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Proof step 41 from proof abs-unshared-h (previous step)

tree1[p1] = fnode(fid), (( ( ((fid.ino => mkdir(# entries, tree0[p0].meta, m0, m1, entries) * abs(tree0[p0].entries, entries)) * abs|(tree1, ino1, p1, fid.ino)) * abs|(tree0, ino0, p0, fid.ino)) * D) )(dirs), tree0[p0].dir?, ispath(p0, ino0, fid.ino, dirs, files), ispath(p1, ino1, fid.ino, dirs, files), p0 ∈ tree0, p1 ∈ tree1, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 42:
tree1[p1] = fnode(fid), fid = ⌜m⌝ , (( ( ((m => mkdir(# entries, tree0[p0].meta, m0, m1, entries) * abs(tree0[p0].entries, entries)) * abs|(tree1, ino1, p1, m)) * abs|(tree0, ino0, p0, m)) * D) )(dirs), tree0[p0].dir?, ispath(p1, ino1, m, dirs, files), ispath(p0, ino0, m, dirs, files), p1 ∈ tree1, p0 ∈ tree0, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦

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