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SuperNova1250 August 2nd, 2016 03:56 AM

Am I doing this right? (Predicate logic)
 
Hey thanks for checking the thread!

So i have to prove a conclusion is true with predicate logic

I don't have a step by step solution so i tried myself

That's my problem :

Conclusion :
∃x : ¬Q(x)

Premises :
1.∀ : x (P(x) → ∀y : (Q(x) → R(x,y)))

2 ∃x : (P(x) ∧ ∃y : ¬R(x,y))

3. ∃x ∃y [ P(x) ∧ ¬R(x,y)] (pulled out quantifiers)

4. ∃x [ P(x) ∧ ¬R(x,b)] EI(3), not sure if it's right

5. P(a) ∧ ¬R(a,b) EI(4)

6. P(a) Simp(5)

7. ¬R(a,b) Simp(5)

8. ∀x ∀y [ P(x) → Q(x) → R(x,y) ] pulled out quantifiers

9. ∀x [ P(x) → Q(x) → R(x,b) ] UI [8]

10. P(a) → Q(a) → R(a,b) UI[9]

11. All three parts have implication between them therefore i can switch order of operations like this :

( P(a) → Q(a) ) → R(a,b)

12. ( ¬ P(a) ∨ Q(a) ) → R(a,b) solved the inner parentheses with Impl1 rule

13. ¬ ( ¬ P(a) ∨ Q(a)) ∨ R(a,b) same rule for outer one

14. ( P(a) ∧ ¬Q(a) ) ∨ R(a,b) De Morgan's + double negation, keeping the second part the same.

15. ( R(a,b) v ¬Q(a) ) ∧ ( R(a,b) v P(a)) distributive law

16. R(a,b) v ¬Q(a) Simp(15)

17. R(ab) v P(a) Simp(15)

18. ¬Q(a) disjunctive syllogism (16,7)

19 ∃x : ¬Q(x) EG[1,8]

First, If you had time to follow all this thank you in advance!

Now i'd like to know whether this is solved correctly?

Did I make aNY mistakeS and if I did, please show me where and point me in the right direction!

SuperNova1250 August 3rd, 2016 11:51 PM

Can anyone please help me with this problem, i'm stuck!

Thanks!


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