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Proof step 8 from proof abs-paths-eq (previous step)

ispath(p1, ino0, dirs, files) ↔ p1 ∈ tree, abs(tree, ino0)(dirs), ispath(p1, ino0, ino1, dirs, files), ispath(p0, ino0, ino1, dirs, files), p0 ∈ tree, ino1 ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ p0 = p1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
abs(tree, ino0)(dirs), ispath(p0, ino0, ino1, dirs, files), ispath(p1, ino0, ino1, dirs, files), ino1 ∈ dirs, p0 ∈ tree, p1 ∈ tree, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ p0 = p1

Rule argument

Additional information

Used simplifier rules: ⊦ ispath(p, n, m, dirs, files) → ispath(p, n, dirs, files); (s)