Reason and Argument Chapter 6 (2/3)

A symbol for the exclusive ‘or’ We will use ұ for the exclusive ‘or’ Strictly speaking, this connective is not necessary. We could just as easily use ‘((p v q) & ~(p & q))’ for ‘p ұ q’ I shall generally avoid use of the exclusive ‘or’, though it be in the book. pqp ұ q TTF TFT FTT FFF

Conditionals Conditionals are statements of the form ‘If _______ then _______’ where the blanks are filled with sentences that express propositions. Again, the first blank is called the antecedent and the second blank is called the consequent.

The kinds of conditionals that are allowed to be ‘  ’ There is some dispute as to whether there are any material conditionals (the kind that are expressed by ‘  ’) in English. In any case, we shall use ‘  ’ for any and all conditionals in the PRESENT tense, ACTIVE voice, INDICATIVE mood.

Subjunctive Conditionals Consider the sentence “If the Germans had won WWII, then we would all be speaking German” A statement of the form ‘p  q’ would involve two separate propositions, and would connect them in the appropriate way. ‘The Germans had won WWII’ does not really express a proposition all by itself, nor does ‘we would all be speaking German’. Further, in the kind of conditional we are concerned with, someone does not commit themselves to the truth or falsity of the antecedent. In a counterfactual conditional as above, one is committed to the falsity of some particular state of affairs (in this case that ‘the Germans do not win WWII’). The above, therefore, is simply ‘p’, not ‘p  q’

The (seemingly) wacko truth table for conditionals: Assume p is “The pitcher throws a fastball” and q is “The batter hits a home run” Line 2 is very straightforward. If Bob bets you that if the pitcher throws a fastball then the batter will hit a home run, Biff will lose his bet if things turn out as on Line 2. But what about the others? pq p  q TT? TFF FT? FF?

Conditional Truth Table Line 1 seems equally straightforward. Biff wins his bet by virtue of saying something true, just as on line 2 he would lose his bet by virtue of saying something false. But what happens when the antecedent is false? pq p  q TTT TFF FT? FF?

Conditional Truth Table Imagine you’re watching the game and Biff makes his bet. You accept, and you see the pitcher throw a curve ball (i.e. NOT a fastball, making the proposition ‘p’ false) yet the batter still hits a home run (making the proposition q true, as on line 3). The best way to interpret this is that the bet is neither won nor lost, and no money changes hands. This would also be the case if the batter has swung at and missed the curveball (line 4). But since we still have to assign one or the other truth values to ‘if p then q’ what do we do? pq p  q TTT TFF FT? FF?

Do I really need to type the title for this slide again? We give the conditional phrase the benefit of the doubt. We have better reasons for saying that the conditional is not false than we have reasons to say it’s not true. Also there are more weird problems that result from taking the conditional to be false in lines 3 and 4 than result from taking it to be true. This comes through more clearly when dealing with conditionals that are not predictions. “If it is raining then the ground is wet” is a true conditional even if it’s not raining. pq p  q TTT TFF FTT FFT

One of those reasons: Material Implication If we focus on the second line of the truth table for conditionals, which was a clear case, we can see that having a true conditional must mean that it is not the case that the antecedent is true and the consequent false. Formalized, that looks like this: –a true conditional (p  q) implies that it is not the case that (p is true and q is false) –or: ~(p & ~q) –By DeMorgan’s Law, ~p v q (read as “It is not the case that p unless q is the case”) is equivalent to the above

If p  q, ~(p & ~q), and ~p v q are equivalent: pq p  q ~(p & ~q)~p v qp & ~q~q~p TTTTTFFF TFFFFTTF FTTTTFFT FFTTTFTT

meet the MODI Modus Ponens p  q p____ q Modus Tollens p  q ~q___ ~p

Valid Moduses: P2CP1 pq p  q TTT TFF FTT FFT P1P2C pq p  q ~q~p TTTFF TFFTF FTTFT FFTTT

A couple common fallacies: (and trouble with conditionals in general) Affirming the consequent p  q q____ p Denying the antecedent p  q ~p___ ~q

Hypothetical Syllogism (Chain Argument) p  q q  r p  r Valid?

Another reason for conditionals to be true when antecedent is false: If lines 3 and 4 of the truth table for conditionals are replaced by either the value ‘F’ or ‘N’ for ‘neither’, we get some highly counterintuitive results: –Modus Tollens is such that the premises are never true at the same time –Denying the Antecedent has the same status as Modus Tollens –Affirming the Consequent comes out valid!

Procedure for using truth tables to find validity: 1.Create the reference columns (one per propositional variable, in alpha. order) 2.Create one column for each logical connective (v,&,~,  ) 3.Fill in reference columns: a)# of rows = 2 n where n = # of propositional variables b)Fill first half of leftmost column’s rows with value ‘T’, the rest with value ‘F’ c)Fill next column with T and F, beginning with T and having exactly half as many consecutive iterations of T and F as occurs in column leftward. d)Repeat (c) until reference columns are filled in. 4.Fill in the rest of the table 5.Check for any case in which the premises are all true and the conclusion is false. If such a case is on the table, the argument form is invalid, valid otherwise.

Example: Ex. 24 #5 Argument is: p  q q  r ~r__ ~p Step 1: Create the reference columns: pqr

Step 2, One column for each connective: pqr p  qq  r ~r~p

Step 3: Fill In Reference Columns pqr p  qq  r ~r~p TTT TTF TFT TFF FTT FTF FFT FFF

Step 4: Fill in remainder pqr p  qq  r ~r~p TTTTTFF TTFTFTF TFTFTFF TFFFTTF FTTTTFT FTFTFTT FFTTTFT FFFTTTT

Check for Validity: Premise 1Premise 2Premise 3Conclusion pqr p  qq  r ~r~p 1TTTTTFF 2TTFTFTF 3TFTFTFF 4TFFFTTF 5FTTTTFT 6FTFTFTT 7FFTTTFT 8FFFTTTT