Propositional Logic Symposia - [8] – Modus Ponens & Modus Tollens

[VideCorSpoon]

Caveat Logi

First, it should be noted that the inference and replacement rules that I am going to show you are described in different ways by different people. For example, Modus Ponens (the standard name) goes by many other aliases such as “Conditional Introduction” and “Arrow Out.” The same goes for all the other inference and replacement rules. In fact, some call the inference and replacement rules just plain “derivation rules.” So you can use any name you want for the rules because the execution of the rule is universally accepted; only the name is different.

Deductive Instructive

As I had said at the end of symposium 7, inference rules are basically deductive-argumentative methods that you can expressly infer to derive a certain step of your proof. These rules are basically plug-in arguments (arguably) that will help you derive the conclusion from the premises in your proof. There are for all intensive purposes 10 inference, replacement and sub-proof rules. This thread deals with the first two of the four basic inference rules.

Modus Ponens (MP)

Because anything said and Latin sounds profound, Modus Ponens translates as “that which affirms by affirming.” Basically think of Modus Ponens as a simple deductive argument set out in front of you where you are presented with; A) conditional argument, B) an antecedent affirming the consequent of your conditional, and B) a conclusion which affirms the consequent of your conditional.

Take this argument for example.

If John is walking, then Mary is walking. John is walking. Thus, Mary is walking.

There is no doubt that you have used this type of reasoning before. It is relatively straight forward to figure out the dynamics of the argument and thus Modus Ponens.

This is a chart of Modus Ponens in general.



Think of it like this. Modus Ponens requires 2 premises in order to derive the conclusion. One premise has to have a conditional (i.e. if a, then b). The other premise has to include the premise, or the antecedent, of the conditional (i.e. the “a” in “If a, then b.”) You do not need to have the conditional and the antecedent displayed conditional, antecedent, thus conclusion. You can have the antecedent, conditional, and then the conclusion. So you do not need to have a particular arrangement in order to infer Modus Ponens. The only thing that will be constant is the conclusion, which always comes at the end of the inference.

Modus Ponens in Proofs

Now keep in mind the previous chart where we observe the conditional, antecedent of the conditional, and the conclusion of the conditional. We will need that to remind us of how to infer Modus Ponens. Let’s try to see if we can do something with Modus Ponens in a sample proof to see how it can be utilized correctly.

The very first thing to remember is the way we set up a proof. If you do not remember, go back to symposium 7 and review how to do it correctly before you go any further than this. We start with a formed proof which is basically step 5 of “How to set up a proof to solve.”


Step 1. Set up your proof. Again, the steps to get to this point are in symposium 7.
Step 2. Identify the conditional on line 1 and the antecedent of the conditional on line 2. Note: A conditional is comprised of a premise (A) and a conclusion (B) which looks like this in logic (A-->B). An antecedent is basically a premise which precedes the consequent, or the conclusion, of the conditional.
Step 3. Now that we have the essential elements for modus ponens to be evoked, we make sure that modus ponens can be successfully put down by identifying the same identical variable(s) for the antecedent of the conditional in line one (A&B) and the same statement on line 2 (A&B).
Step 4. No that we have the foundation for Modus Ponens to be evoked, we can now infer modus ponens on line three. We basically put down the consequent of the conditional on line three. That’s it. Modus Ponens has been inferred to further your proof. But remember, it is ESSENTIAL to put write down the inference rule to the right of your proof on the same line to show your work, essentially proving the path of the argument. When you write down the citation you can either write Modus Ponens, or the abbreviation MP. Also remember to write down the lines you inferred it from.

NOTE: If you look in the top right corner of all the step boxes, you see squares and circles. That is a visual way to remember how the Modus Ponens pattern follows. It helped me remember the rule a lot, so I put it down in case it helps anyone else. The square is the antecedent and the circle is the consequent.

RECAP!!! Modus Ponens is; (A) a conditional, + (B) an identical antecedent of the conditional on a different line = (C) The consequent of the conditional inferred by Modus Ponens.


Modus Tollens (MT)

Modus Tollens is much like Modus Ponens, except it essentially goes the opposite way. Instead of “affirm by affirming”,” we are “denying by denying.” It is an indirect way of inferring a conclusion compared to the direct nature of Modus Ponens.

Like modus Ponens, you need a conditional as part of the inference. But that’s where the similarities end. The other part of the inference requires a negation of the consequent. The conclusion is the negation of the antecedent.



Basically, look for a conditional, a negation of the consequent, and then you can infer the negation of the antecedent.


