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I want to rewrite the following types of propositions without the simple or double implication:

  1. $p \land \lnot q \to r$
  2. $p \land \lnot q \to r \land q$
  3. $(p \to r) \leftrightarrow (q \to r)$

So we have to write these propositions without any implication, for example the first proposition like $p \land \lnot q \land r$ or is something else meant?

pingu
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  • It means to produce an equivalent formula without $\to$. Your formula is not equivalent to 1. – Mauro ALLEGRANZA May 18 '20 at 12:01
  • $A \to B$ is logically equivalent to $\lnot A \lor B$, and $A \leftrightarrow B$ is logically equivalent to $(A \to B) \land (B \to A)$. So, ... – Taroccoesbrocco May 18 '20 at 12:02
  • $Α \to B$ is true when $A$ is true and it implies that $B$ is true. So why is it equivalent to $\lnot A \lor B$ ? – pingu May 18 '20 at 12:18
  • @Pingu You can find a list of detailed answers regarding that here: https://math.stackexchange.com/questions/48161/in-classical-logic-why-is-p-rightarrow-q-true-if-both-p-and-q-are-false – user400188 May 18 '20 at 12:35
  • @user400188 I have a question about the case when p is false and q is true. Why is in this case $p \to q$ true? – pingu May 18 '20 at 16:49
  • What you are asking about here is known as the principal of explosion. The justification for it is that $p\rightarrow q$ should only be false, when the premises is true, and the consequence is false. By virtue of this definition, the principal of explosion states that anything can be proved by assuming falsehood. A great example of this can be found here: https://math.stackexchange.com/questions/2972412/starting-with-a-false-statement-how-can-one-prove-anything-is-true/2972430#2972430 – user400188 May 19 '20 at 11:55
  • If you wish to have a discussion of this property of material implication, I recommend you ask about it in the logic room (https://chat.stackexchange.com/rooms/44058/logic). In this chatroom you can discuss questions as opposed to just asking them, which in my experience is the best way to learn about implication. – user400188 May 19 '20 at 11:58

1 Answers1

2

Hints:

  • $p\rightarrow q$ is equivilent to $\lnot p\lor q$.
  • $p\leftrightarrow q$ is equivilent to $(p\rightarrow q)\land(q\rightarrow p)$

Using these two definitions, it should be easy to write the propositions $(1~-~3)$ without any implication or biconditional symbols.

Answer:

\begin{align}1.&~~~\lnot(p \land \lnot q) \lor r\\2.&~~~\lnot(p \land \lnot q) \lor (r \land q)\\3.&~~~\big((\lnot p \lor r) \land (\lnot q \lor r)\big)\lor \big((p \land\lnot r) \land (q \land\lnot r)\big)\end{align}

Answers $(1~-~3)$ could be improved by simplification using the distribution laws if you feel inclined.

user400188
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  • I have found the first two expressions with the equivalences you stated above, too. As for the third proposition, I have found that it is equivalent to

    $$((p \land \lnot r) \lor ( \lnot q \lor r)) \land ((q \land \lnot r) \lor (\lnot p \lor r)) $$

    Is it also right? Is it equivalent to your result?

     – pingu May 18 '20 at 19:59
  • I must apologise. My original answer contained a mistake, I had written the same term twice $((\lnot p\lor r)\land(\lnot q\lor r))\lor((\lnot p\lor r)\land(\lnot q\lor r))$. Your answer was equivalent to my mistaken answer, but like mine it was not correct. I have since edited the answer to something that is correct. – user400188 May 19 '20 at 11:46
  • After having made again calculations, I have found the result

    $$((p \land \lnot r) \lor (\lnot q \lor r)) \land ((q \land \lnot r) \lor (\lnot p \lor r)) $$

    Is it right? Is it equaivalent to yours?

     – pingu May 21 '20 at 17:50
  • Hey @pingu, that equation is the same as the one you posted before. Did you make an error typing it? – user400188 May 23 '20 at 06:27