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Proof step 42 from proof recovery (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
⊦ n ∈ files ∧ abs(tree, ROOT_INO)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (⌜n⌝ ∈ tree.fids ↔ links(n, dirs) ≠ ∅)
Proof step 44:
⌜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 ∧ ¬ ino0 ∈ orphans(dirs0, files1)) ∧ (⌜ino0⌝ ∈ fs → fs[⌜ino0⌝ ] = abs-files(files1 \ orphans(dirs0, files1))[⌜ino0⌝ ])

Rule argument

Additional information

Used simplifier rules: file-fid: ⊦ n ∈ files ∧ abs(tree, ROOT_INO)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (⌜n⌝ ∈ tree.fids ↔ links(n, dirs) ≠ ∅);% ⊦ ino ∈ files \ ns ↔ ino ∈ files ∧ ¬ ino ∈ ns; (s)% ⊦ ⌜ino⌝ ∈ abs-files(files) ↔ ino ∈ files; (s)