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Proof step 2 from proof abs-tree (previous step)

((abs(tree1, ino) * D1))(dirs), ((abs(tree0, ino) * D0))(dirs), ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ tree0 = tree1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ tree0 = tree1 ↔ (∀ p. (p ∈ tree0 ↔ p ∈ tree1) ∧ (p ∈ tree0 → (tree0[p].dir? ↔ tree1[p].dir?) ∧ (tree0[p].dir? ⊃ tree0[p].meta = tree1[p].meta;tree0[p].fid = tree1[p].fid)))
Proof step 4:
((abs(tree0, ino) * D0))(dirs), ((abs(tree1, ino) * D1))(dirs), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files ⊦ ∀ p. (p ∈ tree0 ↔ p ∈ tree1) ∧ (p ∈ tree0 → (tree0[p].dir? ↔ tree1[p].dir?) ∧ (tree0[p].dir? ⊃ tree0[p].meta = tree1[p].meta;tree0[p].fid = tree1[p].fid))

Rule argument

spec: directory-tree name: tree-ext seq: ⊦ tree0 = tree1 ↔ (∀ p. (p ∈ tree0 ↔ p ∈ tree1) ∧ ( p ∈ tree0 → (tree0[p].dir? ↔ tree1[p].dir?) ∧ (tree0[p].dir? ⊃ tree0[p].meta = tree1[p].meta ;tree0[p].fid = tree1[p].fid ))) subst: [tree1, tree0] → [tree1, tree0] flags: <all paths>

Additional information

Used terms: tree1, tree0