Prove valid or invalid. Show truth table please p-(r-s) Solution Hi It\'s really

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Prove valid or invalid. Show truth table please

p-(r-s)

Explanation / Answer

Hi It's really hard to construct truth tables on here, but I'll try my best to help. a. (p v q) > r, p, therefore r This argument is valid: 'r' is true on every assignment of truth values to 'p', 'q' and 'r' on which '(p v q) > r' and 'p' are both true. b. p > (q v r), p, therefore r This argument is invalid: there is an assignment of truth values to 'p', 'q' and 'r' on which 'p > (q v r)' and 'p' are both true but 'r' is false. If we constructed a truth table to test whether this argument is valid, we'd find that on the line on which 'p' were true, 'q' were true and 'r' were false, 'p > (q v r)' and 'p' would both be true, but 'r' would be false. c. a > b, ~(b & c), therefore a > ~c This argument is valid: 'a > ~c' is true on every assignment of truth values to 'a', 'b' and 'c' on which 'a > b' and '~(b & c)' are both true. d. I'm not sure whether this argument should read 'p v (q & r), r > s, therefore ~q' or 'p v (q & r), r s, therefore ~q', but either way, it's invalid. Construing it as 'p v (q & r), r > s, therefore ~q', we'll find that when 'p' is true, 'q' is true, 'r' is true and 's' is true, 'p v (q & r)' and 'r > s' are both true, but '~q' is false. Construing it as 'p v (q & r), r s, therefore ~q', we'll find that when 'p' is true, 'q' is true, 'r' is false and 's' is false, 'p v (q & r)' and 'r s' are both true, but '~q' is false. We can use natural deduction to prove that arguments (a) and (c) are valid: (1) 1. (p v q) > r Premise (2) 2. p Premise (2) 3. p v q 2 vI (1,2) 4. r 1,3 MP (1) 1. a > b Premise (2) 2. ~(b & c) Premise (3) 3. a Assumption (4) 4. c Assumption (1,3) 5. b 1,3 MP (1,3,4) 6. b & c 4,5 &I (1,2,3,4) 7. (b & c) & ~(b & c) 2,6 &I (1,2,3) 8. ~c 4,7 RAA (1,2) 9. a > ~c 3,8 CP I hope that this helps a little

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