# How do I counter this argument (and truth-table help)?

``````I will pass the exam or I will not pass the exam.

If I will pass the exam, then I will pass the exam even if I don't study.

If I will pass the exam even if I don't study, then studying is pointless.

If I will not pass the exam, then I will not pass the exam even if I study.

If I will not pass the exam even if I study, then studying is pointless.

Therefore, studying is pointless.
``````

I pass the exam : A
I study : B
Studying is pointless: C

A v ~A
A → (~B → A)
(~B → A) → C
~A → (B → ~A)
(B → ~A) → C
C

``````+---+----+----------------+----------------+-----+
| A | B  |  (~B → A) → C  V  (B → ~A) → C  |  C  |
+---+----+----------------+----------------+-----+
| 1 | 1  |                1                |  1  |
| 1 | 0  |                1                |  1  |
| 0 | 1  |                1                |  1  |
| 0 | 0  |                1                |  1  |
+---+----+----------------+----------------+-----+
``````

I'm not sure how to put this argument into a truth-table (I tried; maybe there are 8 rows instead of 4?) and would appreciate it if someone showed me the correct form. Please tell me whether the following statements which I came up with are correct:

Premises 1,2 and 4 are tautologous.
The argument is valid.
The argument is unsound as premises 3 and 5 are untrue. Can I say something like how you would learn stuff by studying even if you fail the exam?

I'm doing computer science and I'm trying to learn some logic as it's helpful there. Thanks for the help!

• Hint: Your statement list can be rewritten as `(A v ~A) ^ (A -> (~B -> A)) ^... -> C` (I'd recommend reducing before putting it into a truth table). In terms of arguments against, this doesn't take into account the situation in which if you do study you pass (`B -> A`). Feb 26, 2016 at 21:55
• Yes; if you want to use truth table, you have to build it for all the sentential letters involved: A,B,C and thus eight rows. The argument is valid if the conclusion has 1 (TRUE) in every row where all the premises have 1, i.e. in every case where all the premises are TRUE also the conclusion is. Feb 27, 2016 at 8:56
• For the truth-table, it would have to be 8 rows, with other headings such as "A ^ ~B", "~A ^ B" and "[(A ^ ~B) v (~A ^ B)] → C" (this one is the conclusion). Is this correct? Feb 27, 2016 at 12:43

I think "I will pass the exam even if I don't study" should be formulated as a conjunction and not an implication. "~B → A" would be wrong since the statement doesn't say that if you don't study then you will pass the exam, but rather that you will pass the exam despite the fact that you don't study, that is, despite being in conjunction with that fact ("even if" is used in ordinary language to hint at an apparent contradiction, but it has no logical implications, so to speak). So it should be "A Λ ~B" (and then premises 2 and 4 aren't tautologies).

As for the truth table, you should always have 2^n lines, where n is the number of variables. In your case, there are 3 variables, so there should be 8 lines. You should start like this (note the pattern):

``````+---+---+---+
| A | B | C |
+---+---+---+
| 1 | 1 | 1 |
| 1 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 0 | 0 |
| 0 | 1 | 1 |
| 0 | 1 | 0 |
| 0 | 0 | 1 |
| 0 | 0 | 0 |
+---+---+---+
``````
• Ah, then the one case in which A is true and B is true makes "A Λ ~B" false so premises 2 and 4 aren't tautologies. They seemed like tautologies so I thought I couldn't counter the argument using those 2 premises. However, I believe the argument is still valid given the very first tautologous premise of "A v ~A". It proceeds to use a dilemma to show that both ways lead to "C", although the "if,then statements" are questionable. Am I correct? Feb 27, 2016 at 12:39
• Validity can be simply shown using the truth table -- if you find a case where all the premises are true (1) and the conclusion if false (0), then the argument is invalid, otherwise it is valid.
– E...
Feb 27, 2016 at 12:48
• The premises in this case are [A v ~A], [A → A ^ ~B], [~A → ~A ^ B], and the conclusion is [[[A ^ ~B] v [~A ^ B]] → C]. I've just finished drawing the truth table out and there appears to be no case in which the premises are all "1" but the conclusion is "0". So it's valid. Correct me if I'm wrong. Feb 27, 2016 at 12:53
• @Chthonic Zyceus It seems correct.
– E...
Feb 27, 2016 at 12:55
• Thanks! This has helped tremendously with my understanding of truth-tables and also cleared up some confusion regarding the "even if" connector, which was pretty misleading for me. Feb 27, 2016 at 12:57

Premise 2 has the appearances of a tautology because of its phrasing, but it is not a tautology. In English, "I will pass the exam even if I don't study" is a hypothetical statement about a potentially fictional world where you did not study. It would not be assigned the the same predicate as "I will pass the exam" in the first half of the premise. Premise 4 is not tautological for the same reason.

If you wished, you could use A → (~B → A) and ~A → (B → ~A) as the formal notation for those two premises, as you did. If so, you would arrive at the conclusion you arrived at. However, I would not consider those two formally notated statements as good representations of their English equivalents.

If I were to try to convert them, I would consider a universe of worlds, W, and in each world I could have predicates such as PassExam(w) and DidStudy(w). Then I might be able to phrase predicate 2 more accurately as PassExam(w1) → forall(w2){(~DidStudy(w2) → PassExam(w2))}. This would demonstrate the fallacy in the logic. You're assuming that because you can pass the test in this world, you must be able to pass the test in all possible worlds, even the ones you did not study in.

It also may suggest a more accurate tautology: "If forall(w){DidStudy(w)->PassExam(w) and ~DidStudy(w)->PassExam(w)} then forall(w){PassExam(w)}. If I pass the exam if I study, and I pass the exam if I didn't study, then I will pass the exam.

If you put some proper thought into the problem statement you can reduce it to a logically trivial and unchallenging problem. Putting the thought in is a fun challenge nonetheless.

If we look at statement two:

If I will pass the exam, then I will pass the exam even if I don't study.

The statement says that if I will pass the exam, I will pass the exam no matter what. `even if I don't study` is extra "useless" information. You can just as easily say:

``````If I will pass the exam, then I will pass the exam even if I eat an apple.
``````

and everything still holds "true".

Thus, we can reduce the statement to:

`A → A`, which is kind of circular.

Alternatively, you could exchange `even if I don't study` with `whether I study or not` which could be written as:

`A → (B v ~B)` which goes to `A is true`

The same can be applied to statement 4:

`~A → (B v ~B)` which goes to `~A is true`

Following the same principles, statement 3 and 5 can be reduced to:

`3: A → C` and `5: ~A → C`

Because `A` and `~A` lies in the future, we cannot know whether `A` or `~A` will happen and they are mutually exclusive so we should write:

`A v ~A → C` which can be further reduced to `C is true`, thus studying is futile.

The issue is that we assume the absence of causality. Even though we might not know whether we pass the test or not, premise 2 and 4 states the future is already determined and any action taken will not change that.