Step 1. Set up your proof.
Step 2. Identify the conditional like you would for modus ponens. Then identify the negation of the consequent on a separate line.
Step 3. Make sure you can successfully evoke the modus tollens inference by checking the consequent of the conditional and the negation of the conditional on the separate line.
Step4. Infer by way of Modus Tollens ~(A&B) on line 3.

RECAP!!! Modus Tollens is; (A) a conditional, + (B) a negated consequent of the conditional on a different line, = (C) The negated antecedent of the conditional inferred by Modus Tollens.

These are basically the essentials for Modus Ponens and Modus Tollens inferences. This is not the end of the thread though. I am going to post a few sample problems incorporating both in inferences in a single proof structure and show you how attack the problem in a simple way. I’m sure I missed a few things, so let me know if there are things that need to be clarified or need to be completely changed.

As always, if you have any questions don’t hesitate to ask.

[VideCorSpoon]

HOW TO SOLVE A PROOF WITH MODUS PONENS AND MODUS TONENS

The way to solve truth functional deductive logic proofs is kind of awkward. There isn’t any set way to go about it. But that should not deter you, because there is always an easy way to solve any proof.

The best way to show you how to approach a proof is to do a sample problem and show you what to look for, what seems simplest in the long run, etc.

First, remember the two inference rules we are going to use (since at this point we only know of two of them for now).



Step 0. Now this is the sample problem we are going to try with only these two inference rules. We start with an already translated and lined problem with the conclusion at the end ( /~M ) .



Step 1. The first step is to put this proof into a standard proof table to actually begin to do the proof. If you do not remember how to do this, go back to symposium 7 for a more thorough explanation.



Step 2. This is the part of the proof that becomes difficult. There is really no set way to attack a proof, but there are ways of approaching a proof that make the solution a little easier. You may become familiar enough with proofs and the inference and replacement rules to see the patterns right off the bat and infer without a second thought and do it very precisely. But what if you are genuinely stumped and you don’t know in what direction to go?

POINT 1. Do every combination and inference possible and sort it all out later!

This is a very valuable trouble shooting tip. When in doubt, do everything so that you can select what works and what doesn’t. Keep in mind a proof does not have to be short, or solved in a single way. There are numerous ways to solve a single proof. It all depends on what argumentative style you have (what inference rules you favor most).

So let’s try every single combination and see what we come up with.
For me at least, I am most comfortable with Modus Ponens. So with that in mind, let’s go through the argument and find the tell tale signs of a modus ponens. Remember, a Modus ponens needs two things. 1) a conditional, and 2) the antecedent (first letter or compound statement) of the conditional.


Step 3. Now that we derived an inference, we need to go back to the beginning and look for any other instances of either Modus Ponens or Modus Tollens. Keep in mind that whatever you add to the proof can and will be included in the rest of you assumptions to solve the proof. Simply, you do not need to work with just lines 1-4. YOU USE ANY AND ALL LINES YOU HAVE IN BOTH YOUR ARGUMENT AND YOUR DERIVATIONS. It is best to exhaust all possibilities for any combinations in the argument (lines 1-4) before going into the proof for combinations. At this point for our proof, there are no such combinations exclusively in the argument (lines 1-4). But there is one instance using the derived S on line five.


Step 4. Now that we have a new derivation line, we again have to go to the beginning and see if any new combination can be revealed with our newest addition, ~R. And wouldn’t you know it… There is a possible combination.


So you successfully reached a conclusion for your argument with proven argumentative methods and inference rules in order to secure the conclusion of the above argument. Huzzah!!!

IF YOU HAVE ANY PROBLEMS (OR NOTICED THAT I SCREWED UP SOME WHERE) LET ME KNOW BECAUSE I’M HAPPY TO ANSWER ANY QUESTIONS OR COMMENTS.

[VideCorSpoon]

Here is a sample problem which can be solved using just Modus Ponens and Modus Tollens. Highlight the blank section for the answers. But remember, the proof can be solved in a number of different configurations, so you may actually be right. As always, ask if there is a problem or if I am wrong about something. I'll add more problems if if you need them.

Problem

1. M
2. R --> ~Q
3. M --> R
4. T --> ~Q / ~T

Solution (highlight for answer)

1. M
2. R --> ~Q
3. M --> R
4. T --> ~Q / ~T
5. R : 1,3 MP
6.~Q : 2,5 MP
7.~T : 4,6 MT

As some may have noticed, there are other ways to solve this proof using the other two inference rules (i.e. hypothetical syllogism 2,3) I'll be getting to hypothetical syllogism and disjunctive syllogism next.

[Alanocrates]

Hey Spoon, are you going to put up the rest of the derivatives in your tutorials????