Logic Confusing Problem

Question

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?

Answer 1

i think 1 is true, by applying compactness theorem you can show that 1 is true

Answer 2

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

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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 $i > 1$ 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.

Answer 3

Both of them are false, this is due to exact definition:

1.for a formula $\psi$ we say $\vDash\psi$ if for all valuations $w$, $w(\psi)=1$ not for just a valuation $v$.

2.We say a set is consistent if we could find a valuation $v$ such that for all formula $\psi$ in it $v(\psi)=1$. But as you said formula $\varphi_1$ could be a contradiction.

Answer 4

Now, you changed the question, first question is false yet because perhaps for each valuation $v$ only $v(\varphi_{n+1})=1$. But if you define $\Sigma=\{\varphi_1,\ldots,\varphi_n\}$ your proposition would be true.

Even you defined $\Sigma=\{\varphi_1,\ldots,\varphi_n\}$ (and you must do it because $\varphi_1\vee\varphi_2\vee\ldots$ is meaningless) your second question would be false since for all valuation $v$ it can be $v(\varphi_2)=1$ and $v(\varphi_1)=0$.

If you want your second proposition be true, after you defined $\Sigma=\{\varphi_1,\ldots,\varphi_n\}$ you should change it:

For all formula $\varphi_k$ there is a valuation $v$ such that $v(\varphi_k)=1$