Definition 2.5.1. Modus ponens.
The deduction
- \(P\) implies \(Q\) is true, and
- \(P\) is true
- hence \(Q\) must be true.
is called modus ponens.
Section 2.5 Modus ponens and chaining implications
Subsection 2.5.1 Modus ponens
Once we have proved an implication
\begin{gather*}
P \implies Q
\end{gather*}
to be always true, then we would like to make use of it. We will (soon) prove that the implication
If \(n\) is even then \(n^2\) is even
is always true. Since we know it cannot be false, we’ll write down the relevant 3 rows of its truth table and suppress the 1 row corresponding to the implication being false:
| \(P\) | \(Q\) | \(P \implies Q\) |
| T | T | T |
| F | T | T |
| F | F | T |
Notice that if we take a number like \(n=14\text{,}\) then we know it is even and so the hypothesis is true. By the above truth-table we know that the conclusion must also be true — the number \(n^2=14^2\) is even. We don’t have to do any more work; the truth table, and our proof, ensure that the conclusion is true.
More generally, if we have proved that
\begin{align*}
P \amp \implies Q
\end{align*}
is always true, then if we know the hypothesis \(P\) is true, then the conclusion \(Q\) must also be true.
This logical deduction was first formalised by Theophrastus 29 .
Notice that if we have proved the implication \(P \implies Q\) to be true, but the hypothesis \(P\) is false, then we cannot conclude anything about the truth value of the conclusion. When the hypothesis is false, the truth value of the conclusion doesn’t matter — the implication is still true. We can verify that by considering the relevant two rows from the truth-table of the implication.
| \(P\) | \(Q\) | \(P \implies Q\) |
| F | T | T |
| F | F | T |
Similarly notice that if we have proved the implication \(P \implies Q\) to be true, and we have proved the conclusion \(Q\) to be true then we cannot conclude anything about the truth value of the hypothesis \(P\text{.}\) Again, this is easily verified by considering the relevant two rows from the truth-table of the implication.
| \(P\) | \(Q\) | \(P \implies Q\) |
| T | T | T |
| F | T | T |
There is, however, one more instance in which we can make a valid conclusion. Consider again the truth-table when the implication \(P \implies Q\) is true but the conclusion \(Q\) is false:
| \(P\) | \(Q\) | \(P \implies Q\) |
| F | F | T |
Here the only possibility is that the hypothesis must be false. This allows us to make another valid deduction.
Definition 2.5.2. Modus tollens.
The deduction
- \(P\) implies \(Q\) is true, and
- \(Q\) is false
- hence \(P\) must be false.
is called modus tollens.
So when we know (back to our silly example) that
If he is Shakespeare then he is dead.
then we can conclude that any live person is not Shakespeare. We will come back to modus tollens a little later in Section 2.6 when we examine the contrapositive.
Subsection 2.5.2 Affirming the consequent and denying the antecedent
Misapplication of modus ponens is a frequent source of logical errors. An extremely common one is called “affirming the consequent”.
Warning 2.5.3. A common logical error.
The false deduction
- \(P\) implies \(Q\) is true, and
- \(Q\) is true,
- and hence \(P\) must be true
is called affirming the consequent.
The flow of logic is wrong — check the truth table. Also notice that our arrow notation for the implication, \(\implies\text{,}\) helps to remind us that truth should flow from the hypothesis to the conclusion and not the other way around.
To see just how wrong this can be, consider again the true implication
If he is Shakespeare then he is dead.
If we were to affirm the consequent, then any dead man must be Shakespeare.
Affirming the consequent does occasionally get used as a rhetorical technique (especially by purveyors of nonsense):
If they are Galileo then they are supressed. I’m being supressed, so I must be Galileo.
or (when being a bit sorry for oneself)
If they are a great artist then they are misunderstood. I’m misunderstood so I must be a great artist.
and the social-media comment section fallacy (with thanks to this comic 30 ):
If I tell the truth then I will offend people. I am offending people, so I must be telling the truth.
So be careful of affirming the consequent — it shows up a lot and is always fallacious.
A very similar logical error is called “denying the antecedent”
Warning 2.5.4. Another common logical error.
The false deduction
- \(P\) implies \(Q\) is true, and
- \(P\) is false,
- and hence \(Q\) must be false
is called denying the antecedent and is a misapplication of modus tollens.
Here are some examples:
If I have been to Toronto then I have visited Canada. I have not been to Toronto. So I have not visited Canada.
If he is Shakespeare then he is dead. Abraham Lincoln is not Shakespeare, so he must be alive.
If tastes bad then it must be healthy. This tastes good, so it must be unhealthy.
Subsection 2.5.3 Chaining implications together
When we construct a proof that \(P \implies Q\) is true, we don’t do it in one big leap. Instead we break it down into a sequence of smaller (and easier) implications that we can chain together. To see how this works, consider the following:
Result 2.5.5.
