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I know that by conversion theorem, $ r → q = \lnot r \vee q $. So if $¬r → q$ will be equal to $r \vee q$ and not $r \vee ¬q$ right?

Let's say we know that $q$ and $p$ are true. Given that $ r → s; ¬r → q; p \vee r;$ are true.

How do we know that whether $r$ is true or false?

Michael Hardy
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BEX
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    \vee instead of v – JMoravitz May 02 '18 at 04:49
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    (1) Yes $(\neg r)\implies q$ is equivalent to $r\vee q$ and this statement is different and unequivalent to $r\vee \neg q$. (2) Suppose $r,q,p$ and your three other statements are all true. No contradictions with your hypotheses. Suppose $\neg r, q, p$ and your three other statements are true. Still no contradictions. Ergo, we don't know then whether $r$ is true or false. – JMoravitz May 02 '18 at 04:52
  • What if there is a conclusion (p∨q)Λr, so is this argument valid? Can we prove by contradiction method? – BEX May 02 '18 at 04:58
  • If $(p\vee q)\wedge r$ is said to be true that implies that both $(p\vee q)$ is true and $r$ is true., which in particular implies that $r$ is true... – JMoravitz May 02 '18 at 04:59
  • If you are asking whether the given hypotheses imply that $(p\vee q)\wedge r$ is true, no they do not. Consider the situation where $p,q,\neg r, s$ are all true. The hypotheses are all true then but the conclusion is false. – JMoravitz May 02 '18 at 05:01
  • hmm cause I realize that r can be both true and false. if r is false, the conditions given above are still true; if r is true, s must be true since r→s is a true statement. Therefore I'm not sure that the argument(conclusion) is valid or invalid. – BEX May 02 '18 at 05:04
  • "The argument" What argument... you haven't provided one. Regardless... considering $p,q,\neg r, s$ and the three other hypothetical statements listed in the question, we have the conclusion $(p\vee q)\wedge r$ is false., meanwhile if we consider $p,q,r,s$ and the three other hypothetical statements listed in the question, we have the conclusion above is instead true. As a result with the given hypotheses both outcomes are possible and any argument which attempts to prove that the conclusion is either true or attempts to prove the conclusion is false must be flawed and invalid. – JMoravitz May 02 '18 at 05:06
  • The argument is the conclusion which is (p∨q)∧r. By the way thanks. – BEX May 02 '18 at 05:09

1 Answers1

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Given

$$\begin{array}\\ \lnot\ r\implies q\\ p\ \lor\ r\\ \hline (r\ \lor\ q)\ \land (p\ \lor\ r) \end{array}$$

Since you said $p$ and $q$ are true, whether or not $r$ is true is not important/ it won't affect the truth of the conclusion/ it's unknown/ it can be either.

And in your title it's $\large\lnot\ r\implies q \color{blue}{\equiv} r\ \lor\ q$, not $\color{red}{=}$.

If you want to prove that $r$ is false, try to find $t$ s.t.

$$\begin{array}\\ r\implies t\\ t\ \land\ s\equiv\bot\ ,\\ \end{array}$$

which $\bot$ means a contradiction, then

$$\begin{array}\\ r\implies s\\ r\implies t\\ t\ \land\ s\equiv\bot\\ \hline \lnot\ r \end{array}$$

linear_combinatori_probabi
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