This is a legacy page, which will be removed during 2021. Answers to IFL1 exercises are no longer updated/corrected. If your lecturer is directing you here, ask them to start moving to use instead the freely downloadable IFL2 and its exercises.
PDFs of answers to most of the exercises in IFL1 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
25 thoughts on “IFL1: Answers to exercises”
Find a contrary proposition to: “Someone loves both Jack and Jill.”
Proposed answer: “No woman loves both Jack and Jill.”
Suppose Jerry and only Jerry loves both Jack and Jill.
The original proposition and the proposed contrary proposition are both true.
Perhaps a better answer might be “No one loves Jack and no one loves Jill.”
In this comment ‘(x)’ stands for universal quantifier, ” stands for existential quantifier.
In 24.1, you say that ‘(x)(Fx->Lxy)’ is equivalent to ‘(x)(Fx->Lxy)’. Why?
Suppose ‘Fx’ means x is a man, and Lxy means x loves y. Doesn’t the former sentence mean that all man love someone, while the latter sentence means Someone is loved by every man?
Hello, I have a problem about Exercises 33 B. The question says that the domain of the quantification is the positive whole numbers. But the constant ‘m’ denote number 0. Is this fine? I thought a constant can only denote what is in the domain. If I am wrong, please tell me. Thank you so much!
By the way, I am looking forward to the second edition of the IFL. Thank you so much for having writing this wonderful book.
Hello, I have a problem about the exercises 24 question B6. The answer translates the wff in question into English as “Any even number is equal to twice some odd number.” I wonder if the answer is wrong. I think the answer is “Every positive number x is such that at least one number is such that the former is even, the latter is odd, and twice the latter is equal to the former.” And doesn’t this mean “For all positive number x, it is even, and there is at least one odd number y such that the twice of y is x?”
If I’m wrong please tell me why. Thank you very much. By the way IFL is such a great book that I enjoy it very much.
Yes you are right! I imagine that I’d intended a different example with a conditional in it!
Sir, Would it be possible to obtain Answers to the Exercises at the end of Chapter 28 of IFT? Thank you.
I’m afraid I don’t seem to have such answers on file. Sorry! Eventually, there will be new answers on line to the new exercises in IFL2
Hello, I was just looking at Exercise 17 answers , C 2. and I think the contradiction markers are wrong. For example on the right tree there are contradiction markers under ~R with no R assumed true earlier in the path in the tree.
Hello,I have a problem about the exercise 2 question 10. I think if some set S are false,but P is right, maybe this question is false.But I’m so silly that maybe I don’t understand this. Anyway, It’s a good book.Thank you very much!
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, if the 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”.
Also in chapter 3: Exercise 5 of the solutions is different from the book.
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”.
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).
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.
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.
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).
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…
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,
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!
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.