Set Theory Proof: P => (Q ν R) ≡ (P => Q) ν (P => R)?

Question

If you could help me write up a formal proof for this, it would help me a lot!

Also, I am unsure how to determine whether [(P => Q) Λ (P => R)] => (Q => R), if you could show me a formal way to determine this, I would be very thankful!

Thank you so very much!

Answer 1

/- (P > (Q v R)) <> ((P > Q) v (P > R))



(1) 1. P > (Q v R) Assumption

(2) 2. ~((P > Q) v (P > R)) Assumption

(3) 3. P Assumption

(1,3) 4. Q v R 1,3 MP

(5) 5. Q Assumption

(6) 6. ~R Assumption

(3,6) 7. P & ~R 3,6 &I

(3,5,6) 8. Q & (P & ~R) 5,7 &I

(3,5,6) 9. Q 8 &E

(5,6) 10. P > Q 3,9 CP

(5,6) 11. (P > Q) v (P > R) 10 vI

(5) 12. ~R > ((P > Q) v (P > R)) 6,11 CP

(2,5) 13. ~~R 2,11 MT

(2,5) 14. R 13 DNE

(15) 15. R Assumption

(1,2,3) 16. R 4,5,14,15,15 vE

(1,2) 17. P > R 3,16 CP

(1,2) 18. (P > Q) v (P > R) 17 vI

(1,2) 19. ((P > Q) v (P > R)) & ~((P > Q) v (P > R)) 18,2 &I

(1) 20. ~~((P > Q) v (P > R)) 2,19 RAA

(1) 21. (P > Q) v (P > R) 20 DNE

(-) 22. (P > (Q v R)) > ((P > Q) v (P > R)) 1,21 CP

(23) 23. (P > Q) v (P > R) Assumption

(24) 24. P > Q Assumption

(3,24) 25. Q 3,24 MP

(3,24) 26. Q v R 25 vI

(27) 27. P > R Assumption

(3,27) 28. R 3,27 MP

(3,27) 29. Q v R 28 vI

(3,23) 30. Q v R 23,24,26,27,29 vE

(23) 31. P > (Q v R) 3,30 CP

(-) 32. ((P > Q) v (P > R)) > (P > (Q v R)) 23,31 CP

(-) 33. (P > (Q v R)) <> ((P > Q) v (P > R)) 22,32 <>I



If P is false, then P > Q and P > R are both true. Hence, if P is false, then (P > Q) & (P > R) is true. If Q is true and R is false, then Q > R is false. Hence, on the supposition that P is false, Q is true, and R is false, (P > Q) & (P > R) is true, but Q > R is false. Thus, ((P > Q) & (P > R)) > (Q > R) is not a logical truth.

Answer 2

The easy second one first. I'll work within the set of positive integers.



Let P be {this integer is a multiple of 6}

Let Q be {this integer is a multiple of 3}

Let R be {this integer is a multiple of 2}



P => Q

Also P => R

So [(P => Q) Λ (P => R)]

You tell me. Does Q => R? No way.



Every multiple of 6 is a multiple of both 3 and 2 , but multiples of 3 need not be multiples of 2.



Obviously, since there are infinitely many primes (meaning prime concepts) (2 and 3 are primes), let's call them p1 and p2 and * is multiplication.



(pi and p2) is true implies p1 is true and p2 is true but p1 and p2 are unrelated.



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First problem



Let p∈P and P=>(Q v R). Then p ∈ (Q v R).



Case (1) Assume that p ∉ Q. Since p ∈ (Q v R) ~ Q, p ∈ R



Case (2) Assume that p ∉ R. Since p ∈ (Q v R) ~ R, p ∈ Q



Therefore p ∈ (Q v R) => (p ∈ R) v (p ∈ R), hence we go from left to right.



Right to left is trivial - you should do that one - okay?



If not, and you allow email ask me and I'll do it but it's so easy that you can certainly do it.



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Answer 3

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