# 7.2: Basic Rules of Implication

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

Okay, time to get a bit more complex. I’m going to introduce four basic rules so that we can do the derivation we just did in a more complex way. It turns out that those premises entail that conclusion in more ways than one!

Here’s the first rule everyone learns. It’s super straightforward. Don’t overthink it:

##### Modus Ponens

(Latin for “Mode of Affirming” or “Mode of Positing”)

If I find a conditional all by itself on one line, and then on another line I find the antecedent to that same conditional, then I am allowed to put the consequent on a new line and cite “MP” (modus ponens) as my justification.

\begin{align*}{} & P \rightarrow Q \\ & \underline{P \ \ \ \ \ \ \ \ }\\ & Q \end{align*}

I know what you’re thinking. “But that’s just what conditionals are!!!” They just mean that if the antecedent is true then so is the consequent. If you were thinking that. Great! That is in fact just what conditionals mean. Here, we just expand the meaning of a conditional out into an argument pattern.

Suppose someone reasoned as follows: “If he says “yes” to go to the dance with me, then I’ll be so excited!” and then hours later, after asking this guy, says “He said “yes”! So now I’m so excited!” It would make perfect sense. There’s a true conditional stating that if A happens, then B will happen, and then we confirm that A happened, so it follows that B will happen.

Modus Ponens allows us to break the consequent off of a conditional.

There are many rules for manipulating conditional propositions,

but this is the most basic: if you want the consequent, find the antecedent.

## How to read the rules of implication

Modus Ponens above looks like an argument, but is also how we write the first set of rules we learn: the RULES OF IMPLICATION. The rules of implication are all one-directional rules that are based on valid inference patterns—mostly inference patterns involving more than one premise. So we take these simple arguments that we know are valid and then turn them into rules that allow us to make moves in natural deduction.

The rules of implication have some set of premises above a line and a conclusion below the line. To read a rule of implication, you should interpret the formulas or patterns above the line as the formulas you need to find in the numbered propositions you have already derived. Then you look below the line for the corresponding formula you should write on a new line. Remember, every rule tells you something like “find x, write y.” So if you find what’s above the line, you then get to write what’s below the line.

Furthermore, each rule is typically written using letters, but those letters are stand-ins for any formula at all. As a result, any of the following could be instances of Modus Ponens:

• \begin{align*}{} & \sim [(G \leftrightarrow P) \vee (C \bullet D)] \supset [S \leftrightarrow (G \vee P)] \\ & \underline{\sim [(G \leftrightarrow P) \vee (C \bullet D)] \ }\\ & [S \leftrightarrow (G \vee P)] \end{align*}
• \begin{align*}{} & \sim N \rightarrow \sim (X \vee \sim S) \\ & \underline{\sim N \ \ \ \ \ \ \ \ \ \ } \\ & \sim (X \vee \sim S) \end{align*}
• \begin{align*}{} & (J \vee Q) \rightarrow [X \vee (\sim R \rightarrow (Z \equiv W)] \\ & \underline{ (J \vee Q) \ \ \ \ \ \ \ \ \ \ \ }\\& [X \vee (\sim R \rightarrow (Z \equiv W)] \end{align*}

What matters is that you can find one formula where the main operator is an arrow/horseshoe and then you can find another formula that is identical to the antecedent or left side of the arrow formula. If you find those, you can write down the right side or consequent of the arrow on a new line by itself.

It’s helpful to some to think of the rules not in terms of variable letters, but instead in terms of shapes or abstract symbols that stand in for any well-formed formula at all. Here’s how we might write Modus Ponens instead:

\begin{align*}{} &\Delta \rightarrow \square \\ & \underline{\Delta \ \ \ \ \ \ \ } \\ & \square \end{align*}

Perhaps it’s easier to think of a triangle (delta) as a stand-in for any proposition at all. Then one can more easily recognize that these are all instances of this general structure or pattern:

If this is easier for you, then I encourage you to write the rules in your notes in symbols.

