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Proof step 5 from proof absc-path (previous step)

((abs|(tree, ino, str + p0, ino0) * D))(dirs), ispath(str + p0, ino, ino1, dirs, files), str + p0 ∈ tree, ∀ tree, ino, ino0, D, dirs, ino1, files. ((abs|(tree, ino, p0, ino0) * D))(dirs) ∧ ispath(p0, ino, ino1, dirs, files) ∧ p0 ∈ tree → ino0 = ino1 ⊦ ino0 = ino1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
((abs|(tree, ino, str ' + p0, ino0) * D))(dirs), ispath(p0, dirs[ino].entries[str], ino1, dirs, files), str ' ∈ tree, p0 ∈ tree[str '], ino ∈ dirs, str ∈ dirs[ino].entries, ∀ tree, ino, ino0, D, dirs, ino1, files. ((abs|(tree, ino, p0, ino0) * D))(dirs) ∧ ispath(p0, ino, ino1, dirs, files) ∧ p0 ∈ tree → ino0 = ino1 ⊦ ino0 = ino1

Rule argument

Additional information

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