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Proof step 32 from proof recovery (previous step)

tree0.dir?, dom of1 < ino, abs(tree0, ROOT_INO)(dirs0), afs-cons(dirs0, files1 \ orphans(dirs0, files1)), afs-cons(dirs0, files1), fids-cons(tree0, abs-files(files1), abs-of(of1)), of-cons(∅, dirs0, files1 \ orphans(dirs0, files1)), of-cons(of1, dirs0, files1), posix-crash(tree0, abs-files(files1), abs-of(of1), tree, fs, ofh), ∀ ino. ¬ ino ∈ files1 ∨ ¬ ino ∈ dirs0, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files1 ⊦ ofh = ∅ ∧ fs = abs-files(files1 \ orphans(dirs0, files1)) ∧ abs(tree, ROOT_INO)(dirs0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
tree0.dir?, dom of1 < ino, abs(tree0, ROOT_INO)(dirs0), afs-cons(dirs0, files1), afs-cons(dirs0, files1 \ orphans(dirs0, files1)), fids-cons(tree0, abs-files(files1), abs-of(of1)), of-cons(of1, dirs0, files1), of-cons(∅, dirs0, files1 \ orphans(dirs0, files1)), posix-crash(tree0, abs-files(files1), abs-of(of1), tree, fs, ofh), ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files1 ⊦ ofh = ∅ ∧ fs = abs-files(files1 \ orphans(dirs0, files1)) ∧ abs(tree, ROOT_INO)(dirs0)

Rule argument

Additional information

Used simplifier rules: ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs → ¬ ino ∈ files; (s)