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First of all, I am no math guy, so if there are formal or wording errors I am sorry ^^

I: Say you got a differential equation: $x'(t) = 1$.
The solution for that equation is: $x(t) = c_1 + t$.
Where $t$ is the time and $t \geq 0$.

II: In addition, say you got a constraint $f = x >= 5$

My question:

Is there a way to express I and II together in a logical formula?

The tricky part here is, that the value of $x$ can change and you can come back to the DiffEQ part.

So for example, a "run" is:

$(x=0) \rightarrow \mathit{DiffEQ} \rightarrow (x = x\cdot 2) \rightarrow \mathit{DiffEQ} \rightarrow ...$

What would happen in my case is, that a logical formula is built of the form:

$(x=0) \wedge \mathit{DiffEQFormula} \wedge (x = x\cdot 2) \wedge \mathit{DiffEQFormula} \wedge (...)$

That logical formula is checked for satisfiability.
If it evaluates to true, the forbidden constraint is satisfied and we have found an Error.

So what is $\mathit{DiffEQFormula}$?

My idea would be something like this: $x = x + t \quad \wedge \quad 0 \leq t \leq \infty \quad\wedge\quad f$

For simplicity, let us assume that I only use First-order linear ODEs

I hope you understand my question and thanks in advance

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    If $x\geqslant 5$, then $x$ can't be less than 5. Please update your question. – 2017-02-25
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    It is not easy to express it in FOL, because we need at least to sorts of variables: one for (real) *numbers* and the other for *functions*. – 2017-02-25
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    In additon, we have to express the fact that the function $f$ and the function $f'$ has a certain "link" between them. – 2017-02-25
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    I) $\exists f' \forall t \ (f'(t)=1)$. – 2017-02-25
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    II) $\exists f \exists c_1 \forall t \ (f(t)=c_1+t)$. – 2017-02-25

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