Re: [tlaplus] Defining functions in TLAPS

Try adding `3 \in S` to your assumptions (along with the commented-out rule about P's behavior `\A s \in S : P[s] = IF s \in {1, 2} THEN TRUE ELSE FALSE`).

Since S is unconstrained, there is no way for you (or TLAPS) to prove or disprove either of the following:

3 \in S
3 \notin S

If `3 \in S`, then we can instantiate the rule about P's behavior to show `P[3] = IF 3 \in {1, 2} THEN TRUE ELSE FALSE`, implying `P[3] = FALSE`.

If `3 \notin S` then we can't say anything about what P[3] equals. In this case P[3] could be TRUE, even given all your other assumptions.

Calvin

On Fri, Dec 2, 2022 at 1:58 PM jack malkovick <sillymouse333@xxxxxxxxx> wrote:

I'm experimenting again with TLPAS in order to learn stuff, and I guess this is the open/closed world problem.

I was trying to see if it's possible to define predicate P as TRUE only for value 1 and 2 and FALSE implicitly for any other value. Of course, none of the options below worked.
Is the any way to do this?

THEOREM TestFunction ==
ASSUME
NEW S,
NEW P \in [S -> BOOLEAN],
\* \A s \in S : P[s] = IF s \in {1, 2} THEN TRUE ELSE FALSE
\* P = (1 :> TRUE @@ 2 :> TRUE)
P[1], P[2]
PROVE
P[3] => FALSE
PROOF
OBVIOUS

--
You received this message because you are subscribed to the Google Groups "tlaplus" group.
To unsubscribe from this group and stop receiving emails from it, send an email to tlaplus+unsubscribe@xxxxxxxxxxxxxxxx.
To view this discussion on the web visit https://groups.google.com/d/msgid/tlaplus/c86392a7-d2d4-4b62-935b-265874627159n%40googlegroups.com.