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Proof step 46 from proof abs-nlinks-h (previous step)

abs(tree, ino0)(dirs), ino1 ∈ files, ino0 ∈ dirs, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ # links(ino1, dirs) = 0, links(ino1, dirs) = ∅, ⌜ino1⌝ ∈ tree.fids

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 47:
⊦ ls = ls0 ↔ (∀ alink. alink ∈ ls ↔ alink ∈ ls0)
Proof step 48:
abs(tree, ino0)(dirs), ino0 ∈ dirs, ino1 ∈ files, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ # links(ino1, dirs) = 0, ⌜ino1⌝ ∈ tree.fids, ∀ alink. alink ∈ links(ino1, dirs) ↔ alink ∈ ∅

Rule argument

spec: set-basic, inst: alink-set name: Extension seq: ⊦ ls = ls0 ↔ (∀ alink. alink ∈ ls ↔ alink ∈ ls0) subst: [ls0, ls] → [∅, links(ino1, dirs)] flags: <all paths>

Additional information

Used terms: ∅, links(ino1, dirs)