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Proof step 77 from proof abs-paths-eq (previous step)

tree[p3].dir?, tree[p3 + p].dir?, abs(tree, ino0)(dirs), ispath(p3, ino0, ino1, dirs, files), ino0 ∈ dirs, p ∈ tree[p3], p3 ∈ tree, ino1 ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∃ m0. ispath(p3, ino0, m0, dirs, files) ∧ ispath(p, m0, ino1, dirs, files) ⊦ ε = p

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 78:
tree[p3].dir?, tree[p3 + p].dir?, abs(tree, ino0)(dirs), ispath(p, m, m, dirs, files), ispath(p3, ino0, m, dirs, files), ino0 ∈ dirs, p ∈ tree[p3], p3 ∈ tree, m ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ ε = p

Rule argument

Additional information

Used simplifier rules: path-in-dirs: p ≠ ε ⊦ ispath(p, n, m, dirs, files) → n ∈ dirs; (forward) from specification afs-base%path-inj: ⊦ ispath(p, n, m0, dirs, files) ∧ ispath(p, n, m1, dirs, files) → m0 = m1; (forward) from specification afs-base