martinjoconnor
commented
2 years ago Take a look at the language SWRL FAQ here:
https://github.com/protegeproject/swrlapi/wiki/SWRLLanguageFAQ
SWRL only allows conjunction ('and'), but you can usually work around that limitation by writing multiple rules.
Here, you can do what you want to do directly in the language itself without using built-ins, e.g.,
A(?letter) -> C(?letter)
B(?letter) -> C(?letter)
A,B and C is class. I want to show if letter is instance of A or B ,than this letter is instance of C. So I write this sqrl statement, I am confused about whether this statement is correct.
abox:caa(A or B(?letter)->C(?letter)
and I was wondering if 'and' ,'or' could be used in swrl statements.