PDFs of answers to most of the exercises in *IFL* are linked below. (Do let me know of errors, by adding a comment below.)

- Exercises 1
- Exercises 2
- Exercises 3
- Exercises 4
- Exercises 5
- Exercises 7
- Exercises 8
- Exercises 9
- Exercises 10
- Exercises 11
- Exercises 12
- Exercises 13
- Exercises 14
- Exercises 16
- Exercises 17
- Exercises 22
- Exercises 23
- Exercises 24
- Exercises 25
- Exercises 26
- Exercises 27
- Exercises 29
- Exercises 32
- Exercises 33
- Exercises 34
- Exercises 35

Concerning exercise C from chapter 14, It seems to me that the argument 10. from previous exercise can also be considered as problematic when translated using “if … then …” for the reasons explored in chapter 15, as it has “not-P or Q” among its premisses, and the conclusion is in the form “if P then Q”. Consider:

Either Wordsworth didn’t write “The Lord of the Rings”, or he wasn’t a novelist.

It isn’t the case that Wordsworth wasn’t a novelist and that he wrote at least one novel.

So, if Wordsworth wrote “Lord of the Rings”, he wrote no novels.

Here , it seems that if the first premiss is endorsed on the basis that Not-P is asserted, the material conditional in the conclusion won’t be robust.

Just to correct a minor error in your 4th printing (2011) – in your given online answer to Question 7, section D, Exercises 7, I think the ‘Qs’ should be replaced by ‘Ps’. Correct me if this supposed ‘error’ is in fact no such thing!

Hello,

Just wanted to let you know of a mistake in the answers for exercise 29.B part 7 in line (4) you write “Kyn” and then later you rewrite it as “Kxn” I don’t know which of the variables you meant it to be but I’ve tried the tree with Kyn and it leaves an open branch so I’m guessing it must be Kxn.

Thank you for your time,

Great book!

Exercises 16, Question 6, has Q \/R in the textbook, but the answers has Q^R; the difference between ‘and’ and ‘or’ being somewhat important at times, I just thought I’d mention it…

Stay logical,

Isabel

On Exercise 13 C6, if A is a contradiction, there would be no valuation which makes A true (has it false on the answers), since A is false, there would be no valuation B which would allow A|=B.

I had the same problem with this answer, and it seems wrong. Indeed, it seems to contradict the explanation given in the book for (v) “If C is a contradiction, and A1, A2, …, An, B tautologically entail C, then A1, A2, …, An, tautologically entail ¬B.” (p. 121). There it is clearly stated that “if C is a contradiction, C is never true on any valuation.” But then in the answer to exercise 13 C6, Smith says “if A is a contradiction, then there is no valuation which makes A false.” Something is not right here.

Yes, “false” was a typo for “true”. Sigh! Now corrected (and apologies for the oversight in not correcting before in response to FlyingHamster).

Also, in the same set of exercises (16), exercise 5, the last premiss is (~R\/S) at the textbook but the solution given is for (~R^S).

By the way, Prof. Smith, I’ like to say a big ‘thank you’ for your excellent book.

Greetings from a student in Greece.

Respectfully yours,

Philip Papagiannopoulos

In the answers for exercises 27, it says that wff (7) is false on the second q-valuation that is provided. But (7) is true on that valuation. The number 4 is an even number in the domain which is less than every odd number in the domain.

Many thanks. Correction now uploaded.

In the answer to exercise 8A – 6, I think the answer isn’t quite fully expanded, with a not(S) left on the left hand side.

Oops, yes! Thanks for spotting that. Corrected on the server now (the new version may take a little while to show up).

There is a really minor mistake in the solution of the exercise 2 in chapter 3. Instead of “Some humans AND men” should be “Some humans ARE men”.

Also in chapter 3: Exercise 5 of the solutions is different from the book.

On page 37 is proof defined as(When a chain of argument leads from initial permisses to final conclusion via intermediate inferential steps, each one of wich is clearly valid,then we will say that the argument constitues a proof of conclusion fro those permisses).To me it is not sufficient that inferntial steps are valid.Take an example where argument is valid but not sound then least one of the permisses are not true but then can the conclusion either be true or false but to me that can´t be proof of any thing.To avoid any misunderstanding I here asume not sound argument but valid that to say it is not possible that all premisses can be true together or in a truth table tere is no line wich all the permisses are true.

A proof establishes that,

ifthe premisses are true, the conclusion is true. Yep, that’s not an outright demonstration of the conclusion. Rightly or wrongly, that’s a standard logician’s usage of “proof”.