Facing some challenges to material conditional

Facing some challenges to material conditional

Héctor Hernández O.

IIF-UNAM

h2o_mat@hotmail.com

My aim in this work is to propose a pragmatic solution to some challenging problems posed by A. Blum, D. Edgington and others to Truth-Functional Analysis (TFA) of the ordinary indicative conditionals. The accusation against TFA is that it yields the validity of some arguments which areintuitively invalid. I argue that, at least in the case of these challenges, the actual guilty of this “undesirable consequence” is not the material conditional, but it is rather a wrong formalization of the arguments.

Alex Blum (1986) has objected to treating ‘(’ as formalization of ‘if ..then…’ of natural language on the ground of an intuitively persuasive counterexample. The example is basedon the following theorem: (1) (P&Q)(R ├ (P(R( ∨ (Q(R(.

Blum argues that, although (1) is a theorem of classical logic, it may well be true that ‘If both Al and Bill insert their keys, then the vault will open’ and yet be false that ‘Either, if Al inserts his key then the vault will open, or, if Bill inserts his key then the vault will open’. However, this example can be understood in twoways: I or II.

(I) A situation where (a) is true, while (b) and (c) are false:
a. ‘If both Al and Bill insert their keys, then the vault will open’
b. ‘if only Al inserts his key then the vault will open’….. (If P then R)
c. ‘if only[1] Bill inserts his key then the vault will open’…(If Q then R)
Now, since P is ‘Only Al inserts his key’ and Q is ‘Only Bill inserts his key’, then‘If (P&Q) then R’ cannot be (a), but ‘if only Al inserts his key and only Bill inserts his key, then the vault will open’, which is a false conditional because of falsehood of (b) y (c). Therefore, in this case, the counterexample disappears.

(II) Since ‘If P then Q’ and ‘If Q then R’ have to be false in the example considered, it is not sufficient to insert Al’s key nor Bill’s key to open thevault. It is necessary to insert both keys in order to open the vault. In fact, in this context, if we wanted to express the conditional (a) using other words, we would say likely something similar to (d): ‘It is necessary to insert both keys, Al’s key and Bill’s key, to open the vault’. This points out that Blum`s example is actually a conditional of the form ‘Only if (P&Q) then R’ instead of‘If (P&Q) then R’.

More reasons in favour of this conclusion are the following: It is clear that neither ‘Not-B’ follows from ‘If A then B’ and ‘Not-A’ (a fallacy called denial of the antecedent) nor ‘A’ follows from ‘If A then B’ and ‘B’ (a fallacy called affirming of the consequent). However, from ‘Only if A then B’ and ‘Not-A’ follows ‘Not-B’ and, similarly, from ‘Only if A then B’ and ‘B’follows ‘A’. Now, note that, in the situation II, from (a) in conjunction with ‘Not-(P&Q)’ it can be inferred ‘Not-R’, and from (a) and ‘R’ it can be inferred ‘(P&Q)’. Therefore, the behaviour of (a) is just that of ‘Only if (P&Q) then R’. Thus, Blum’s example is not an instance of (1), but it seems to be an instance of (2): “R((P&Q) ├ (P(R( ∨ (Q(R(” which, of course, is invalid not only using ‘ifthen’ but also using ‘(’.

Now, let us consider other challenge. Edgignton (1993) has presented the following intuitively invalid argument which, however, when it is symbolized turns to be a valid one:

(1) If God does not exist, then it is not the case that
if I pray my prayers will be answered by Him. (1) (E ( ((P(A)
(2) I do not pray. (2) (P

Therefore, God exists.( E.

First of all, let us consider why the formalization given above is erroneous. Take (1) to mean (1*): (E((P((A), then if God does not exist the speaker is committed to pray, but this is implausible, for the speaker, by means of premise (2), discarded that possibility. Intuitively, the speaker does not have to pray in order to turn (1) true in case...