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Proof step 2 from proof path-in-eq (previous step)

abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh) ⊦ p ∈ tree ↔ ispath(p, ino, dirs, files)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information