NEXT IS THE COMPLEMENT OF MODUS PONENS:

##### Modus Tollens

(Latin for “Mode of Denying”)

If I find a conditional all by itself on one line, and then on another line I find the negated form of the consequent to that same conditional, then I am allowed to put the negated form of the antecedent on a new line and cite “MT” (modus tollens) as my justification.

\begin{align*}{} & P \rightarrow Q \\ & \underline{\neg Q \ \ \ \ \ \ \ \ }\\ & \neg P \end{align*}

Modus Tollens is a bit harder to understand, but it is still simply based on the definition of a conditional. We can look at the truth table for a conditional for starters:

P Q (P $$\rightarrow$$ Q)
T T T
T F F
F T T
F F T

Now, remember that the premises to the argument form of modus tollens above tell us that P implies Q and that Q is false. So we look at the rows in our table where Q is false, find that there’s only one where “P implies Q” is true (row 4), and then look at what P is on that row. It is false. That’s how we figure out modus tollens on a truth table.

We figure it out more intuitively by thinking that a conditional means “if the antecedent is true, then the consequent must be true.” Another way of saying essentially the same thing is “if the consequent is false, then the antecedent must be false.” Why? Well, if the consequent is false, then a true antecedent would give use “True $$\rightarrow$$ False.” That would be bad, since we already know that “P $$\rightarrow$$ Q” is true and therefore cannot be “True $$\rightarrow$$ False” (P can’t be true while Q is false).

Strategically, modus tollens allows us to break the antecedent off of a conditional, but we have to negate it in the process. So it’s another rule for breaking apart a conditional: if you want the antecedent by itself, you’ll need to find the negated consequent, and then you’ll have to negate the antecedent when you bring it down to its own line. That’s the price we pay for getting the antecedent on its own line.

Think about this: if Bruce Banner (The Hulk) says “If I get angry with you, then you will know it”, then it follows that any time you can’t tell that Banner is made at you is a time that he in fact is not mad at you. If he were mad at you, he’d be all green and scary looking. So if you can’t tell that he’s mad (if the consequent is false), then he certainly isn’t mad at you (the antecedent is also false).

Or suppose it was true that “if you eat a bunch of food, then you’ll be full.” It follows that any time you aren’t full, you won’t have just eaten a bunch of food. Or “if you eat rotten food, you’ll get a stomach ache.” Suppose you don’t have a stomach ache. Good news! You didn’t eat rotten food.

The next rule deals with disjunctions instead of conditionals (we’ve had enough conditionals for one day).

##### Disjunctive Syllogism

(Latin for “Disjunctive Syllogism”)1

If I find a disjunction all by itself on one line, and then on another line I find the negated form of either disjunct to that same disjunction, then I am allowed to put the other disjunct on a new line and cite “DS” (disjunctive syllogism) as my justification.

\begin{align*}{} & P \vee Q \\ & \underline{\neg Q \ \ \ \ \ \ \ \ }\\ & P \end{align*}

Now we might get super technical and say that the rule also involves a different argument form where the second premise is $$\neg$$P and the conclusion is Q. But we don’t have to get too technical in this course. We can operate at a slightly more intuitive level and just recognize that we’re allowed to take a disjunct out of a disjunction if we find the negation of the other disjunct.

Strategically, this rule allows us to break apart a disjunction, by pulling out one disjunct and putting it on a line of its own. If you want a disjunct, find the negation of the other disjunct. Simple, right?

The final rule of the first four we’re introducing here is the following:

##### Hypothetical Syllogism

(“Hypotheticals” and conditionals or implications are the same thing) If I find two conditionals—all by themselves on separate lines—where the antecedent of one is the consequent of the other, then I can “cut out the middle man” and write a new conditional on its own line which has the antecedent and conditional which aren’t identical to one another. I justify this move by writing “HS” (hypothetical syllogism).

\begin{align*}{} & P \rightarrow Q \\ & \underline{Q \rightarrow R \ \ \ }\\ & P \rightarrow R \end{align*}

Again, this one makes some intuitive sense (hopefully). It’s similar to the property of “transitivity” like you might have studied in a high school math class. In the form above, the Q is the “middle man” that we can take out if we find that it’s useful.

