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Proof step 35 from proof recovery (previous step)

ofh = ∅, tree = tree0, 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), ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files1, ∀ fid. fid ∈ fs ↔ fid ∈ tree.fids ∧ fid ∈ abs-files(files1), ∀ fid. fid ∈ fs → fs[fid] = abs-files(files1)[fid] ⊦ 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 36:
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)), ∀ fid. fid ∈ fs → fs[fid] = abs-files(files1)[fid], ∀ fid. fid ∈ fs ↔ fid ∈ tree0.fids ∧ fid ∈ abs-files(files1), ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files1 ⊦ fs = abs-files(files1 \ orphans(dirs0, files1))

Proof step 35 from proof recovery (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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