p=1, q=0, r=1

Evaluate:

p ^ q => r

My solution:

1 ^ 0 => 1

0 (because a conjunction is only true if both conjuncts are true) =>1

Therefore false

(but according to the grading this is incorrect)

**Best answer by hamster**

Okay, so lets start this by saying its been a while since I did formal logic. But hopefully I can help you out.

The above image is a truth table. We can clearly see that P ^ Q (p AND q) is true

**if and only if**P is true and Q is true.

Now lets look at implication (P => Q) Notice that on the third line is says that:

- P => Q is true when:
- P is false.
- Q is True

**implication is not causation!**The reason for this is because causation is not "truth functional" and by this we simply mean that if both P and Q are true it is still unclear whether P cause Q. for example:

- let P equal "The world is round"
- let Q equal "Salmon is pink"
- Therefore: Salmon is pink because the shaped of the world caused them to be pink.

Basically, the entire point of deductive logic is that IF the truth status of the premises are known then we can be 100% certain the the conclusion is correct. And this holds for any P and for any Q. In the case of causation the conclusion is not certain to follow from the premises. And so therefore deductive logic abandons the concept.

While I'm here the forth line is (false => false = true) is the idea is that anything logically follows from a falsehood. (see: https://en.wikipedia.org/wiki/Principle_of_explosion ). Thats not relevant to your problem but I thought I'd flag it up.

So thats the first thing you need to know to solve this homework problem. The next thing you need to know is:

- "scope"
- that letters can represent complex functions.

- R v T v ¬X ^ ¬ P

- (4 + 4) * 0
- 4 + (4 * 0)

- (¬P) v Q
- ¬ (P v Q)

Okay, with everything I have said here I believe you should be able to solve the homework problem. Good luck! View original