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Proof step 1 from proof entries-any (conclusion)

sdt ≠ ∅ ⊦ ((abs(sdt, sdi) * D))(dirs) ↔ (∃ tree, ino, str, sdi0, sdt0. ¬ str ∈ sdt0 ∧ sdt = sdt0[str, tree] ∧ ¬ str ∈ sdi0 ∧ sdi = sdi0[str, ino] ∧ ((abs(tree, ino) * abs(sdt0, sdi0) * D))(dirs))

System applied rule normalize

And obtained the following new goals

Proof step 2:
⊦ ((abs(sdt, sdi) * D))(dirs) ↔ (∃ tree, ino, str, sdi0, sdt0. ¬ str ∈ sdt0 ∧ sdt = sdt0[str, tree] ∧ ¬ str ∈ sdi0 ∧ sdi = sdi0[str, ino] ∧ ((abs(tree, ino) * abs(sdt0, sdi0) * D))(dirs)), sdt = ∅

Proof step 1 from proof entries-any (conclusion)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information