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Proof step 23 from proof abs-tree (previous step)

((abs(tree1, m) * D1))(dirs), D0(dirs), m ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files ⊦ ¬ tree1.dir? ∧ ⌜m⌝ = tree1.fid

User applied rule constructor cut

And obtained the following new goals

Proof step 24:
tree1 = dnode(md, sdt), ((abs(tree1, m) * D1))(dirs), D0(dirs), m ∈ files, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ ¬ tree1.dir? ∧ ⌜m⌝ = tree1.fid
Proof step 29:
tree1 = fnode(fid), ((abs(tree1, m) * D1))(dirs), D0(dirs), m ∈ files, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ ¬ tree1.dir? ∧ ⌜m⌝ = tree1.fid

Rule argument

var: tree1

Additional information

Used simplifier rules: directory-tree, entries-tree generated by ∅, dnode, fnode, [ ];;