Think in terms of the following formula: P $$\rightarrow$$ Q $$\rightarrow$$ R. If you imagine the arrow just going right through the “Q”, you’ll see the intuition behind hypothetical syllogism: the “middle man” is superfluous and can be removed.

This is a rule for combining conditionals by sticking the antecedent of one and the consequent of another together.

Note: the order of the premises doesn’t matter, so the (Q $$\rightarrow$$ R) can show up first. That’s not important. What is important is that the “middle man” appears on both the left and the right of the two premises. What this means is that the following is not an acceptable instance of hypothetical syllogism:

\begin{align*}{} & P \rightarrow Q \\ & \underline{Q \rightarrow R}\\ & P \rightarrow R \end{align*}

Negative Buzzer Noise! That’s not hypothetical syllogism. That’s something else entirely, and it’s ugly. More importantly, it’s invalid and so won’t work as a rule of inference.

Okay, here are those first four rules again:

### 4 Rules of Implication:

 Modus Ponens \begin{align*}{} &\Delta \rightarrow \square \\ & \underline{\Delta \ \ \ \ \ \ \ } \\ & \square \end{align*} Disjunctive Syllogism \begin{align*}{} &\Delta \vee \square \\ & \underline{\neg\Delta \ \ \ \ \ \ \ } \\ & \square \end{align*} Modus Tollens \begin{align*}{} &\square \rightarrow \Delta \\ & \underline{\neg\Delta \ \ \ \ \ \ \ } \\ & \neg\square \end{align*} Hypothetical Syllogism \begin{align*}{} &\square \rightarrow \Delta \\ & \underline{\Delta \rightarrow \star} \\ & \square \rightarrow \star \end{align*}
##### First Four Rules in Words:

Modus Ponens (MP): If you find the left side of an arrow, you get to write the right side

Modus Tollens (MT): If you negate the right side of an arrow, you get to negate the left side

Disjunctive Syllogism (DS): If you negate one side of a disjunction, you get to write the other disjunct.

Hypothetical Syllogism (HS): If you have two conditionals with a “middle man”, you can take the middle man out.

Now let’s go back to the beginning and play the original problem again, but this time we’ll turn up the difficulty just a bit.

1. $$\neg$$ B $$\wedge$$ (B $$\rightarrow$$ F)

2. $$\neg$$F /$$\neg$$ B

I cheated and already told you about simplification, which comes later on in our study of natural deduction, but we’re going to use is again here.

1. $$\neg$$ B $$\wedge$$ (B $$\rightarrow$$ F)

2. $$\neg$$F /$$\neg$$ B

3. (B $$\rightarrow$$ F) 1, simp

Okay, now where are we? We’ve got a “B $$\rightarrow$$ F” and we want “$$\neg$$ B.” Does that sound familiar? We want the negated form of the antecedent to a conditional. Look at the rules and try to figure out which rule we use next...

I’ll wait...

Okay, you got it right. It’s modus tollens! Here’s how the completed derivation looks:

1. $$\neg$$ B $$\wedge$$ (B $$\rightarrow$$ F)

2. $$\neg$$F /$$\neg$$ B

3. (B $$\rightarrow$$ F) 1, simp

4. $$\neg$$ B 2, 3, MT

Again, I know I’m done when the last line I derived and the conclusion are identical:

1. $$\neg$$ B $$\wedge$$ (B $$\rightarrow$$ F)

2. $$\neg$$F /$$\boxed{\neg \text{B}}$$

3. (B $$\rightarrow$$ F) 1, simp

4. $$\boxed{\neg \text{B}}$$ 2, 3, MT

Let’s walk through a few more together, eh?

