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Proof step 24 from proof abs-nlinks-h (previous step)

dirs[m0] = mkdir(m, md, m1, m2, entries), tree[p0 + str '] = fnode(⌜ino1⌝ ), links(ino1, dirs) = ∅, (_∃(λ ino1. abs|(tree, ino0, p0, ino1) * abs(tree[p0], ino1)))(dirs), ispath(p0, ino0, m0, dirs, files), str ∈ entries, m0 ∈ dirs, ino0 ∈ dirs, str ' ∈ tree[p0], ino1 ∈ files, p0 ∈ tree, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, entries[str] ∈ dirs ∨ entries[str] ∈ files ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
⊦ p ∈ tree ∧ ((abs|(tree, ino, p, ino0) * D))(dirs) ∧ ispath(p, ino, ino1, dirs, files) → ino0 = ino1
Proof step 26:
dirs[m0] = mkdir(m, md, m1, m2, entries), tree[p0 + str '] = fnode(⌜ino1⌝ ), links(ino1, dirs) = ∅, ((abs|(tree, ino0, p0, m0) * abs(tree[p0], m0)))(dirs), ispath(p0, ino0, m0, dirs, files), str ∈ entries, m0 ∈ dirs, ino0 ∈ dirs, str ' ∈ tree[p0], ino1 ∈ files, p0 ∈ tree, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, entries[str] ∈ dirs ∨ entries[str] ∈ 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)(dirs) ↔ (∃ ino. nD(ino)(dirs)); (s)