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[tlaplus] Re: Defining linearity of integrals using Temporal Logic of Actions



As stated in the other thread, TLAPM does not yet support checking proofs involving real numbers. However, if you are just looking to write the proof in the TLA+ structured proof formalism without machine-checking its correctness, that can certainly be done.

Andrew Helwer

On Tuesday, November 19, 2024 at 5:42:18 PM UTC-5 marta zhango wrote:
I would like to make a proof using Temporal Logic of Actions (TLA) to show the linearity of the Fourier Transform.  I want to define the linearity of integrals so that I can using as a conditions in the proof.  

Perhaps with a let using some letter to define the action, say `A`.  Then say that `by A` means by the linearity of integrals at the end of the line  How would it look like?

For instance I can define the action

$\mathcal{I}(h(t)) \triangleq \int_a^b h(t) \, dt$

and stating linearity of integrals as an axiom

$\text{Linearity} : \quad \mathcal{I}(a f(t) + b g(t)) = a \mathcal{I}(f(t)) + b \mathcal{I}(g(t))$

The proof would look like

$\mathcal{F}\{h(t)\} = \int_{-\infty}^{\infty} \big(a f(t) + b g(t)\big) e^{-i\omega t} \, dt$

$\mathcal{F}\{h(t)\} = a \int_{-\infty}^{\infty} f(t) e^{-i\omega t} \, dt + b \int_{-\infty}^{\infty} g(t) e^{-i\omega t} \, dt \;\;\;$ by [linearity]

What should I put instead of `by [linearity]` in my proof and how would the linearity statement be defined?

$\rule{8cm}{0.3mm}$

Have proceeded as follows

LET: $\;\;\; h(t) \triangleq a \cdot f(t) + b \cdot g(t)$

$\;\;\;\;\;\;\; \mathcal{I}(h(t)) \triangleq \int_a^b h(t) \, dt = a \int_a^b  f(t) \, dt + b \int_a^b g(t) \, dx$

$\;\;\;\;\;\;\; \mathcal{F}\{h(t)\} \triangleq \int_{-\infty}^{\infty} h(t) \, e^{-i\omega t} \, dt$

$\langle 1 \rangle \;\; \mathcal{F}\{h(t)\} = \int_{-\infty}^{\infty} (a f(t) + b g(t)) \, e^{-i\omega t} \, dt$

$\quad\quad\quad\quad \;\;\; = \int_{-\infty}^{\infty} a \left( f(t) e^{-i\omega t} \right) + b \left( g(t) \, e^{-i\omega t} \right) \, dt$

$\quad\quad\quad\quad \;\;\; = a \int_{-\infty}^{\infty} f(t) e^{-i\omega t} \, dt + b \int_{-\infty}^{\infty} g(t) e^{-i\omega t} \, dt \;\;\;\;\; [\mathcal{I}(h(t))]$

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