Symbolic Logic

[Cole]

I wasn't sure where else this topic could go. I have a minor in symbolic logic and recently I cracked open my books again to look at some problems. There's one that I just can't solve and was wondering, first, if there's anyone here who is familiar with symbolic logic problems.

[drew]

I wasn't sure where else this topic could go. I have a minor in symbolic logic and recently I cracked open my books again to look at some problems. There's one that I just can't solve and was wondering, first, if there's anyone here who is familiar with symbolic logic problems.

Go ahead, ask (didn't know you could minor in symbolic logic, but it's a common topic in computer science, which I minored in).

[Cole]

Well, it's a minor in philosophy, but all I did was take a bunch of logic courses so that's why I say that

With the following premises prove the conclusion:/ I v K

1. G ⊃ (H⋅I)
2. J ⊃ (H⋅K)
3. ((L⊃~G)⋅M) ⊃ N
4. (M⊃N) ⊃ (L⋅J)

Conclusion (as above) is I v K

[drew]

Well, it's a minor in philosophy, but all I did was take a bunch of logic courses so that's why I say that

With the following premises prove the conclusion:/ I v K

1. G ⊃ (H⋅I)
2. J ⊃ (H⋅K)
3. ((L⊃~G)⋅M) ⊃ N
4. (M⊃N) ⊃ (L⋅J)

Conclusion (as above) is I v K

I don't know how to solve it off-hand, but I'd like to try to figure it out.

Just to make sure I have the symbols correct, the dot means 'AND', right? The first premise says "If G, then H and I"?

[Cole]

Hey again drew,

Yes the dot is and. The "v" is or and yes that sideways horseshoe sentence DOES read "If G then (H and I).

[Major Tom]

1. G ⊃ (H⋅I)
2. J ⊃ (H⋅K)
3. ((L⊃~G)⋅M) ⊃ N
4. (M⊃N) ⊃ (L⋅J)

Conclusion (as above) is I v K
Let's go in the proof by contradiction path, so we're assuming I v K is false, which can only happen if both I and K are false. We'll substitute both for false in the four assumption and see where that leads us.

1 becomes: G ⊃ false (which is equivalent to G being false, so we'll substitute G for false as well).
2 becomes: J ⊃ false (which is equivalent to J being false, so we'll substitute J for false as well).
3 becomes: ((false ⊃ false)⋅M) ⊃ N, which is equivalent to M ⊃ N, so in 4 we'll substitute M ⊃ N by true.
4. true ⊃ (L ⋅false), equivalent to true ⊃ false, which is false.

So, assumption 4 fails if we assume the conclusion to be false, QED.

[drew]

1. G ⊃ (H⋅I)
2. J ⊃ (H⋅K)
3. ((L⊃~G)⋅M) ⊃ N
4. (M⊃N) ⊃ (L⋅J)

Conclusion (as above) is I v K
Let's go in the proof by contradiction path, so we're assuming I v K is false, which can only happen if both I and K are false. We'll substitute both for false in the four assumption and see where that leads us.

1 becomes: G ⊃ false (which is equivalent to G being false, so we'll substitute G for false as well).
2 becomes: J ⊃ false (which is equivalent to J being false, so we'll substitute J for false as well).
3 becomes: ((false ⊃ false)⋅M) ⊃ N, which is equivalent to M ⊃ N, so in 4 we'll substitute M ⊃ N by true.
4. true ⊃ (L ⋅false), equivalent to true ⊃ false, which is false.

So, assumption 4 fails if we assume the conclusion to be false, QED.

Nice job! I didn't think to use contradiction, but that's so much simpler than where I was headed.

[Cole]

Hey MT thanks for that explanation!

I actually used the contradiction earlier to prove that (i.e. with truth table logic) but unfortunately this exercise calls for usage of classic substitution rules and stuff. (i.e. using logic tools like modus ponens, modus tollens, hypothetic syllogism etc. etc.) You're supposed to use those premises and change them to their logical equivalents step by step in order to come to the conclusion. In the below argument, if I could prove J, then I could conclude (H⋅K) by modus ponens. Then I could simplify that to just K (through simplification) and then I could conclude I v K through the rule of addition. The problem is getting J.

This is how far I've gotten with my proof (justification included for each line)

1. G ⊃ (H⋅I) Prove:/ I v K
2. J ⊃ (H⋅K)
3. ((L⊃~G)⋅M) ⊃ N
4. (M⊃N) ⊃ (L⋅J)
-------------------
5. (L⊃~G)⊃(M⊃N) (3, exportation)
6. (L⊃~G)⊃ (L⋅J) (5,4 Hypothetical syllogism)

That's about as far as I can get. I have to be able to isolate the J from premise 6, but the only way to do that is to prove (L⊃~G), which I haven't been able to do.

[Cole]

One thing I was thinking:

With the rule of addition, you can tack on as many "ors" as you want. So for example, if I know A is true, I can add A v B v C all I want because I know it won't change the truth value. Also:

A⊃B is equivalent to ~AvB through the rule of conditional exchange. Looking at the above problem, if I can isolate ~G, then I can easily add and make (~Lv~G) which would then convert to (L⊃~G). From then on, it becomes a piece of cake. Getting there is my problem right now

[drew]

...this exercise calls for usage of classic substitution rules and stuff.

Why?