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Proof step 5 from proof path-fold (previous step)

((abs|(tree, ino0, p, ino1) * abs(tree[p], ino1)))(dirs), p ∈ tree ⊦ (_∃(λ ino1. abs|(tree, ino0, p, ino1) * abs(tree[p], ino1)))(dirs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
((abs|(tree, ino0, p, ino1) * abs(tree[p], ino1)))(dirs), p ∈ tree ⊦ ∃ ino. ((abs|(tree, ino0, p, ino) * abs(tree[p], ino)))(dirs)

Rule argument

Additional information

Used simplifier rules: ⊦ (_∃ nD)(dirs) ↔ (∃ ino. nD(ino)(dirs)); (s)