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Proof step 13 from proof abs-meta-h (previous step)

tree[p].dir? ↔ m ∈ dirs, ((_∃(λ ino1. abs|(tree, ino, p, ino1) * abs(tree[p], ino1)) * D))(dirs), ispath(p, ino, m, dirs, files), p ∈ tree, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ meta(p, tree, abs-files(files)) = meta(ε, m, dirs, files)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
⊦ p ∈ tree ∧ ((abs|(tree, ino, p, ino0) * D))(dirs) ∧ ispath(p, ino, ino1, dirs, files) → ino0 = ino1
Proof step 15:
tree[p].dir? ↔ m ∈ dirs, (((abs|(tree, ino, p, m) * abs(tree[p], m)) * D))(dirs), ispath(p, ino, m, dirs, files), p ∈ tree, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ meta(p, tree, abs-files(files)) = meta(ε, m, dirs, files)

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 * D))(dirs) ↔ (∃ ino. ((nD(ino) * D))(dirs)); (s)