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Proof step 3 from proof path-fold (previous step)

((abs|(tree, ino0, p, ino1) * abs(tree[p], ino1)))(dirs), p ∈ tree ⊦ abs(tree, ino0)(dirs)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ p ∈ tree → abs(tree, ino0) = _∃(λ ino1. abs|(tree, ino0, p, ino1) * abs(tree[p], ino1))
Proof step 5:
((abs|(tree, ino0, p, ino1) * abs(tree[p], ino1)))(dirs), p ∈ tree ⊦ (_∃(λ ino1. abs|(tree, ino0, p, ino1) * abs(tree[p], ino1)))(dirs)

Rule argument

spec: <toplevel> name: path-unfold seq: ⊦ p ∈ tree → abs(tree, ino0) = _∃(λ ino1. abs|(tree, ino0, p, ino1) * abs(tree[p], ino1)) subst: [p, ino0, tree] → [p, ino0, tree] flags: <all paths>

Additional information

Used terms: p, ino0, tree