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Proof step 32 from proof abs-acyclic (previous step)

tree[p] = dnode(md0, sdt), entries[str] = ROOT_INO, (( ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], ROOT_INO)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ sdt, str ∈ entries ⊦ sdt[str].dir?, (⌜ino0⌝ ∈ tree.fids → ino0 ∈ files) ∧ ispath(p + str ', ROOT_INO, ROOT_INO, dirs, files)

User applied rule case distinction

And obtained the following new goals

Proof step 33:
⊦ ⌜n⌝ ∈ tree.fids ∧ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) → n ∈ files
Proof step 34:
entries[str] = ROOT_INO, tree[p] = dnode(md0, sdt), (( ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], ROOT_INO)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ entries, str ∈ sdt ⊦ sdt[str].dir?, ispath(p + str ', ROOT_INO, ROOT_INO, dirs, files)

Rule argument

pos: right 2

Additional information

Used simplifier rules: fid-file: ⊦ ⌜n⌝ ∈ tree.fids ∧ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) → n ∈ files;