Modus tollens with respect to the right subjunction.
Proof 01. 1 ⊢ A ↔ B, hypo. 02. 2 ⊢ ¬A, hypo. 03. 3 ⊢ B, hypo. 04. 1, 3 ⊢ A, rsubj_elim 1 3. 05. 1, 2, 3 ⊢ ⊥, neg_elim 2 4. 06. 1, 2 ⊢ ¬B, neg_intro 5. rsubj_tollens. ⊢ (A ↔ B) → ¬A → ¬B, subj_intro_ii 6.