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Proof step 8 from proof absc-drop (previous step)

tree[parent(p)].dir?, parent(p) ∈ tree, str + q0 ⊑ p, ∀ tree, p, tree0, ino0, ino1, D, dirs. tree[parent(p)].dir? ∧ parent(p) ∈ tree ∧ q0 ⊑ p → (((abs|(tree[p, tree0], ino0, q0, ino1) * D))(dirs) ↔ ((abs|(tree, ino0, q0, ino1) * D))(dirs)) ∨ p = ε ⊦ ((abs|(tree[p, tree0], ino0, str + q0, ino1) * D))(dirs) ↔ ((abs|(tree, ino0, str + q0, ino1) * D))(dirs), p = ε

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
tree[parent(p)].dir?, parent(p) ∈ tree, str ' + q0 ⊑ p, ∀ tree, p, tree0, ino0, ino1, D, dirs. tree[parent(p)].dir? ∧ parent(p) ∈ tree ∧ q0 ⊑ p → (((abs|(tree[p, tree0], ino0, q0, ino1) * D))(dirs) ↔ ((abs|(tree, ino0, q0, ino1) * D))(dirs)) ∨ p = ε ⊦ ((abs|(tree[p, tree0], ino0, str ' + q0, ino1) * D))(dirs) ↔ ((abs|(tree, ino0, str ' + q0, ino1) * D))(dirs), p = ε

Rule argument

Additional information

Used simplifier rules: ⊦ str + p = str ' + p; (s)