Shantila's Inside Logic #12

Biconditionals and Logical equivalence

We have seen that conditionals sometimes only go "one way." For example, "If Sally lives in Cleveland, she lives in Ohio" can be represented as C>O, and it certainly should not be interpreted as meaning O>C. She may live in Ohio without living in Cleveland. Similarly C>O does not guarantee ~C>~O. In general, a conditional P>Q does not entail either Q>P or ~P>~Q. That is, from P>Q it is not valid to conclude either Q>P or ~P>~Q.

Here is another example illustrating the same point from simple arithmetic.

If a number is larger than 5, then it is larger than 2.

This is a truth about simple arithmetic. From this true statement it obviously does not follow that "if a number is larger than 2 it is larger than 5". For example, 3 is larger than 2 but it is not larger than 5. It also does not follow that "If a number is not larger than 5, then it is not larger than 2." Think again of 3.

But sometimes it turns out that a conditional does go "both ways,' that is both P>Q and Q>P are true. Recall the example from Britney's discussion of her clothes. She told Larry King she always wears her gold heels and her gold blouse together. In English we can say, "she wears the gold heels if, and only if, she wears the gold blouse." Let us examine both ideas that are expressed here:

(a) she wears the gold heels if she wears the gold blouse. B>H

(b) she wears the gold heels only if she wears the gold blouse. ~B>~H

Statement (a) is the usual sort of "if" statement we have been looking at since the the first day of this logic course, that is, B>H. This sentence means that her wearing the gold blouse guarantees (or, as is sometimes said, is sufficient for) her wearing the gold heels.

On the other hand, statement (b) tells us that her wearing the gold blouse is necessary for her wearing the gold heels, and we have represented it as ~B>~H: it means that if she doesn't wear the gold blouse she doesn't wear the gold heels.

Now one interesting feature of the statement ~B>~H is that it means just the same thing as H>B. Think about the meanings of these two statements in the example. We can show that they mean the same thing by proving both of the following sequents (as you are asked to do in the Practice exercises below; notice that this basically is review because we already proved 37 in an earlier chapter and 57 is equally straightforward using CP).

37 P>Q } ~Q>~P

57 ~Q>~P } P>Q

In a situation like this, the two sentences P>Q and ~Q>~P can be proved to mean the same thing (assuming one we can prove the other, and vice versa).

Since H>B and ~B>~H mean the same thing, H>B is an equally good interpretation of the English sentence (b) "she wears the gold heels only if she wears the gold blouse."

Now consider again the "if & only if" sentence: "she wears the gold heels if, and only if, she wears the gold blouse."This construction is so important we can give it a special symbol.

We will use <> so that P<> Q is called a biconditional and says "P if and only if Q" and we regard P<>Q serves as an abbreviation for (Q>P)&(P>Q).

Taking (a) and (b) together in the example about Britney's clothes, where the conditional goes both ways, we can use our new symbol <> to abbreviate the "if and only if" sentence simply as B<>H.

To consider another example of a biconditional, recall Britney's reasoning about the pregnancy test. She initially used the conditional, "if you are pregnant, the indicator will turn blue," P>B. One's being pregnant guarantees the indicator will turn blue. Of course such tests, when working properly, are such that ~P>~B also is true, "the indicator will turn blue only if you are pregnant" (that is, if you are not pregnant, the indicator will not turn blue) -- which (as we saw just above in the sequents 37 and 57) means the same thing as B>P. So if the test indicator is working properly, one's being pregnant is both "necessary and sufficient" for the indicator turning blue. That is, P<>B.

Since the new symbol <> is introduced so that P<>Q is an abbreviation for (P>Q)&(Q>P), our two rules for <> are easy and very straightforward.

Here are two examples in which the new Rule is used.

58 P<>Q } P>Q

59 P<>Q } Q>P

Notice that in these two examples 58 and 59, we can prove a conditional without using the CP strategy. Here is another example in which the new Rule is used, as well as CP.

60 P<>Q } ~P>~Q

The new <>E rule is used in an obvious way here on line 3 and CP is used on line 5.

Now here is the "Introduction" rule for biconditionals.

Here are a couple of examples in which this Rule <>I is used.

61 P>Q, Q>P} P<>Q

61 P>Q, ~P>~Q } P<>Q

The key here is to see that we can derive P<>Q using <>I if we have both P>Q (our first premise) and Q>P, which we derive on line 7 using CP (assuming Q on line 3 and deriving P on line 6).

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Logical equivalence

Earlier we discussed H>B and ~B>~H having the same meaning; a more precise way of expressing this idea is to say that the two sentences are equivalent.

We will say that two sentences R and S are logically equivalent just in case R<>S is a logical truth (as discussed in the previous chapter). This means of course that R and S are logically equivalent just in case the sequent } R<>S is valid.

So, for example, the sentences P&Q and Q&P are logically equivalent, since the following sequent is valid, as is easily proven (see Practice exercise 12.4 below).

68 } (P&Q) <> (Q&P)

Given the valid sequents 37 and 57 discussed above, we also can prove the following, which means that P>Q and ~Q>~P are equivalent.

69 } (P>Q) <> (~Q>~P)

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*Practice

12.1 Prove the following sequents using the Rules of our system. (These are review.)

37 P>Q } ~Q>~P

57 ~Q>~P } P>Q

12.2 In reasoning with a biconditional like P<>B the fallacies Bad/AC and Bad/DA are not possible because the conditional goes both ways in that case. In reasoning with P<>B, as in the pregnancy test example, either P or B can be used to derive the other (likewise for ~P and ~B). Show that these statements are true by using the Rules of our system (including the new rules <>E and <>I) to prove each of the following sequents:

58 P<>B, P } B

59 P<>B, B } P

60 P<>B, ~P } ~B

61 P<>B, ~B } ~P

12.3 Prove each of the following sequents using the Rules of our system.

62 P<>Q } (P>Q)&(Q>P)

63 P<>Q } (P>Q)&(~P>~Q)

64 (P>Q)&(Q>P) } P<>Q

65 P<>Q, P>R } Q>R

66 P<>Q, Q<>R } P<>R

67 P>R, ~P>~Q, R>Q} P<>Q

12.4 (a) Prove that (P&Q) and (Q&P) are logically equivalent by proving the following sequent.

68 } (P&Q) <> (Q&P)

(b) Prove that (P>Q) and (~Q>~P) are logically equivalent by proving the following sequent. (Hint: use the proofs of 37 and 57 above in 12.1).

69 } (P>Q) <> (~Q>~P)

12.5 Symbolize this argument and construct a derivation to show that it is valid. I will go if and only if Sally goes. I will drive if and only if Joe does not go. Both Sally and Joe are going. So I will go but I will not drive.