matt grime said:
Also, roger, when I say something in this argument is "obviously true" I mean "I obviously want it to be true because that is nice" or soemthing. I don't mean it has an easy proof, indeed it has no proof really since it is one of those axioms we are going to have in our system.
Well, not to confuse things - you're talking about math and I about logic, but here are the proofs in two natural deduction systems, with minimal explanation. I'm quite curious to see if they make sense to anyone.
In the system that I'm used to, the argument to be proven is actually one of the rules, so the proof is quite simple (anyone can look up the names I use
here - Edit: there are explanations of them further down the page):
1) P -> Q [premise]
2) Q -> R [premise]
3) P -> R [1, 2, Hypothetical Syllogism/QED]
This system has no conditional proof (you can't introduce assumptions out of nowhere) and I can't see a way to prove the exact argument without the HS rule, so I'm adding another premise: P.
1) P -> Q [premise]
2) Q -> R [premise]
3) P [premise]
4) Q [1, 3, Modus Ponens]
5) R [2, 4, Modus Ponens/QED]
This just shows that (P -> Q) and (Q -> R) connect in such a way that if P is true, then R must also be true.
The
other system, which I'm not familiar with, does have a conditional proof rule, so I can prove that [((P -> Q) & (Q -> R)) -> (P -> R)] is a tautology by deriving it from nothing, i.e., the empty set of premises (and hopefully not make any technical blunders).
1)) (P -> Q) & (Q -> R) [hypothesis - I assume this in order to derive the tautology, note that it is nested]
2)) P -> Q [1, Conjunction Elimination]
3)) Q -> R [1, Conjunction Elimination]
4))) P [hypothesis - I assume this in order to derive (P -> R), nested again]
5))) Q [2, 4, Modus Ponens]
6))) R [3, 5, Modus Ponens]
7)) P -> R [4, 6, Conditional Proof]
8) ((P -> Q) & (Q -> R)) -> (P -> R) [1, 8, Conditional Proof/QED]
(I edited some unnecessary steps I used the first time.) This says the same thing: Given (P -> Q) and (Q -> R), if we assume that P is true, (P -> Q) gives us Q. Now that we have Q, (Q -> R) gives us R. So (P -> Q) and (Q -> R) give us (P -> R). Pretty straightforward, I think. The formalism may turn some people off, but it was comforting to me to see everything laid bare. Of course, I'm quite a strange cookie.
