I'm struggling to understand this proof. The question reads as:
Consider the following formula, where f represents a function defined on an open interval $I$: \begin{equation} \forall_{y>0} \exists_{x \in I} [f(x) =y] \end{equation} Pick a value of $a$ so that the formula can be proved in the case when $I=(a, \infty)$ and $f$ is the function \begin{equation} f(x) = x+1 \end{equation} and prove the formula.
I chose $a$ to be equal to $-1$. Then the equation becomes:
\begin{equation} \forall_{y>0} \exists_{x \in (-1, \infty)} [x+1 =y] \end{equation}
So my notes say that $\forall_{\alpha (x)} [\beta (x)]$ is an abbreviation of $\forall_{x} [\alpha (x) \Rightarrow \beta (x)]$.
And $\exists_{\alpha (x)} [\beta (x)]$ is an abbreviation for $\exists_{x} [\alpha (x) \wedge \beta (x)]$
So according to me, the question is: \begin{equation} \forall_{y} [[y>0] \Rightarrow [ \exists_{x} [ [x\in (-1, \infty)] \wedge [x+1 = y]]]] \end{equation}
Hopefully, I'm correct so far...
My proof below:
\begin{align} 1&.\quad \quad \forall_{y} [[y>0] \Rightarrow [ \exists_{x} [ [x\in (-1, \infty)] \wedge [x+1 = y]]]] \\ 2&. \quad \quad y \tag{initialise}\\ 3&. \quad \quad y>0 \tag{1} \\ 4&. \quad \quad x=y-1 \tag{choose} \\ 5&. \quad \quad y-1 \in (-1, \infty) \tag{3} \\ 6&. \quad \quad x \in (-1, \infty) \tag{5} \\ 7&. \quad \quad x+1 = y-1+1 \tag{4} \\ 8&. \quad \quad x+1=y \tag{7} \\ 9&. \quad \quad [x \in (-1, \infty)] \wedge [x+1 = y] \tag{7,8} \\ 10&. \quad \quad \exists_{x} [ [x\in (-1, \infty)] \wedge [x+1 = y]] \tag{4 $\to$ 9} \\ 11&. \quad \quad [y>0] \Rightarrow [ \exists_{x} [ [x\in (-1, \infty)] \wedge [x+1 = y]]] \tag{3 $\to$ 10} \\ 12&. \quad \quad \forall_{y} [[y>0] \Rightarrow [ \exists_{x} [ [x\in (-1, \infty)] \wedge [x+1 = y]]]] \tag{2$\to$11} \end{align}
There are a couple of things in my proof that doesn't make sense to me. First, in line two we initialise $y$. But then we assume that $y>0$, so I write that down again.
In fact, I don't really know what we are trying to prove.
It seems like we are trying to prove that every $y$-value, for $y>0$ has an $x$-value associated. But when I look at my proof, I don't see that at all.
Can somebody please help me identify what I'm doing wrong?
EDIT:
Kavi Rama Murthy Pointed out that it would be better to choose $x$ as $y-1$. I think it makes sense to me now. Can somebody please verify my changes to the proof?
Your choice of $a$ is good. To complete the proof simply note that if $y>0$ and $x=y-1$ then $x\in (-1,\infty)=(a,\infty)$ and $f(x)=x+1=y$. This completes the proof.