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Proof step 2 from proof abs-has-rootpath (
previous step
)
abs(tree, ino0)(dirs), ino1 ∈ dirs ⊦ ∃ p. ispath(p, ino0, ino1, dirs, files)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
abs(tree, ino0)(dirs), ino1 ∈ dirs ⊦ ∃ p. ispath(p, ino0, ino1, dirs, files)
Rule argument
Additional information