1
Conditional P->Q: If P is True, then Q is also True.
Truth table
P Q P->Q
T T T
T F F
F T T
F F T

BiConditional P<->Q: If P is True, then Q is also True; If Q is True, 
then P is also True.
P Q P<->Q
T T T
T F F
F T F
F F T

The motivation for the truth-value assignments made for the con-ditional is the fact that, as intuitively understood, P->Q is true if Q is deducible from P in some sort of way.

So, if P is true and Q is false, we want P->Q to be false, which accounts for the second line of the table.

My understanding of this is as follows: We are asserting that IF P is TRUE then Q also is TRUE; therefore if Q happens to be False for a True P, then our assertion is FALSE. But remember here that we are giving/inserting real values to P and Q and then determining/evaluating the validity of P->Q (our assertion).

Next, suppose that Q is true. Then, independently of P and its truth value, it is plausible to assert that P->Q is true. This reasoning sug- gests the assignments made in the first and third lines of the table.

Wo! How?? What reasoning!!! He hasn't reasoned anything! The weasel! He just made a supposition and then a bloody assertion!

To justify the fourth line, consider the statement (P A Q) -> P. We expect this to be true regardless of the choice of P and Q. But, if P and Q are both false, then P A Q is false, and we are led to the conclusion that if both antecedent and consequent are false, a conditional is true.

What? He's saying: IF (P AND Q is TRUE) then P is also true. How the heck can that statement be independent of the values assigned to P and Q?? Still he says "We expect this to be true.." I don't expect it!! What's going on!!

I tried thinking of this another way:

P->Q implies: IF P is TRUE then Q also is TRUE
P<->Q implies: IF P is TRUE then Q also is TRUE; AND
               IF Q is TRUE then P also is TRUE

Let's say we make an assertion P->Q = [IF P is TRUE then Q also is TRUE] If we insert values:

 T T then our assertion P->Q is T
 T F then our assertion P->Q is False
 F T Not Applicable since we can't derive a Q value from a False P!
 F F Not Applicable since we can't derive a Q value from a False P!

For the Biconditional (making our assertion in both directions)

T T T
T F F
F T F
F F Not Applicable since we can't determine if both values are False

Why is Stoll right?????

Brian M. Scott
  • 616,228
  • He’s right because that is in fact how the conditional $\to$ is defined in propositional logic. You may feel that he hasn’t given you an adequate explanation of why we define it that way, but that’s a separate issue. (And I can assure you that Bob would not have taken kindly to being called a weasel!) In fact I agree that he could have done a better job of giving an intuitive justification of the third line, but his justification of the fourth line is quite reasonable. – Brian M. Scott Oct 19 '13 at 07:00
  • hopefully he's not in the land of the living and can't strangle me - seriously, look at what he says here (pg 163): " We agree that <-> is the strongest connective (that is, it is to encompass most), and then follows ->. Next in order are V and A, which are assigned equal strength, and then follows -| (me: NOT symbol) the weakest connective." But the Wikipedia says something seemingly different: https://en.wikipedia.org/wiki/Logical_connective#Order_of_precedence AND and OR don't have equal precedence and NOT the weakest connective has the highest precedence! –  Oct 19 '13 at 09:29
  • It's a good book but there are these patches that startle me (like sudden gunfire)! Plus I'm not calling him a weasel - that's silly, just parts of his logic in the book :) –  Oct 19 '13 at 09:31
  • He does say: "These five tables are to be understood as definitions; they arc the customary definitions adopted for mathematics. We have made merely a feeble attempt to make them seem plausible on the basis of meaning." which is why i like the book - though I wish he had placed that at the beginning and not the end., –  Oct 19 '13 at 09:42
  • Bob died a little over 20 years ago, so I expect that you’re safe. :-) – Brian M. Scott Oct 19 '13 at 17:43

3 Answers3

1

You would agree, surely, from your experience with arithmetic that the following is true?

If $x=1$, then $x+1 = 2$

But wait, you don't actually know that $x=1$. Maybe $x$ is 2, or seventeen!

The way I like to justify the truth table is by looking at how we can use truth values to make inferences. Here are some exercises to try given Stoll's values for the truth table, and again for any other definition you choose.

1. Suppose you know that $A \to B$ is true.

  • Can $A$ be false?
  • Can $B$ be false?

