# Thread: Proving implication

1. ## Proving implication

Hi,

I'm hoping someone can help me with proving the following statement: $\forall x \in\mathbb{R}, P(x) \Rightarrow \forall y \in\mathbb{R}, P(y)$.

First, I assume $\forall x \in\mathbb{R}, P(x)$.
Then I say "Let $y\in\mathbb{R}.$"
..... how do I finish this proof?

It is clear to me that this statement is true. I can't say P(y) is true because x = y, since x is not a defined value. I would really appreciate help!!!!

2. ## Re: Proving implication

Originally Posted by otownsend
help me with proving the following statement: $\forall x \in\mathbb{R}, P(x) \Rightarrow \forall y \in\mathbb{R}, P(y)$.
It is clear to me that this statement is true. I can't say P(y) is true because x = y, since x is not a defined value. I would really appreciate help!!!!
To do this I need to use conditional proof, C.P.
$\forall x \in\mathbb{R}, P(x) \Rightarrow \forall y \in\mathbb{R}, P(y)$.
$\forall x \in\mathbb{R}, P(x)$. assunption.
$P(z)$ UI (universal instantiation)
$(\forall y)[P(y)]$ UG (universal generalization).
$(\forall x) [P(x)] \Rightarrow (\forall y)[ P(y)]$ CP (conditional proof).

Now I am with you in that this is a very strange looking proof.
However, almost the exact proof is found in the textbook SYMBOLIC LOGIC by Copi.