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Proof step 4 from proof abs-has-rootpath (
previous step
)
⊦ ((abs(tree, ino0) * {dirs0} ))(dirs) ∧ ino1 ∈ dirs ∧ ¬ ino1 ∈ dirs0 → (∃ p. ispath(p, ino0, ino1, dirs, files))
Lemma
abs-has-rootpath-h