PROP has all the formal proof rules of BOOL. But we need to add Intro and Elim rules for the conditional and biconditional.

PROP is an extension of BOOL, so PROP has all the formal proof rules that BOOL does. But we need to add Intro and Elim rules for the conditional and biconditional.

Reasoning from a conditional is as natural as it gets. For example,

1. If Pia stole diamonds, then Quinn stole rubies.

2. If Quinn stole rubies, then Raquel didn’t steal rubies.

3. Raquel stole diamonds or rubies.

4. Pia stole diamonds.

Here’s how to reason from a conditional: if you have a conditional, and you have the antecedent, you can infer the consequent. Premise 1 says if Pia stole diamonds, then Quinn stole rubies, and premise 4 says that Pia stole diamonds. Hence we can infer that Quinn stole rubies.

**Modus ponens (MP):** from P and if P then Q, you can infer Q.

Basically, conditionals are made for this sort of inference: all the conditional says is if the antecedent is true, then the consequent is true, so anytime you have the antecedent, you can infer the consequent.

That inference is so famous it got a Latin name: * modus ponens*, or

**Informal Proof Guidelines:** whenever you do modus ponens in a proof write “by MP”.

Since modus ponens is a well-known proof method, we can make our informal proofs clear for our audience by writing “by MP” whenever we use it. For example, here’s what an informal proof of that conclusion would look like:

Proof. Since we know Pia stole diamonds (premise 4), and if Pia stole diamonds, then Quinn stole rubies (premise 1), we can infer Quinn stole rubies (by MP). Since we also know that if Quinn stole rubies, Raquel didn’t steal rubies (premise 2), we can infer Raquel didn’t steal rubies (by MP). Since we also know that Raquel stole diamonds or rubies, and it wasn’t rubies, she must have stolen diamonds (shortened proof by cases). Done.

Notice a few things about this proof: we used MP in multiple places, and cited it each time. We also used the shortened version of proof by cases. You could correctly spell it all out explicitly considering each case, if you wanted to.

Now you can practice. Here’s another argument:

1. If Pia is guilty, then Quinn is innocent.

2. If Quinn is innocent, Raquel is innocent.

3. Pia is guilty.

Thus,

4. Raquel is innocent.

Next consider this argument:

1. If Pia is guilty, then Quinn is guilty.

2. Quinn is guilty.

Thus,

3. Pia is guilty.

There are two things to keep in mind when reasoning from conditionals. First, make sure you are actually doing modus ponens! MP is when you have a conditional and the antecedent. That allows you to infer the consequent.

In the argument above, you have a conditional and the consequent. That does you no good whatsoever. Conditionals don’t go in reverse: P→Q is not the same as, and does not entail, Q→P.

So if you just have P→Q and Q, you can’t infer anything special using the conditional. We say “anything special using the conditional” because of course you can infer things like ~~Q, or (P→Q)&Q, but those inferences don’t use the conditional in an important way.

**Affirming the consequent:** P→Q and Q, therefore P. Remember this is a * fallacy*, not a proof method!

Inferring P from P→Q and Q can be tempting, though, which is why it is a well-known fallacy called * affirming the consequent*.

The second thing to keep in mind is that there are many ways of expressing conditionals in English, so make sure you get your conditionals right.

Reasoning from a biconditional is just as easy as reasoning from a conditional. After all, a biconditional just is two conditionals squeezed together.

**With ↔ you can do MP in either direction.**

The only difference is that with a biconditional you can do modus ponens in either direction. From P↔Q and P you can infer Q (by MP), and from P↔Q and Q you can infer P (by MP).

To put it another way, with ↔ there is no fallacy of affirming the consequent, because you have both conditionals; the consequent of one is the antecedent of the other, and hence a valid use of MP.

Here’s a valid argument you figured out long ago:

1. Pia, Quinn or Raquel is guilty.

2. If Pia is guilty, then Quinn is guilty.

3. If Raquel is guilty, then Pia or Quinn is guilty.

Thus,

4. Quinn is guilty.

Now you can give a proof of it.

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