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Proof step 78 from proof abs-paths-eq (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 79:
⊦ p ∈ tree → abs(tree, ino0) = _∃(λ ino1. abs|(tree, ino0, p, ino1) * abs(tree[p], ino1))
Proof step 80:
tree[p3 + p].dir?, tree[p3].dir?, (_∃(λ ino1. abs|(tree, ino0, p3, ino1) * abs(tree[p3], ino1)))(dirs), 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

Rule argument

spec: <toplevel> name: path-unfold seq: ⊦ p ∈ tree → abs(tree, ino0) = _∃(λ ino1. abs|(tree, ino0, p, ino1) * abs(tree[p], ino1)) subst: [p, ino0, tree] → [p3, ino0, tree] flags: <all paths>

Additional information

Used terms: p3, ino0, tree