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Proof step 19 from proof abs-fold-new-p (previous step)

⊦ p ≠ ε ∧ q ⊑ p ∧ parent(p) ∈ tree ∧ tree[parent(p)].dir? → abs|(tree[p, tree0], ino0, q, ino1) = abs|(tree, ino0, q, ino1)

A spec-lemma