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Proof step 2 from proof abs-dir-h (previous step)

((abs(tree, ino) * D))(dirs), ispath(p, ino, ino0, dirs, files), ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ tree[p].dir? ↔ ino0 ∈ dirs

User applied rule structural induction

And obtained the following new goals

Proof step 3:
((abs(tree, ino) * D))(dirs), ispath(ε, ino, ino0, dirs, files), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ tree[ε].dir? ↔ ino0 ∈ dirs
Proof step 11:
((abs(tree, ino) * D))(dirs), ispath(str + p0, ino, ino0, dirs, files), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ tree, ino, D, dirs, ino0, files. ((abs(tree, ino) * D))(dirs) ∧ ispath(p0, ino, ino0, dirs, files) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (tree[p0].dir? ↔ ino0 ∈ dirs) ⊦ tree[str + p0].dir? ↔ ino0 ∈ dirs

Rule argument

indvar: p

Additional information

Used simplifier rules: path freely generated by ε, +;;