If your answer to both of these questions is "no", then what you have in mind is "A and B", not "if A then B".

2. Suppose you know that $A \to B$ is true. Suppose you also know that $B$ is true. Can $A$ be false?

If your answer to this question is "no", then you have the conditional backwards: you have in mind "if B then A", or equivalently "A if B".

3. Suppose you know that $A \to B$ is true. Suppose you also know that $A$ is true. Can $B$ be false?

The answers "yes and yes, yes, no" collectively determine that the truth table must be of the form Stoll describes.

  • umm.. there's no maybe x = 2 because the equality won't be satisfied - so I don't follow that. If A->B is true then A and B CANNOT be false and I can't even begin to grasp what you are implying about "A and B".. sorry.. –  Oct 19 '13 at 09:40
  • @paleywiener: It's okay if $x=1$ isn't satisfied; a lot of the time $x$ really is not one! But those times when $x$ is one, we know that $x+1$ will be $2$: that's what the conditional expresses. –  Oct 19 '13 at 13:37
0

As a comment ( don't have enough points yet), maybe this would help:

http://en.wikipedia.org/wiki/Material_conditional

BFD
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0

A common assumption goes that we can determine the truth value of a compound statement from atomic components. In other words, if we input truth values for its components, then we will get a truth value for the compound statement. For example, if we have some binary connective X and we input truth values for "p" and "q" in (pXq), we will get some truth value for (pXq). Additionally, the other common assumption goes that we have a constant, fixed truth set for both atomic statements and compound statements. Since we assume that we only have two truth values, this means all compound statements will take on some truth value in {true, false}. Without these assumptions, you end up studying something different than what Stoll is talking about.

Now, given the above assumptions what are "<->" and "->". Well, if "(p<->q)" holds true, then "p" and "q" have the same truth value, and if "p" and "q" have the same truth value, then (p<->q) holds true. Consequently, if p holds true, and so does q, then (p<->q) holds true. If p is false, and q is true, then (p<->q) is false. If p is true, and q is false, then (p<->q) is false. If p is false, and q is false, then (p<->q) is true. Thus, what Stoll has <-> makes sense.

Now, you've agreed that if p is true, and q is true, then (p->q) is true. Also, you've agreed that if p is true, and q is false, then (p->q) is false. Consequently, we have four possibilities for ->.

1:

 A  f  t
 f  f  f
 t  f  t

2:

 B  f  t
 f  f  t
 t  f  t

3: 

 C  f  t
 f  t  f
 t  f  t

4: 

 D  f  t
 f  t  t
 t  f  t

Now 3 is <->. So, it can't be ->, since we already think these distinct. 1 is logical conjunction. So, similarly we can reason that it can't be ->. Could -> be 2? Well, if it did work out that way, then (p->p) would NOT be a tautology. So, if you want (p->p) as a tautology, then -> cannot be 2. This leaves us with -> as 4 or operation D.

Doug Spoonwood
  • 11,211
  • Your para-1 is in essence saying that my remark about things being "Not Applicable" is wrong: it's a "composite sentence" built from an "atomic/prime-sentence" and it can have only True or False values and must take a value. I agree! –  Oct 19 '13 at 15:15
  • Now, considering the truth-table for (IFF) <-> , given that I have to insert a True or False value in place of "Not Applicable" what should it be - True or False. Should it be F F F or F F T. I dunno.. flip a coin? Perhaps that's what he did and then adjusted "Conditional" to match! I don't understand your para-2.. "Well, if "(p<->q)" holds true, then "p" and "q" have the same truth value" -- why should they have the same truth value - don't see how your para-1 results in this conclusion. –  Oct 19 '13 at 16:09
  • @paleywiener Stoll didn't just flip a coin. Every single logician agrees with the truth table for these two connectives if we have only 2 truth values. Iff p, thenn q, means that "if p, then q" holds true and "if q, then p" also holds true. Suppose p true. By modus ponens (or detachment) and "if p, then q" as true, q holds true also. Suppose p not true. Then p qualifies as false. But, since "if q, then p" holds true, by modus tollens q is also not true, and thus false. So, q always has the same truth value as p if we have "iff p, thenn q". – Doug Spoonwood Oct 21 '13 at 23:35
  • ah! thank you! https://en.wikipedia.org/wiki/Modus_ponens –  Oct 22 '13 at 10:26