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Proof step 1 from proof abs-paths-eq (conclusion)

abs(tree, ino0)(dirs), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ispath(p0, ino0, ino1, dirs, files), ispath(p1, ino0, ino1, dirs, files), ino1 ∈ dirs ⊦ p0 = p1

System applied rule normalize

And obtained the following new goals

Proof step 2:
abs(tree, ino0)(dirs), ispath(p1, ino0, ino1, dirs, files), ispath(p0, ino0, ino1, dirs, files), ino1 ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ p0 = p1

Proof step 1 from proof abs-paths-eq (conclusion)

System applied rule normalize

And obtained the following new goals

Rule argument

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