$10 paypal for math guru help (proofs, logical equivalence, inference)

[JimRaynor]

aa

 

[GodlessAstronomer]

Just post the problems here, people will help you for nothing. I'll give it a go, but I'm hungover so make no promises.

 

[JimRaynor]

lol ok hold on

 

[JimRaynor]

aa

 

[schneiderguy]

a. p->q
b. r->s
c. (q^s)->(t v u)
d. p
e. r
f. ~t

1. p -> q (premise a)
   p      (premise d)
   therefore q (Modus Ponens)

2. r-> s (premise b)
   r     (premise e)
   therefore s (Modus Ponens)

3. q (1)
   s (2)
   therefore q^s (conjunction)

4. (q^s)->(t v u) (premise c)
   (q^s)	(3)
   therefore (t v u) (Modus Ponens)

5. (t v u) (4)
    ~t   (premise f)
   therefore u (elimination)

Oh you want all of them. There's #4. I don't want to do your homework for you but you should be able to figure it out using that as an example

You have a list of all of the rules of inference, right?

 

[dighn]

#3:

7 - hypothetical syllogism (1, 2)
8 - p ^ r: modus ponens (4, 7)
9 - simplification (8)
10 - disjunctive syllogism (3, 5)
11 - modus tollens (6, 10)
12 - conjunction (9, 11)
13 - addition (12)

that's the only one i'm doing

 

[JimRaynor]

Thank you gentlemen, very much. Impressive.

omg only 2 more questions!!!

 

[schneiderguy]

#5 should be pretty simple. Feel free to PM me if you need help

 

[rudder]

42

 

[rocadelpunk]

have you made truth table before? might help/google

 

[RESmonkey]

Discrete Mathematics is soooo lame. Really boring math theory.

 

[JimRaynor]

truth tables are easy, i'm not really familiar with these proofs without a table and the laws of inference

 

[JimRaynor]

Ok I think I got #5, it did seem easy.

1. p -> (q ^ r) Premise
2. r -> s Premise
3. p Premise
4. (q ^ r) Inference (1,3)
5. r Simplification (4)
6. s Modus Ponens (2,5)

Correct??

Stuck on this one though:

1. p -> q Premise
2. ~q V r Premise
3. ~s -> ~r Premise
4. ~(s ^ ~t) Premise
5. p Premise
6. q Modus Ponens (1, 5)

Have to get to T???

 

[schneiderguy]

Yeah #5 looks right.

For #6 I'll give you a hint and see if you can get it from there.

Since you have q:

~q V r
q
therefore r (elimination)

 

[JimRaynor]

Problem #6

1. p -> q Premise
2. ~q V r Premise
3. ~s -> ~r Premise
4. ~(s ^ ~t) Premise
5. p Premise
6. q Modus Ponens (1, 5)
7. r Elimination (2,6)
8. s Modus Tollens (3,7)
9. (~s ^ t) Distributive (4--can i do this?)
10.

 

[schneiderguy]

~(s ^ ~t) = (~s V t) by DeMorgan's Law

 

[JimRaynor]

Problem #6

1. p -> q Premise
2. ~q V r Premise
3. ~s -> ~r Premise
4. ~(s ^ ~t) Premise
5. p Premise
6. q Modus Ponens (1, 5)
7. r Elimination (2,6)
8. s Modus Tollens (3,7)
9. (~s ^ t) Distributive (4--can i do this?)
10. t Disjunctive Inference (8,9)

does that work?

 

[schneiderguy]

Nope, the correct way to negate something is to negate the variables and switch AND or OR/OR to AND.

So for example:

~(p^q) = ~p V ~q
~(pVq) = ~p ^ ~q

 

[NoReMoRsE]

Xnor xor?