1. P $$\vee$$ (R $$\rightarrow$$ Q)

2. $$\neg$$ P

3. (Q $$\rightarrow$$ G) / (R $$\rightarrow$$ G)

What’s a person to do? Remember, it’s a game: just figure out what you need, look in your toolkit, find the right tool, then use it!

Here, we need to combine 3 with the conditional in 1 so that we can get our conclusion. We know that our last step will be to use Hypothetical Syllogism. I like to work backwards like this in my head and then write down the steps going forwards. So we’re now (in our heads) trying to figure out how we might get the

“(R $$\rightarrow$$ Q)” all by itself so that we can do an HS. Remember that HS only works when we have both conditionals all by themselves (when the arrows/horsehoes are the main operators of their respective formulas).

Look at 1 and 2. How do we break apart that disjunction?

Oh yeah! Disjunctive Syllogism!

What else do we need for a disjunctive syllogism?

Oh yeah! The negation of the other disjunct.

Where is that “$$\neg$$ P”?

Oh yeah! Line 2!

Now we’re ready to do the whole thing going forwards:

1. P $$\vee$$ (R $$\rightarrow$$ Q)

2. $$\neg$$ P

3. (Q $$\rightarrow$$ G) /(R $$\rightarrow$$ G)

4. (R $$\rightarrow$$ Q) 1, 2, DS

5. (R $$\rightarrow$$ G) 3, 4, HS

We’re done! Check out the last line and the conclusion that comes after the slash: there’re exactly the same. Bravo!

##### Important Lesson:

It doesn’t matter, for any rule of implication, which order the premises come in.

The lines we’re looking for in our derivation (our natural deduction) don’t have to match the lines in the rule of implication in terms of order. They only have to match in terms of structure or form.

See how 3 and 4 are the premises for our HS, but they don’t match the order of the premises as they’re written in the rule? Who cares? As long as we have the premises we need. We don’t have to worry about their order. Modus Ponens premises can look like this:

1. D

2. D $$\rightarrow$$ H

And Disjunctive Syllogism premises can look like this:

1. $$\neg$$(Z $$\vee$$ K)

2. (Z $$\vee$$ K) $$\vee$$ Q

The only thing that matters is that we’ve matched the form of the premises exactly.

Another important note: the P’s and Q’s in our rules above can stand for anything. So Modus Tollens premises can look like this:

1. $$\neg$$$$\neg$$(D $$\vee$$ F)

2. W $$\rightarrow$$ $$\neg$$ (D $$\vee$$ F)

And Hypothetical Syllogism premises can look like this:

1. $$\neg$$$$\neg$$(D $$\vee$$ F) $$\rightarrow$$ ((Z $$\rightarrow$$ F) $$\leftrightarrow$$ Q)

2. $$\neg$$W $$\rightarrow$$ $$\neg$$$$\neg$$ (D $$\vee$$ F)

See how the antecedent of one is equivalent to the consequent of the other? That means we can do a Hypothetical Syllogism.

Let’s try another together. This time I’ll turn the difficulty up to 11.

1. [$$\neg$$$$\neg$$(D $$\vee$$ F) $$\rightarrow$$ ((Z $$\rightarrow$$ F) $$\leftrightarrow$$ Q)]

2. [$$\neg$$W $$\rightarrow$$ $$\neg$$$$\neg$$ (D $$\vee$$ F)]

3. [$$\neg$$W $$\rightarrow$$ ((Z $$\rightarrow$$ F) $$\leftrightarrow$$ Q)] $$\rightarrow$$ ($$\neg$$$$\neg$$T $$\vee$$ $$\neg$$F)

4. $$\neg$$$$\neg$$F

5. (G $$\rightarrow$$ $$\neg$$T) / $$\neg$$G

“Woah! That wacky fool thinks we can do THAT!!?!?!??!”

Yes, I do. And I’ll thank you not to call me a wacky fool.

“Sorry, I’m just intimidated by the brackets and all of the negations and FIVE premises!?!?! Eeeeek!”

