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Proof step 2 from proof abs-has-rootpath-h (previous step)

((abs(tree, ino0) * {dirs0} ))(dirs), ino1 ∈ dirs ⊦ ino1 ∈ dirs0, ∃ p. ispath(p, ino0, ino1, dirs, files)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information