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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
fs[ri[inodekey(ino)]].nsubdirs = 0, fs[ri[inodekey(ino)]].directory, afs-cons(abs-dirs(fs, ri), abs-files(fs, ri)), inodekey(ino) ∈ ri ⊦ # subdirs(ino, abs-dirs(fs, ri)) = 0, subdirs(ino, abs-dirs(fs, ri)) = ∅

Rule argument

Additional information

Used simplifier rules: size-zero: ⊦ # strings ≠ 0 → strings ≠ ∅; (forward) from specification set-basic[stringset]%size-zero-01: ⊦ strings ≠ ∅ → # strings ≠ 0; (forward) from specification set-basic[stringset]% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ n ≥ m ↔ m ≤ n; (s)%⊦ ino ∈ abs-dirs(fs, ri) ↔ inodekey(ino) ∈ ri ∧ fs[ri[inodekey(ino)]].directory; (s)