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Proof step 21 from proof abs-tree (previous step)

((abs(tree1, fid.ino) * D1))(dirs), D0(dirs), ∀ ino0. ⌜ino0⌝ = fid → ino0 ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files ⊦ ¬ tree1.dir? ∧ fid = tree1.fid

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
fid = ⌜m⌝ , ((abs(tree1, m) * D1))(dirs), D0(dirs), ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino0. ⌜ino0⌝ = fid → ino0 ∈ files ⊦ ¬ tree1.dir? ∧ fid = tree1.fid

Rule argument

spec: file-id-data name: elim seq: ⊦ m = fid.ino ↔ fid = ⌜m⌝ subst: [fid, m] → [fid, m]

Additional information

Used simplifier rules: ⊦ m = fid.ino ↔ fid = ⌜m⌝ ; (elim) Used terms: fid, m