The following material is obtained form How to prove it by Daniel J Velleman. The book has asked to check the validity of the following argument without using truth table. The argument goes like this:
Either John or Bill is telling the truth. Either Sam or Bill is lying. Therefore, either John is telling the truth or Sam is lying.
What I think:
In the first statement either John or Sam is telling the truth. In the next statement we see Bill on Sam is lying.So what I am trying to say is does the statement regarding Bill matter for the given argument to be valid since conclusion does not have anything to do with Bill. Now assuming Both premises true , can't we tell John is telling the truth or Sam is lying making the argument valid?
Is the argument below valid??
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In the first statement either John or Bill is telling the truth. In the next statement we see Bill on Sam is lying.So what I am trying to say is does the statement regarding Bill matter for the given argument to be valid since conclusion does not have anything to do with Bill. Now assuming Both premises true , can't we tell John is telling the truth or Sam is lying making the argument valid?
You wish to determine the validity for the argument: ${j\vee b, \neg s\vee\neg b\vdash j\vee \neg s}$
In classical logic we can make an argument by cases, since it is axiomatic that Bill is either lying or telling the truth.
Assuming Bill is lying, then the first premise means John is telling the truth (by disjunctive syllogism), or Sam is lying (by disjunction introduction). $$\begin{split}\neg b, j\vee b & \vdash j\\ j&\vdash j\vee\neg s\\ \hline \neg b, j\vee b &\vdash j\vee\neg s \end{split}$$
Assuming Bill is telling the truth, then the second premise means Sam is lying (by disjunctive syllogism), or John is telling the truth (by disjunction introduction)$$\begin{split}b, \neg s\vee \neg b & \vdash \neg s\\ \neg s&\vdash j\vee\neg s\\\hline b, \neg s\vee \neg b &\vdash j\vee\neg s \end{split}$$
Either case derives that either John is telling the truth or Sam is lying.
$$\begin{split} &\vdash \neg b\vee b\\ \neg b, j\vee b&\vdash j\vee\neg s\\b, \neg s\vee \neg b & \vdash j\vee\neg s\\\hline j\vee b, \neg s\vee\neg b& \vdash j\vee \neg s\end{split}$$
You're correct, and here's an easier way to see it. "(J truth) or (S lie)" is equivalent to "If not (J truth), then (S lie)". So, let's assume "not (J truth)". By the first statement, we obtain (B truth), which combines with the second statement to imply (S lie).
We have now proven "(J truth) or (S lie)".