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Proof step 1 from proof entries-dom (conclusion)
⊦ ((abs(sdt, sdi) * D))(dirs) → (str ∈ sdi ↔ str ∈ sdt)
System applied rule normalize
And obtained the following new goals
Proof step 2
:
((abs(sdt, sdi) * D))(dirs) ⊦ str ∈ sdi ↔ str ∈ sdt
Rule argument
Additional information