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Proof step 2 from proof afs-cons-abs-dir-size (previous step)

afs-cons(abs-dirs(fs, ri), abs-files(fs, ri)), ino ∈ abs-dirs(fs, ri) ⊦ abs-dirs(fs, ri)[ino].entries = ∅, fs[ri[inodekey(ino)]].size ≥ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
fs[ri[inodekey(ino)]].size = 0, fs[ri[inodekey(ino)]].directory, afs-cons(abs-dirs(fs, ri), abs-files(fs, ri)), inodekey(ino) ∈ ri, dentrykey(ino, str) ∈ ri ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ n ≥ m ↔ m ≤ n; (s)% ⊦ ¬ str ∈ ∅; (s)%⊦ entries = abs-children(ino, fs, ri) ↔ (∀ str. (str ∈ entries ↔ dentrykey(ino, str) ∈ ri) ∧ ( str ∈ entries → entries[str] = fs[ri[dentrykey(ino, str)]].ino)); (ss)% ⊦ mkdir(m, md, m0, m1, entries).entries = entries; (s)%⊦ abs-dir(ino, fs, ri) = mkdir (fs[ri[inodekey(ino)]].size, fs[ri[inodekey(ino)]].meta, fs[ri[inodekey(ino)]].nlink, fs[ri[inodekey(ino)]].nsubdirs, abs-children(ino, fs, ri)); (s)%⊦ inodekey(ino) ∈ ri ∧ fs[ri[inodekey(ino)]].directory → abs-dirs(fs, ri)[ino] = abs-dir(ino, fs, ri); (s)%⊦ ino ∈ abs-dirs(fs, ri) ↔ inodekey(ino) ∈ ri ∧ fs[ri[inodekey(ino)]].directory; (s)