Calm down, dear student, all will be well. We just need to take it one step at a time.

Again, we’ll work our way backwards in our head and then work forwards on paper.

Okay, if we’re working backwards, we should start with the conclusion, right? We need a “$$\neg$$G” and looking through the premises, we don’t find one. Instead, the only G in the whole problem is in premise 5:

5. $$\require{enclose}\enclose{circle}{(\text{G }} \rightarrow \neg$$T) / $$\neg$$G

Which rule (out of the 4 we’ve officially learned so far) allows us to grab the antecedent out of a conditional?

You’re right! It’s Modus Tollens. Well done.

Okay, so we’ll eventually do an MT to get our conclusion “$$\neg$$G”. What would we need to actually be able to perform the MT?

Ummmmm... I dunno.

Remember, Natural Deduction is one step at a time. Don’t get intimidated by trying to think about the whole problem at once. Instead, just focus on the individual step. Here, we’re looking at Modus Tollens and trying to figure out what would complete the pattern.

\begin{align*}{} &\enclose{circle}{:)} \rightarrow \Delta \\ & \underline{\neg\Delta \ \ \ \ \ \ \ } \\ & \neg \enclose{circle}{:)} \end{align*}

I’ve replaced the letters with symbols here because sometimes it can be easier to think about replacing a symbol with a complex formula. Here’s what we know so far:

The first line is this: (G $$\rightarrow$$ $$\neg$$T), which means that the happy face is G and the delta/triangle is...

Uhhhh... T?

Close! But not quite. The first line of MT just deals with whatever the antecedent is negations and all and whatever the consequent is negations and all. So the delta or triangle symbol is actually $$\neg$$T. It’s whatever comes after the arrow.

So...that would mean $$\neg \Delta$$ is $$\neg$$$$\neg$$T!

You got it! See? This isn’t so scary. Okay, so we need to find $$\neg$$$$\neg$$T somewhere in the premises.

Ooooooo! I found it:

3. [$$\neg$$W $$\rightarrow$$ ((Z $$\rightarrow$$ F) $$\leftrightarrow$$ Q)] $$\rightarrow$$ ($$\boxed{\neg \neg \text{T}}$$ $$\vee$$ $$\neg$$F)

Yep! Premise 3 has a $$\neg$$$$\neg$$T in its consequent. Let’s just focus in on that consequent for a second:

($$\neg$$$$\neg$$T $$\vee$$ $$\neg$$F)

How do we break off on disjunct from a disjunction? Disjunctive syllogism! What else do we need to complete a disjunctive syllogism!

Ummm.... I’m stumped again.

A DS has the following form:

\begin{align*}{} &P \vee Q \\ & \underline{\neg Q \ \ \ \ \ \ \ } \\ & P \end{align*}

Here, our P is $$\neg$$$$\neg$$T and our Q is $$\neg$$F, which means our $$\neg$$Q is $$\neg$$$$\neg$$F. Following so far?

Yeah, I think so. If we just substitute $$\neg$$F in for all of the Q’s, then we get this, right?

\begin{align*}{} &\textbf{P} \vee \neg F \\ & \underline{\neg \neg F \ \ \ \ \ } \\ & \neg \textbf{P} \end{align*}

Right-o! Well done. And then if we put in $$\neg$$$$\neg$$T for the P’s, then we get this:

\begin{align*}{} & \neg \neg T \vee \neg F \\ & \underline{\neg \neg F \ \ \ \ } \\ & \neg \neg T \end{align*}

So, now we’re looking for a $$\neg$$$$\neg$$F to complete our DS. And it turns out that premise 4 just is $$\neg$$$$\neg$$F. Bravissimo! So we know we can do our DS when the time comes. Remember, though, that we were looking at the consequent of a conditional premise. How do we break the consequent out of a conditional?

Modus Tollens!

Yes! You got it. What do we need to find in order to complete Modus Tollens?

The...Er...Um...Antecedent?

