MHB Prove that the expression is a valid argument using the deduction method

[stan1992]

(∃x)[P(x) → Q(x)]∧(∀y)[Q(y) → R(y)]∧(∀x)P(x) → (∃x)R(x)

 

[Evgeny.Makarov]

Please write the complete problem statement in the body of the message and not in the thread title.

Also, please give the definitions of "valid argument" and "deduction method" since these concepts differ between textbooks.

Finally, http://mathhelpboards.com/rules/ ask you to show some effort. What exactly is your difficulty in solving this problem?

 

[stan1992]

Please write the complete problem statement in the body of the message and not in the thread title.

Also, please give the definitions of "valid argument" and "deduction method" since these concepts differ between textbooks.

Finally, http://mathhelpboards.com/rules/ ask you to show some effort. What exactly is your difficulty in solving this problem?

Prove that the expression below is a valid argument using the deduction method (that is using equivalences and rules
of inference in a proof sequence)

(∃x)[P(x) → Q(x)]∧(∀y)[Q(y) → R(y)]∧(∀x)P(x) → (∃x)R(x)

1.(∃x)[P(x) → Q(x)] prem
2.(∀y)[Q(y) → R(y)] prem
3.P(x)→Q(x) 1,ui
4.P(x) 2,ui
5.(∀x)Q(x) 5,ug
6.??
7.??
n.(∃x)R(x)

I don't know how to finish this or if I'm even on the right track

 

[Evgeny.Makarov]

I have bad news for you: rules of inference differ between textbooks, too, as stated in this https://driven2services.com/staging/mh/index.php?threads/29/. (Smile) Therefore, you'll have to list the rules and equivalences or at least give the book reference.

(∃x)[P(x) → Q(x)]∧(∀y)[Q(y) → R(y)]∧(∀x)P(x) → (∃x)R(x)

1.(∃x)[P(x) → Q(x)] prem
2.(∀y)[Q(y) → R(y)] prem

It would make sense to add the third premise (∀x)P(x).

Informally, the reasoning is as follows. $P(x)$ holds for all $x$, and for at least one of them $P(x)$ implies $Q(x)$. Therefore, $Q(x)$ holds for some $x$. Further, $Q(y)$ always implies $R(y)$, including when $y$ equals the $x$ found earlier. Thus, $R(x)$ holds for that $x$.