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Proof step 6 from proof path-unfold (previous step)

p ∈ tree ⊦ ((abs(tree, ino0) * D))(dirs) ↔ (∃ ino. (((abs|(tree, ino0, p, ino) * abs(tree[p], ino)) * D))(dirs))

User applied rule case distinction

And obtained the following new goals

Proof step 7:
((abs(tree, ino0) * D))(dirs), p ∈ tree ⊦ ∃ ino. (((abs|(tree, ino0, p, ino) * abs(tree[p], ino)) * D))(dirs)
Proof step 26:
p ∈ tree, ∃ ino. (((abs|(tree, ino0, p, ino) * abs(tree[p], ino)) * D))(dirs) ⊦ ((abs(tree, ino0) * D))(dirs)

Rule argument

pos: right 1

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