MHB Rules of Inference: Get Help with Hypothesis 1 & 2

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Hi,

I am havinga bit of challenge with the following question.
What seems confusing to me is th relationship between hypothesis 1 and 2.
I will appreciate all help. View attachment 9403
 

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Welcome to the forum.

Eluki said:
I am having a bit of challenge with the following question.
It would help if you write exactly what the problem is asking you to do. And I mean as precisely as possible: saying "Prove that the hypotheses imply the conclusion" is not precise enough because the word "imply" can have several different meanings in mathematical logic.

Eluki said:
What seems confusing to me is the relationship between hypothesis 1 and 2.
Would you be puzzled if I said that I find confusing the relationship between the following statements: "All my pets are dogs or cats" and "If a pet is a cat, it is a feline"? I suspect you would because it is not even clear what I mean by a relationship: these are two separate statements, nobody claims that they have to be consistent, or follow from each other, or anything like that. What relationship did you find or want to find between hypotheses 1 and 2, and why is this relationship confusing?
 
Here's a proof (fill out the details).
1. Ax (Px v Qx)
2. Ax (~Qx v Sx)
3. Ax (Rx ->~Sx)
4. Ex ~Px
5. Pa v Qa , universal, 1.
6. ~Qa v Sa universal, 2.
7. Ra ->~Sa , universal, 3.
8. ~Pb, existential, 4.
9.Qb ,5,8 DS.
10. ~~Qb, 9, DN.
11. Sb 7,6, DS.
12. ~~Sb ->~Rb 7, MT.
13. ~Rb , 11,12, MP.
14. Ex ~Rx, QED.
 
Eluki said:
Hi,

I am havinga bit of challenge with the following question.
What seems confusing to me is th relationship between hypothesis 1 and 2.
I will appreciate all help. View attachment 9403
proof:
1) $\forall x(P(x)\vee Q(x)$

2)$\forall x( \neg Q(x)\vee S(x))$

3)$\forall x (R(x)\rightarrow\neg S(x))$

4)$\exists x(\neg P(x))$

5) $(P(x)\vee Q(x)$.......from one and using universal elimination UE

6)$ \neg Q(x)\vee S(x)$..........from two UE

7)$ R(x)\rightarrow\neg S(x)$..........from (3),UE

6)$(\neg P(y))$...............hypothesis for existential elimination EE

7)$\neg(P(x)\rightarrow Q(x)$..........(5) using material implication

8).$ \neg\neg Q(x)\rightarrow S(x)$.........(6) using material imlication

9)$Q(x)$...................hypothesis for conditional proof

10)$\neg Q(x)$................hypothesis for contradiction

11)$Q(x)\wedge\neg Q(x)$..............(9)and (10) using AI (addition introdaction)

12)$\neg\neg Q(x)$.................(10)to (11) contradiction

13)$S(x)$....................(8),(12) using Modus Ponens(MP)

14)$Q(x)\rightarrow S(x)$.........from (9) to(13) and using conditional proof

15)$\neg P(x)$.............hypothesis for conditional proof

16) $Q(x)$................using (15), (7) and MP

17)$S(x)$................using (14) and (16) and MP

18)Repeat process from steps (10) to (12) to end up with $\neg\neg S(x)$

19)$\neg\neg S(x)\rightarrow\neg R(x)$......(7) and using contrapositive

20) $\neg R(x)$.................(18),(19) using MP

21)$\neg P(x)\rightarrow\neg R(x)$.......from (15) to(20) and using conditional proof

22)$\forall x(\neg P(x)\rightarrow\neg R(x))$......from (21) and using universal introduction (UI)

23)$\neg P(y)\rightarrow\neg R(y))$............from(22) and using UE where we put x=y

24)$\neg R(y)$.................. (6),(24) and using MP

25)$\exists x(\neg R(x))$................from (24) and using existensial introduction EI

26) $\exists x(\neg R(x))$................from (4) and (6) to (25) and using EE

As you can see the general plan of the proof is to prove 1st $\forall x(\neg P(y)\rightarrow\neg R(y))$ and then using $\neg P(y)$. to prove $\exists\neg R(x)$ using EI and EE
1) Notice the changing of the variables from x to y and then back to x

Now to test your understanding start your own proof by hypothising $\neg P(x)$
and use my proof as help
You may use different rules of inference if you like
 
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