Logic Confusing Problem I Read one logic book, can my two conclusion are true?

1- Suppose 



for each valuation v, we have such n that

 


  
    can we say we have such n that
    
    
  


2- Suppose 



for each valuation v, we have such n that

 

can we say  set {φ1,φ1∨φ2,φ1∨ φ2∨φ3,...} is consistent? 

i think the (2) is not always true?  for (1) and (2) any description?
This is a logic "confused" problem ...

We approach it "by steps".

Question 1

We assume that the meaning of the first part of the question is :


  for a valuation $v$ there is some $n$ for which : $v(\varphi_n)=1$.


Is so, due to the fact that : $v(\varphi_1 \lor \ldots \lor \varphi_n) = max_i \{ v(\varphi_i) \}$, we have that for the valuation $v$ :


  $v(\varphi_1 \lor \ldots \lor \varphi_n) = 1$.


Conclusion : if for a valuation $v$ there is a formula $\varphi_n$ such that $v(\varphi_n)=1$, then the conjunction $φ_1 \lor φ_2 \lor \ldots \lor φ_n$ is consistent (or satisfiable).

The symbol :


  
    $\vDash \varphi_1 \lor \ldots \lor \varphi_n$
  


means that the formula $\varphi_1 \lor \ldots \lor \varphi_n$ is valid (or a tautology).

The above formula is valid when we have $v(\varphi_1 \lor \ldots \lor \varphi_n) = 1$ for all valuation $v$.

Comclusion : the fact that for every valuation $v$ there is a $n$ such that ... does not implies that there is an $n$ such that for every valuation $v$ ...

In other terms, the condition :


  for a valuation $v$ there is some $n$ for which : $v(\varphi_n)=1$.


is not enough to conclude with $\vDash \varphi_1 \lor \ldots \lor \varphi_n$.



Question 2

Assuming a set $\Sigma = \{ φ_1, φ_2, \ldots \}$ of formulae, if $\Sigma$ is consistent, then :


  $\Gamma = \{ φ_1, φ_1∨φ_2, φ_1∨φ_2∨φ_3, \ldots \}$ is always consistent.


Being $\Sigma$ consistent, there is a valuation $v$ such that $v(\varphi_i)=1$, for all $i$.

Thus :


  $v(φ_1)=v(φ_1∨φ_2)=v(φ_1∨φ_2∨φ_3) = \ldots=1$


i.e. the valuation $v$ satisfy all the formulae in $\Gamma$ and thus $\Gamma$ is consistent.

If $\Sigma$ is not consistent, this means that for each valuation $v$ there is a formula $\varphi_i$ such that $v(\varphi_i)=0$; but it is enough that $i &gt; 1$ and $\Gamma$ will be still consistent.

In fact, if for the valuation $v_0$, $v_0(\varphi_1)=1$, then again :


  
    $v_0(φ_1)=v_0(φ_1∨φ_2)=v_0(φ_1∨φ_2∨φ_3) = \ldots=1$.
  


If the formula $\varphi_1$ is unsatisfiable (e.g. a contradiction, like : $p \land \lnot p$) then for each valuation $v$, $v(\varphi_1)=0$; in this case, the set $\Gamma$ is not consistent.

By Compactness Theorem a set $\Gamma$ of formulae is consistent iff every finite subset of $\Gamma$ is consistent; thus, if $\{ \varphi_1 \} \subseteq \Gamma$ is not consistent, also $\Gamma$ is not.
i think 1 is true, by applying compactness theorem you can show that 1 is true