Let \(P,Q\) and \(R\) be statement. Then the following statement is always true:
\begin{gather*}
\Big( (P \implies R) \land (R \implies Q) \Big) \implies (P \implies Q).
\end{gather*}
Since we are going to need to refer to this piece of mathematics a few times in this section, we have take the trouble to format it clearly and given it a number.
This result is an example of a tautology, a statement that is always true. We will come back to tautologies later in the text. To show that it is always true we could either build the truth-table, or we can do some reasoning. Both of these methods are proofs, but we won’t be quite so formal until the next chapter. The truth-table is not hard to construct but a bit tedious; since each of \(P,Q,R\) can either be true or false, there are \(2^3=8\) rows to consider:
| \(P\) | \(Q\) | \(R\) | \(P\implies R\) | \(R\implies Q\) | \(P\implies Q\) | The statement Result 2.5.5 |
| T | T | T | T | T | T | T |
| T | T | F | F | T | T | T |
| T | F | T | T | F | F | T |
| T | F | F | F | T | F | T |
| F | T | T | T | T | T | T |
| F | T | F | T | T | T | T |
| F | F | T | T | F | T | T |
| F | F | F | T | T | T | T |
The above is a perfectly reasonable way to show that the statement is always true. However, one can do the same just by a little reasoning; it also has the benefit of improving our understanding of the statement. We’ll present the argument in dot-point form:
- The statement is an implication with hypothesis \(((P \implies R) \land (R \implies Q))\) and conclusion \((P \implies Q)\text{.}\) An implication is false when the hypothesis is true but the conclusion is false, and otherwise the implication is true.
- Since the conclusion, \((P\implies Q)\text{,}\) is itself an implication, it can only be false when its hypothesis is true and its conclusion is false. So we must have \(P\) is true, but \(Q\) is false.
- In order for the hypothesis to be true, both implications, \((P\implies R)\) and \((R \implies Q)\text{,}\) must be true (since a conjunction of two statements is only true when both statements are true).
- Since \(P\) is true, and we require \((P\implies R)\) to be true, we must have \(R\) is true.
- Since \(Q\) is false and we require \((R \implies Q)\) to be true, we must have \(R\) is false.
But since \(R\) is a statement it cannot be true and false at the same time. - So there is no way for us to make the statement false. Since it is never false, it must always be true.
So back to the statement \((P \implies Q)\text{.}\) We’d like to show it is always true, but we cannot do it in one big leap. Instead, assume that we can make two smaller steps and prove that the two implications \((P \implies R)\) and \((R \implies Q)\) are always true. The conjunction of two implications, \((P \implies R) \land (R \implies Q)\text{,}\) is also true and is exactly the hypothesis in Result Result 2.5.5. Since the implication in Result Result 2.5.5 is always true and its hypothesis is true — modus ponens — its conclusion must be true.
So while we could try to prove \((P \implies Q)\) is true in one big leap, it is sufficient to instead prove it is true in two smaller steps \((P \implies R)\) and \((R \implies Q)\text{.}\) More generally when we prove \((P \implies Q)\text{,}\) we will instead prove a sequence of implications:
\begin{align*}
P &\implies P_1 & \text{and}\\
P_1 & \implies P_2 & \text{and}\\
P_2 & \implies P_3 & \text{and}& \dots\\
& \vdots\\
P_n &\implies Q
\end{align*}
where each of these intermediate implications is easier to prove.
Once we have done that, consider what happens if \(P\) is true or \(P\) is false:
- \(P\) is true
- If \(P\) is true, the first implication tells us \(P_1\) is true (modus ponens). Then since \(P_1\) is true, the next implication tells us \(P_2\) is true (again modus ponens). Since \(P_2\) is true, \(P_3\) is true, \(P_4\) is true, and so on until we can conclude \(Q\) is true. Since \(P\) is true and \(Q\) is true, the implication \(P \implies Q\) is true.
- \(P\) is false
- On the other hand, if \(P\) is false then we know, just by looking at the implication truth-table, that the implication \(P \implies Q\) is true.
Notice that when \(P\) is false, the fact that \((P \implies Q)\) is true is immediate and simply relies on the truth-table of the implication; we don’t have to do any work or any reasoning. On the other hand, when \(P\) is true, we do need to work to show that \((P \implies Q)\) is true. For this reason almost all of our proofs will start with the assumption that \(P\) is true. We generally leave the case “\(P\) is false” unstated, assuming that our reader knows their truth-tables.
Theophrastus was a student of Plato, a contemporary of Aristotle, and wrote on everything from botany to logic. Given the fragmented historical record, it is perhaps safer to write that the first record that we have of the formal statement of modus ponens comes from Theophrastus.
www.smbc-comics.com/comic/the-offensive-truth
