Show that (¬𝑃 ⟶ 𝑅) ∧ (𝑄 ↔ 𝑃) ⟺ (𝑃 ∨ 𝑄 ∨ 𝑅) ∧ (𝑃 ∨ ¬𝑄 ∨ 𝑅) ∧ (𝑃 ∨ ¬𝑄 ∨ ¬𝑅) ∧ (¬𝑃 ∨ 𝑄 ∨ 𝑅) ∧ (¬𝑃 ∨ 𝑄 ∨ ¬𝑅).

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It follows that

$(¬𝑃 \to 𝑅) ∧ (𝑄 ↔ 𝑃)$

$⟺(\neg\neg𝑃 \lor 𝑅) ∧ (𝑄 \to 𝑃) ∧ (P \to Q)$

$⟺(𝑃 \lor 𝑅) ∧ (\neg 𝑄 \lor 𝑃) ∧ (\neg P \lor Q)$

$⟺(𝑃 \lor F\lor 𝑅) ∧ (\neg 𝑄 \lor 𝑃\lor F) ∧ (\neg P \lor Q\lor F)$

$⟺(𝑃 \lor (Q\land \neg Q)\lor 𝑅) ∧ (𝑃\lor\neg 𝑄 \lor (R\land\neg R)) ∧ (\neg P \lor Q\lor (R\land\neg R))$

$⟺(𝑃 \lor Q\lor 𝑅) ∧ (𝑃 \lor \neg Q\lor 𝑅) ∧(𝑃\lor\neg 𝑄 \lor R) ∧(𝑃\lor\neg 𝑄 \lor \neg R) ∧ (\neg P \lor Q\lor R) ∧ (\neg P \lor Q\lor \neg R)$

$⟺ (𝑃 ∨ 𝑄 ∨ 𝑅) ∧ (𝑃 ∨ ¬𝑄 ∨ 𝑅) ∧ (𝑃 ∨ ¬𝑄 ∨ ¬𝑅) ∧ (¬𝑃 ∨ 𝑄 ∨ 𝑅) ∧ (¬𝑃 ∨ 𝑄 ∨ ¬𝑅).$