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Proof step 5 from proof path-split-wand (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 6:
abs(tree, ino0)(dirs), p ∈ tree ⊦ ∃ ino. ((abs(tree[p], ino) * (abs(tree[p], ino) -* abs(tree, ino0))))(dirs)
Proof step 65:
p ∈ tree, ∃ ino. ((abs(tree[p], ino) * (abs(tree[p], ino) -* abs(tree, ino0))))(dirs) ⊦ abs(tree, ino0)(dirs)

Rule argument

pos: right 1

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