I have some questions to ask.
- Suppose I want to define some sequence of propositional formulas $\{\varphi_{j}\}_{j\in\mathbb{N}}$. First, I define it this way. Fix an enumeration $p_{1},p_{2},\ldots$ of propositional variables. For any $j\in\mathbb{N}$, define $\varphi_{j}$ recursively as follows. \begin{align*} \varphi_{1}&=p_{1}\\ \varphi_{j+1}&=\varphi_{j}\wedge p_{j+1}. \end{align*} Now, I don't want to specify it recursively. This time I want to say simply this way: $$\mbox{For each}\ j\in\mathbb{N},\ \mbox{define}\ \varphi_{j}=p_{1}\wedge p_{2}\wedge\cdots\wedge p_{j}.$$ The question is: is the latter way of defining $\{\varphi_{j}\}_{j\in\mathbb{N}}$ ambiguous? While one can easily get a sequence $p_{1},p_{1}\wedge p_{2},p_{1}\wedge p_{2}\wedge p_{3},\ldots$ after sticking to the former way of defining, I wonder if it is possible that one may misinterpret the latter way of defining---that it produces something different from what the former does. Which way of defining---recursively, or bluntly---is more acceptable?
- From the definition of $\{\varphi_{j}\}_{j\in\mathbb{N}}$ defined recursively above, I want to show that, for each $j\in\mathbb{N}$, $\varphi_{j+1}\models\varphi_{j}$ but $\varphi_{j}\not\models\varphi_{j+1}$. (Here $\models$ means logically implies and $\not\models$ means does not logically imply). By quick inspection, one can see that $\varphi_{j+1}\models\varphi_{j}$ since a conjunction must logically imply one of its constituents. And one can easily arrive at $\varphi_{j}\not\models\varphi_{j+1}$ too since $p_{j+1}$ does not appear in $\varphi_{j}$, and so there is some truth assignment satisfying $\varphi_{j}$ but not $p_{j+1}$. However, the point is: do I have to use proof by induction? I have tried, but the proof is not complicated and it does not even rely on the induction hypothesis. The induction hypothesis is never needed. Does this mean I can prove the claim directly? Or what?)
Thank you in advance.
1) Both ways are perfectly acceptable. If you don't like to use dots (which could be considered ambiguous but are generally admitted), you can write $$\varphi_j=\bigwedge_{1\leq i\leq j} p_i.$$
2) You don't need the proof by induction, you can just fix any $j\in\mathbb N$, and explain your reasoning at this level. For the first way it is the rule of conjunction elimination, and for the second you describe the truth assignment satisfying $\varphi_{j}$ but not $\varphi_{j+1}$.