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Proof step 82 from proof abs-paths-eq (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 83:
p = str + p0, ((abs|(tree, ino0, p3, m) * abs(tree[p3], m)))(dirs), tree[p3].dir?, tree[p3 + p].dir?, ispath(p, m, m, dirs, files), ispath(p3, ino0, m, dirs, files), p3 ∈ tree, ino0 ∈ dirs, p ∈ tree[p3], m ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ ε = p

Rule argument

var: p

Additional information

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