Say it with confidence!

The antecedent!

Now we’re getting somewhere. What’s the antecedent to premise 3?

This mess?

3. $$\boxed{[ \neg W \rightarrow ((Z \rightarrow F) \leftrightarrow Q)]}$$ $$\rightarrow$$ ($$\neg$$$$\neg$$T $$\vee$$ $$\neg$$F)

Yes, the main operator in premise 3 is the conditional that comes after the section you’ve circled. And whatever comes before the main conditional operator is the antecedent. Looking at that antecedent all by itself:

[$$\neg$$W $$\rightarrow$$ ((Z $$\rightarrow$$ F) $$\leftrightarrow$$ Q)]

What sort of formula is this? In other words, what’s the main operator?

It’s another conditional, right?

Yep! Let’s look at premises 1 and 2:

1. [$$\neg$$$$\neg$$(D $$\vee$$ F) $$\rightarrow$$ ((Z $$\rightarrow$$ F) $$\leftrightarrow$$ Q)]

2. [$$\neg$$W $$\rightarrow$$ $$\neg$$$$\neg$$ (D $$\vee$$ F)]

Do you see how that antecedent above is related to premises 1 and 2? It’s the antecedent from one premise combined with the consequent from another premise.

1. [$$\enclose{circle}{\cancel{\neg \neg (\text{D} \vee \text{F})}} \rightarrow \enclose{circle}{((\text{Z} \rightarrow \text{F}) \leftrightarrow \text{Q})]}$$

2. [$$\enclose{circle}{\neg \text{W }} \rightarrow \enclose{circle}{\cancel{\neg\neg (\text{D} \vee \text{F})}}$$]

That means we’re going to use which of our 4 rules?

Hypothetical syllogism!

Yes! Well done. Now we’ve gone through the whole problem backwards and we’re ready to go through the derivation forwards on paper.

Okay, so I first copy the problem down:

1. [$$\neg$$$$\neg$$(D $$\vee$$ F) $$\rightarrow$$ ((Z $$\rightarrow$$ F) $$\leftrightarrow$$ Q)]

2. [$$\neg$$W $$\rightarrow$$ $$\neg$$$$\neg$$ (D $$\vee$$ F)]

3. [$$\neg$$W $$\rightarrow$$ ((Z $$\rightarrow$$ F) $$\leftrightarrow$$ Q)] $$\rightarrow$$ ($$\neg$$$$\neg$$T $$\vee$$ $$\neg$$F)

4. $$\neg$$$$\neg$$F

5. (G $$\rightarrow$$ $$\neg$$T) / $$\neg$$G

And then I work backwards from the order we just discussed. So, the last thing we talked about was doing a HS on 1 and 2:

6. [$$\neg$$W $$\rightarrow$$ ((Z $$\rightarrow$$ F) $$\leftrightarrow$$ Q)] 1, 2, HS

Excellent!

Then I do a Modus Ponens with 3:

7. ($$\neg$$$$\neg$$T $$\vee$$ $$\neg$$F) 3, 6, MP

Then I’m ready to do a Disjunctive Syllogism:

8. $$\neg$$$$\neg$$T 4, 7, DS

And finally, a Modus Tollens! We’re so close I can taste it!

9. $$\neg$$G 5, 8, MT

And we’re done!

Very nice! Don’t forget to make sure that the last line you’ve derived is identical to the conclusion we were meant to derive.

It is! $$\neg$$G is the same as $$\neg$$G!

See? Even on this behemoth of a problem with 9 total lines (!!!) you were able to think through it step-by-step and solve the derivation. Don’t be intimidated by long or complex problems. This is all pretty mechanical and rote, even when it gets complex. There’s no magic special ability called “logical reasoning” that some of us have and some of us don’t. YOU CAN DO THIS!

[1] But seriously, some people call this “modus tollendo ponens” or “mode of positing by denying”.

This page titled 7.2: Basic Rules of Implication is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Andrew Lavin via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.