## PHL310 Valid Reasoning II, past assignments

Past Assignments27 JanuaryI'd like to see what system you learned, and where you are at. Five problems, just for me to see. The first two are just translations into some kind of formal logic. For problem 1, use a propositional logic (the smallest things in your language will be sentences). For problem two, use a quantified logic with predicates. For 3-5, generate a proof (a syntactic proof, not a truth table). Use whatever rules or proof methods you remember.

- If both Nemo and Jaws are fish, then neither Patrick nor Mr. Krab are.
- All men are mortal.
- Premises: (P → Q), (R → Q), (P v R). Conclusion: Q.
- Conclusion: ((P → Q) → (¬Q → ¬P))
- Premises: ∀x(Fx ↔ Gx), ∃xFx. Conclusion: ∃xGx

3 FebruaryHere's a homework with two tracks, since some of us are catching up and some remember the material well. If remember the propositional logic, do (or at least try) track 2. Else do track 1.

1.Read chapters 1 through 5 ofA Concise Introduction to Logic, and answer the following questions from the book:

- Chapter 2 question 5.
- Chapter 3 question 1a and question 2a.
- Chapter 4 question 1b, d, f, h.
- Chapter 5 questions 1 and 2d.
2.We can make use of a theorem -- or of a special sentence called an "axiom" -- in the following way. If you replace every occurence of a sentence in a theorem or axiom with a different sentence (but the same replacement in each case), you also get a theorem or axiom. So, if ((P → Q) → (¬Q → ¬P)) is a theorem, you can replace P with R, and Q with (S v T) and get the following theorem: ((R → (SvT)) → (¬(SvT) → ¬R)).

It turns out that we can do away with indirect derivation if we allow ourselves certain sentences as axioms (an axiom is a sentence that we assume--it's like a very special premise; but we treat it the way we treat theorems in that we'll allow versions or instances of it).

Prove the following:Premises: (R → S), (¬S v ¬R).But prove it without using indirect derivation, but rather a direct derivation and also by using the following (((P → Q) ^ (P → ¬Q)) → ¬P) as your axiom. (If you feel stuck, first prove it using indirect derivation, and then look at your proof and think of how you can use that axiom....) The way to use the axiom is: any time you like, you can write down either (((P → Q) ^ (P → ¬Q)) → ¬P), or you can write a sentence that is made from that by replacing each P with some sentence, and each Q with some sentence. Your justification on the right is "axiom".

Conclusion: ¬R.

Try this one if you feel ambitious. Using that axiom, can you prove without indirect derivation:Premises: (P → R), (Q → R), (P v Q).You asked for hints. They're not that hard, actually. Consider first whether you might not just get the conclusion from the axiom and modus ponens. In that case, the instance of the axiom you'd want for the first problem is:

Conclusion: R.(((R → φ) ^ (R → ¬ φ)) → ¬R). And for the second would be:(((¬R → ψ) ^ (¬R → ¬ ψ)) → ¬¬R)In each case, what will you use for φ, what will you use for ψ?