 → strings = ∅; (forward) from specification set-basic[stringset] Used terms: ino