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I would like to confirm my proof.

Kenny Lau
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Orange
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    You already wrote (1) and (2) and this is duplicate of them. – choco_addicted Apr 04 '16 at 14:10
  • This is a valid proof. However, for future reference, you have the option of editing your post in (1) from choco's link above and including your attempt, or posting it as an answer in link (2) and closing the question. – Inazuma Apr 04 '16 at 14:12
  • ok I've got it. sorry for my bad ! – Orange Apr 04 '16 at 14:15
  • Another way to verify these arguments is to also draw up a truth table. If the conclusion (A -> D V C) is true every time the premises (A -> B V C) (B -> D) are true, then the statement works.

    However: If there is anywhere where a premise(s) can be false and the conclusion true, then the argument is not valid.

     – Inazuma Apr 04 '16 at 14:21

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$$\begin{array}{} (1)&A\implies(B\lor C)&\text{premise}\\ (2)&B\implies D&\text{premise}\\ (3)&\neg(D\lor C)&\text{assumption}\\ (4)&\neg D\land \neg C&\text{De Morgan's laws, from (3)}\\ (5)&\neg D&\text{directly form (4)}\\ (6)&\neg C&\text{directly form (4)}\\ (7)&\neg B&\text{modus tollens, from (2) and (5)}\\ (8)&\neg B \land \neg C&\text{combining (6) and (7)}\\ (9)&\neg (B \lor C)&\text{De Morgan's laws, from (8)}\\ (10)&\neg A&\text{modus tollens, from (1) and (9)}\\ (11)&\neg (D\lor C)\implies\neg A&\text{combining (3) and (10)}\\ (12)&A\implies D\lor C&\text{modus tollens, from (11)}\\ \blacksquare \end{array}$$

Kenny Lau
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