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May 4, 2020, 2:15:08 AM5/4/20

to Calculational Mathematics

Hi Jeremy,

Would it be fair to view

P

!= {...}

Q

as a proof of ¬[¬(P=Q)] ?

Agape,

Eric

May 4, 2020, 8:09:13 AM5/4/20

to calculationa...@googlegroups.com

Well, that would mean “somewhere P == Q”, no? Which roughly means that for some point x in the underlying space, we have P.x == Q.x .

I believe my notation was indicating that the equality was dependent on the scalar context. For example, for punctual, conjunctive f , we can calculate:

|[ Context: x => y

f.x

!= { x == x /\ y , f is punctual }

f.(x /\ y)

== { f is conjunctive }

f.x /\ f.y

]|

which allows us to conclude:

[ (x => y) => (f.x => f.y) ] ,

or that f is punctually monotonic. There’s no way to carry out that first step using the conventions in Chapter 4 (I believe) of Dijkstra and Scholten’s Predicate Calculus and Program Semantics .

It’s been a long, long time since I wrote those JAWs but I’ll re-read the note and let me know if you have more questions.

+j

On May 4, 2020, at 02:15, Eric Macaulay <elih...@gmail.com> wrote:

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