Post History
#3: Post edited
by
kouty
·
2024-12-18T04:13:11Z (about 13 hours ago)
- I think that now I understood a couple of points: The following statement is now clearer.
- > Let $i(P_1) = 0$ and suppose $\phi(1)$ does not hold. That is, there is a finite subset $\Gamma'$ of $\Gamma$ which has no model in which $P_1$ takes the value $i(P_1) = 0$. Then we define $i(P_1) = 1$ and show that $\phi(1)$, i.e., every finite subset of $\Gamma$ has a model in which $P_1$ takes the value $i(P_1) = 1$.
If, e.g., when $i(P_1) = 0$, $\phi(1)$ does not hold, then no formula that is true only when $i(P_1)$ is selected for assembling subsets. Then we'll build an union of subsets of formulas such as $i(P_1) = 1$. One of those is $\Gamma$'. In the case in which at least one subset contains contradictory formulas, it's not Satisfiable, i.e. has no model.How to build 'the line' in the whole possibly infinite truth table? This is at least one line corresponding to each formula P. Say $P_1$ $\land$ ..., $\land$ $P_n$.
- I think that now I understood a couple of points: The following statement is now clearer.
- > Let $i(P_1) = 0$ and suppose $\phi(1)$ does not hold. That is, there is a finite subset $\Gamma'$ of $\Gamma$ which has no model in which $P_1$ takes the value $i(P_1) = 0$. Then we define $i(P_1) = 1$ and show that $\phi(1)$, i.e., every finite subset of $\Gamma$ has a model in which $P_1$ takes the value $i(P_1) = 1$.
- 1. If, e.g., when $i(P_1) = 0$, $\phi(1)$ does not hold, then no formula that is true only when $i(P_1)$ is selected for assembling subsets. Then we'll build an union of subsets of formulas such as $i(P_1) = 1$. One of those is $\Gamma$'. In the case in which at least one subset contains contradictory formulas, it's not Satisfiable, i.e. has no model.
- 2. How to build 'the line' in the whole possibly infinite truth table? This is at least one line corresponding to ewach formula P. Say $P_1$ $\land$ ..., $\land$ $P_n$.
#2: Post edited
by
kouty
·
2024-12-17T22:01:55Z (about 20 hours ago)
- I think that now I understood a couple of points: The following statement is now clearer.
- > Let $i(P_1) = 0$ and suppose $\phi(1)$ does not hold. That is, there is a finite subset $\Gamma'$ of $\Gamma$ which has no model in which $P_1$ takes the value $i(P_1) = 0$. Then we define $i(P_1) = 1$ and show that $\phi(1)$, i.e., every finite subset of $\Gamma$ has a model in which $P_1$ takes the value $i(P_1) = 1$.
- If, e.g., when $i(P_1) = 0$, $\phi(1)$ does not hold, then no formula that is true only when $i(P_1)$ is selected for assembling subsets. Then we'll build an union of subsets of formulas such as $i(P_1) = 1$. One of those is $\Gamma$'. In the case in which at least one subset contains contradictory formulas, it's not Satisfiable, i.e. has no model.
How to build 'the line' in the whole possibly infinite truth table? This is at least one line corresponding to each formula P. Say $P_1$ $\land$ ..., $\land$ $P_n.
- I think that now I understood a couple of points: The following statement is now clearer.
- > Let $i(P_1) = 0$ and suppose $\phi(1)$ does not hold. That is, there is a finite subset $\Gamma'$ of $\Gamma$ which has no model in which $P_1$ takes the value $i(P_1) = 0$. Then we define $i(P_1) = 1$ and show that $\phi(1)$, i.e., every finite subset of $\Gamma$ has a model in which $P_1$ takes the value $i(P_1) = 1$.
- If, e.g., when $i(P_1) = 0$, $\phi(1)$ does not hold, then no formula that is true only when $i(P_1)$ is selected for assembling subsets. Then we'll build an union of subsets of formulas such as $i(P_1) = 1$. One of those is $\Gamma$'. In the case in which at least one subset contains contradictory formulas, it's not Satisfiable, i.e. has no model.
- How to build 'the line' in the whole possibly infinite truth table? This is at least one line corresponding to each formula P. Say $P_1$ $\land$ ..., $\land$ $P_n$.
#1: Initial revision
by
kouty
·
2024-12-17T22:01:03Z (about 20 hours ago)
I think that now I understood a couple of points: The following statement is now clearer. > Let $i(P_1) = 0$ and suppose $\phi(1)$ does not hold. That is, there is a finite subset $\Gamma'$ of $\Gamma$ which has no model in which $P_1$ takes the value $i(P_1) = 0$. Then we define $i(P_1) = 1$ and show that $\phi(1)$, i.e., every finite subset of $\Gamma$ has a model in which $P_1$ takes the value $i(P_1) = 1$. If, e.g., when $i(P_1) = 0$, $\phi(1)$ does not hold, then no formula that is true only when $i(P_1)$ is selected for assembling subsets. Then we'll build an union of subsets of formulas such as $i(P_1) = 1$. One of those is $\Gamma$'. In the case in which at least one subset contains contradictory formulas, it's not Satisfiable, i.e. has no model. How to build 'the line' in the whole possibly infinite truth table? This is at least one line corresponding to each formula P. Say $P_1$ $\land$ ..., $\land$ $P_n.