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Proof step 80 from proof abs-paths-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 81:
⊦ p ∈ tree ∧ ((abs|(tree, ino, p, ino0) * D))(dirs) ∧ ispath(p, ino, ino1, dirs, files) → ino0 = ino1
Proof step 82:
((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

Rule argument

Additional information

Used simplifier rules: absc-path: ⊦ p ∈ tree ∧ ((abs|(tree, ino, p, ino0) * D))(dirs) ∧ ispath(p, ino, ino1, dirs, files) → ino0 = ino1;%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap]%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]% ⊦ (_∃ nD)(dirs) ↔ (∃ ino. nD(ino)(dirs)); (s)