# if $s \implies \lnot w$, does $\lnot s \implies w$?

Quick question. I'm given a set of logical statements:

1) $l \lor s$
2) $\lnot c$
3) $l \rightarrow b$
4) $s \rightarrow c$
5) $s \rightarrow \lnot w$

Now $l, s, c, b$ and $w$ represent certain statements. I won't type out what they are since it's not important for what I'm asking, but suffice to say that the question wants us to deduce that:

$\lnot c \rightarrow \lnot s$ by Modus Tollens (4) (and $\lnot c$ is in (2))
$\therefore l$ (Disjunctive Syllogism of (1)
$\therefore b$ (Modus Ponens and (4))

Which when I read through it makes perfect sense.

But see (5). When we get to deducing $\lnot s$ can we not assume $\lnot s \rightarrow w$ therefore deducing $w$?

• As for the question in the title: no, it doesn't. As for the question in the body, I don't follow at all. What does (5) have to do with anything? In particular, as you have already deduced $\lnot s$, (5) plays no further role in any deduction (besides possibly deducing a contradiction, but that's not the case here either). Jun 18, 2011 at 13:06
• If it is sunny, it doesn't rain. But if it doesn't rain, it doesn't mean it has to be sunny. --Edit: ah, I hadn't seen Didier's answer below ;P Jun 18, 2011 at 14:57
• There is a Wikipedia article on this fallacy: en.wikipedia.org/wiki/Denying_the_antecedent Jun 20, 2011 at 1:04

You cannot go from S implies not W to not S implies W. You can use truth tables to validate this. What you can do is use contraposition and go from S implies not W to W implies not S.

So $P \Rightarrow Q \not\equiv \neg{}P \Rightarrow \neg{}Q$

But $P \Rightarrow Q \equiv \neg{}Q \Rightarrow \neg{}P$

• I'm not sure how to LaTeX this because this really is my first reply on a stack exchange website :-)
– user12205
Jun 18, 2011 at 13:05
• @Jeroen: welcome! If you already know how to use LaTeX, then just enclose the mathematics expressions in dollar signs. For example: enclosing S \implies \neg W between a pair of dollar signs give $S\implies \neg W$. Jun 18, 2011 at 13:08
• Oh, so it just works like regular inline equations :-) I tried that but I got no result in the WYSIWYG. I guess it gets parsed when you hit the add comment button! Thanks.
– user12205
Jun 18, 2011 at 13:11
• @Jeroen: it didn't work in the WYSIWYG (preview pane)? It is supposed to, though sometimes it take about 5 seconds to start parsing. If it doesn't work, please drop a note on meta including your operating system and web browser of choice. Thanks! Jun 18, 2011 at 13:19
• @Willie: I thought the preview pane was only for answers and not comments. Do you really have a preview pane for the comments? Jun 18, 2011 at 16:02

Example: s = It is midnight, w = The sun shines.

More seriously, an easy way is to reduce everything to elementary statements. Here "p implies q" is equivalent to "not-p or q" hence "s implies not-w" is equivalent to "not-s or not-w" while "not-s implies w" is equivalent to "s or w" and these are not equivalent.

I just wanted to clarify what you were able to deduce and chime in with the others about your question.

The premises you're working with:

1) $l \lor s$
2) $\lnot c$
3) $l \rightarrow b$
4) $s \rightarrow c$
5) $s \rightarrow \lnot w$

You mention that the exercise you were working with "wants us to deduce" some statements; first you start with

$\lnot c \rightarrow \lnot s\quad$ by Modus Tollens (4) (and $\lnot c$ is in (2))

• This should probably be written as two steps (as in 6, 7 below). (See 6, 7, 8, 9 based on your logic, with additional justifications.)

6) $\;\lnot c \;\rightarrow \;\lnot s \quad$ (Modus Tollens, 4)
7) $\;\lnot s\;\qquad\qquad\quad$ (Modus Ponens, by (2) & (6))
8) $\;\therefore \;l\qquad\qquad\quad$ (Disjunctive syllogism of (1) & (7))
9) $\;\therefore \;b\qquad\qquad\quad$ (Modus ponens by (4) & (8))

I realize the question in your post didn't concern your deduction of the above steps, but I just wanted to make sure you weren't taking "shortcuts" and possibly omitting a step and references to previous statements used to justify those statements.

Now, note that, up to this point premise (5) is irrelevant/superfluous to your conclusions above...

You knew that it was appropriate and smart to apply Modus Tollens to conclude statement 6) above; I'm assuming now that others have clarified your question, you realize that given $\lnot s$, one cannot apply Modus Tollens to $s \rightarrow \lnot w$ to conclude therefore $w$, as that is not equivalent to Modus Tollens.

That said:

• If you had $w$, you could deduce therefore $\lnot\lnot w$, and by Modus Tollens, get $\lnot s$, (by $\lnot \lnot w$ and premise (5), with Modus Tollens); but why would you want to do that? ...you already have $\lnot s$!
• One other point: you could have actually constructed the 5th premise from the first 4 premises, along with your subsequent deduction of $\;\lnot s$. For example:

7) $\lnot s\qquad\qquad\qquad$(...reasoning above)
7+) $\lnot s \lor \lnot w\quad\;$ (Disjunction Introduction/Addition from (7): that is, if we have established
$\qquad\qquad\qquad\qquad\quad\lnot s$, then we've established $\lnot s \lor$(anything))
5) $s \rightarrow \lnot w\qquad\;$ (Given (7+) and the fact that $\lnot s \lor \lnot w \equiv s \rightarrow \lnot w$).