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Proof step 47 from proof recovery (previous step)

⌜ino0⌝ ∈ fs ↔ links(ino0, dirs0) ≠ ∅ ∧ ino0 ∈ files1, 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)), ⌜ino0⌝ ∈ fs → fs[⌜ino0⌝ ] = abs-files(files1)[⌜ino0⌝ ], ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files1 ⊦ (⌜ino0⌝ ∈ fs ↔ ino0 ∈ files1 ∧ (links(ino0, dirs0) ≠ ∅ ∨ ino0 = ROOT_INO)) ∧ (⌜ino0⌝ ∈ fs → fs[⌜ino0⌝ ] = abs-files(files1 \ orphans(dirs0, files1))[⌜ino0⌝ ])

User applied rule case distinction

And obtained the following new goals

Proof step 48:
⌜ino0⌝ ∈ fs ↔ links(ino0, dirs0) ≠ ∅ ∧ ino0 ∈ files1, 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), ⌜ino0⌝ ∈ fs, ⌜ino0⌝ ∈ fs → fs[⌜ino0⌝ ] = abs-files(files1)[⌜ino0⌝ ], ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files1 ⊦ fs[⌜ino0⌝ ] = abs-files(files1 \ orphans(dirs0, files1))[⌜ino0⌝ ]

Rule argument

pos: right 1

Additional information

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