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Proof step 3 from proof absc-drop (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ D = D0 ↔ (∀ D1, dirs. ((D * D1))(dirs) ↔ ((D0 * D1))(dirs))
Proof step 5:
tree[parent(p)].dir?, parent(p) ∈ tree, q ⊑ p ⊦ p = ε, ∀ D1, dirs. ((abs|(tree[p, tree0], ino0, q, ino1) * D1))(dirs) ↔ ((abs|(tree, ino0, q, ino1) * D1))(dirs)

Rule argument

spec: heap-sep, inst: dir-heap name: Eq-P seq: ⊦ D = D0 ↔ (∀ D1, dirs. ((D * D1))(dirs) ↔ ((D0 * D1))(dirs)) subst: [D0, D] → [abs|(tree, ino0, q, ino1), abs|(tree[p, tree0], ino0, q, ino1)] flags: <all paths>

Additional information

Used terms: abs|(tree, ino0, q, ino1), abs|(tree[p, tree0], ino0, q, ino1)