- #1
mathmari
Gold Member
MHB
- 5,049
- 7
Hey!
Let $L=\{+, ' , 0, 1, T, C\}$ be the language where $ ' : \frac{d}{dz}$, $T(x) \Leftrightarrow x \notin \mathbb{C}$ (so $x$ is not constant), $C(x) \Leftrightarrow x \in \mathbb{C}$ (so $x$ is constant)
$x, f, g \in \mathbb{C}[z, e^{\lambda z } \mid \lambda \in \mathbb{C}]. $
(In this ring the differential equations have always a solution.)
We have the following formula
$$\exists x \phi (x , \overline{y}) \text{ where } \phi: \displaystyle{\bigwedge_{j=1}^n \phi_j} \text{ with } \phi_j : \left\{\begin{matrix}
Lx=f \\
Lx \neq g
\end{matrix}\right.$$
I want to find an equivalent quantifier-free formula. Some examples are the following:
First by induction on the number of equations $n$ we show that a formula $\phi: \displaystyle{\bigwedge_{j=1}^n L_j x=f_j}$ can be reduced to the formula $Lx=f \land \psi_1$ where $\psi_1$ doesn't contain any $x$.
Then in the same way we show that a formula $\phi: \displaystyle{\bigwedge_{j=1}^n L_j x\neq g_j}$ can be reduced to the formula $Lx\neq g \land \psi_2$ where $\psi_2$ doesn't contain any $x$.
And then $\phi: \displaystyle{\bigwedge_{j=1}^n \phi_j}$ with $\phi_j : \left\{\begin{matrix}
Lx=f \\
Lx \neq g
\end{matrix}\right.$ can be reduced to the formula $L_1 x=f \land L_2 \neq g \land \psi$, where $\psi$ doesn't contain any $x$.
Is my idea correct? (Wondering)
Let $L=\{+, ' , 0, 1, T, C\}$ be the language where $ ' : \frac{d}{dz}$, $T(x) \Leftrightarrow x \notin \mathbb{C}$ (so $x$ is not constant), $C(x) \Leftrightarrow x \in \mathbb{C}$ (so $x$ is constant)
$x, f, g \in \mathbb{C}[z, e^{\lambda z } \mid \lambda \in \mathbb{C}]. $
(In this ring the differential equations have always a solution.)
We have the following formula
$$\exists x \phi (x , \overline{y}) \text{ where } \phi: \displaystyle{\bigwedge_{j=1}^n \phi_j} \text{ with } \phi_j : \left\{\begin{matrix}
Lx=f \\
Lx \neq g
\end{matrix}\right.$$
I want to find an equivalent quantifier-free formula. Some examples are the following:
- $$\begin{pmatrix}
x'+x=f\\
x''+x'=g
\end{pmatrix} \Leftrightarrow \begin{pmatrix}
x'+x=f\\
f'=g
\end{pmatrix}$$
- $$\begin{pmatrix}
x'+x=f\\
x''+2x'=g
\end{pmatrix} \overset{\text{Gauss elimination}}{\underset{\text{Linearity of differentiation}}{\Leftrightarrow}} \begin{pmatrix}
x'+x=f\\
f'=g
\end{pmatrix} \Leftrightarrow \begin{pmatrix}
x=f-g+f'\\
x'=g-f'
\end{pmatrix} \Leftrightarrow \begin{pmatrix}
x=f-g+f'\\
f-g'+f''=g-f'
\end{pmatrix}$$
- $$\begin{pmatrix}
x'+x=f\\
x''+2x' \neq g
\end{pmatrix} \Leftrightarrow \begin{pmatrix}
x'+x=f\\
x'\neq g-f
\end{pmatrix} \Leftrightarrow \begin{pmatrix}
x'+x=f\\
x \neq g-2f
\end{pmatrix} \Leftrightarrow \begin{pmatrix}
\begin{pmatrix}
(g-2f)'+(g-2f) \neq f\\
x'+x=f
\end{pmatrix}\\
\begin{pmatrix}
(g-2f)'+(g-2f)=f\\
x'+x=f\\
\text{kern}(x'+x) \neq \{0\}
\end{pmatrix}
\end{pmatrix} \overset{\text{In this ring it always stand that kern}(x'+x) = \{0\}}{\Leftrightarrow } \begin{pmatrix}
(g-2f)'+(g-2f) \neq f\\
x'+x=f
\end{pmatrix}$$
First by induction on the number of equations $n$ we show that a formula $\phi: \displaystyle{\bigwedge_{j=1}^n L_j x=f_j}$ can be reduced to the formula $Lx=f \land \psi_1$ where $\psi_1$ doesn't contain any $x$.
Then in the same way we show that a formula $\phi: \displaystyle{\bigwedge_{j=1}^n L_j x\neq g_j}$ can be reduced to the formula $Lx\neq g \land \psi_2$ where $\psi_2$ doesn't contain any $x$.
And then $\phi: \displaystyle{\bigwedge_{j=1}^n \phi_j}$ with $\phi_j : \left\{\begin{matrix}
Lx=f \\
Lx \neq g
\end{matrix}\right.$ can be reduced to the formula $L_1 x=f \land L_2 \neq g \land \psi$, where $\psi$ doesn't contain any $x$.
Is my idea correct? (Wondering)