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Proof step 4 from proof abs-path-h (previous step)

((abs(tree, ino) * D))(dirs), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ tree, ino, D, dirs, files. ((abs(tree, ino) * D))(dirs) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (ispath(p0, ino, dirs, files) ↔ p0 ∈ tree) ⊦ ispath(str + p0, ino, dirs, files) ↔ str + p0 ∈ tree

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
((abs(tree, ino) * D))(dirs), ∀ tree, ino, D, dirs, files. ((abs(tree, ino) * D))(dirs) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (ispath(p0, ino, dirs, files) ↔ p0 ∈ tree), ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ ino ∈ dirs ∧ str ∈ dirs[ino].entries ∧ ispath(p0, dirs[ino].entries[str], dirs, files) ↔ str ' ∈ tree ∧ p0 ∈ tree[str ']

Rule argument

Additional information

Used simplifier rules: ⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p]; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ valid-dir-ino(ino, dirs, files) ↔ ino ∈ dirs; (s)%⊦ ispath(str ' + q, n, dirs, files) ↔ valid-dir-ino(n, dirs, files) ∧ str ∈ dirs[n].entries ∧ ispath(q, dirs[n].entries[str], dirs, files); (s)% ⊦ str + p = str ' + p; (s)