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

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To prove that the expression (∃x)[P(x) → Q(x)]∧(∀y)[Q(y) → R(y)]∧(∀x)P(x) → (∃x)R(x) is a valid argument using the deduction method, one must demonstrate that the conclusion logically follows from the premises through a sequence of valid inference rules. A valid argument is one where if the premises are true, the conclusion must also be true. The deduction method involves deriving the conclusion from the premises using logical equivalences and rules of inference. The reasoning indicates that since P(x) holds for all x, and at least one P(x) implies Q(x), it follows that Q(x) must hold for some x, leading to R(x) through the implication of Q(y) to R(y). Therefore, the argument is valid as it adheres to the logical structure required for deduction.
stan1992
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(∃x)[P(x) → Q(x)]∧(∀y)[Q(y) → R(y)]∧(∀x)P(x) → (∃x)R(x)
 
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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?
 
Evgeny.Makarov said:
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
 
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.

stan1992 said:
(∃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 $x$.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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