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Proof step 1 from proof absc-path (conclusion)

⊦ p ∈ tree ∧ ((abs|(tree, ino, p, ino0) * D))(dirs) ∧ ispath(p, ino, ino1, dirs, files) → ino0 = ino1

System applied rule normalize

And obtained the following new goals

Proof step 2:
((abs|(tree, ino, p, ino0) * D))(dirs), ispath(p, ino, ino1, dirs, files), p ∈ tree ⊦ ino0 = ino1

Rule argument

Additional information