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Proof step 29 from proof path-unfold (previous step)

(((abs|(tree, ino0, str + p0, ino) * abs(tree[str + p0], ino)) * D))(dirs), str + p0 ∈ tree, ∀ tree, ino0, ino, D, dirs. (((abs|(tree, ino0, p0, ino) * abs(tree[p0], ino)) * D))(dirs) ∧ p0 ∈ tree → ((abs(tree, ino0) * D))(dirs) ⊦ ((abs(tree, ino0) * D))(dirs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
((abs|(tree, ino0, str ' + p0, ino) * abs(tree[str ' + p0], ino) * D))(dirs), str ' ∈ tree, p0 ∈ tree[str '], ∀ tree, ino0, ino, D, dirs. (((abs|(tree, ino0, p0, ino) * abs(tree[p0], ino)) * D))(dirs) ∧ p0 ∈ tree → ((abs(tree, ino0) * D))(dirs) ⊦ ((abs(tree, ino0) * D))(dirs)

Rule argument

Additional information

Used simplifier rules: 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)