axiom | Tools for learning | Authentication library

The exception is a bug (please report it, as recommended in the comment), however note that those are not legal OWL axioms. The syntax and semantic specification shows sameAs as requiring two arguments at least.

(Consider that the axiom is supposed to allow definition of synonym individuals; one argument only offers no new information)

If the axioms are generated by an inferred axiom generator, looks like that code has a bug as well.

There is no inferred axiom generator that materialises the sameAs relations. You could write one yourself, based e.g., on InferredPropertyAssertionGenerator, or you can open an issue on the OWLAPI GitHub repo for the functionality to be added to the library.

OWL semantics is defined under open-world assumption, so you can't check if the cardinality for a certain property is exactly N, because there may be other property instances even if not declared.

More precisely, these are the checks that you can do:

You can check if a cardinality is exactly 1 only if you explicitly state that "value1" is the only value for :Ind1. In this case :Ind1 will be part of :Class1.

In FOL:

∀x.(R(Ind1, x) → x = "value1")

In DL:

∃R⁻.{Ind1} ⊑ {"value1"}

In OWL2 (not tested):

seems it's related to your config file

use -X flag with your maven command to see the full stack trace

mvn -X spring-boot:run

One general approach to programming in Coq is to keep dependent types confined to specifications, without using them in programs.

So for instance, a matrix could be represented as a pair of its dimensions and its contents in a nested recursive type

You've almost correctly guessed the syntax, you can write a proof by cases for any predicate with the syntax proof (cases "").

For the example you provided:

It seems to me that a:SSN has an empty lexical space in the example because it isn't itself "the set of finite-length sequences of zero or more characters that match the regular expression [0-9]{3}-[0-9]{2}-[0-9]{4}." Rather, that is the definition of a:SSN. The definition itself was made by constraining a datatype (xsd:string) that does not have an empty lexical space, which is why the section on patterns that you cited applies. That is, the example uses a pattern to constrain a datatype with a non-empty lexical space to define a datatype with an empty lexical space. Accordingly, since "there can be no literals of datatype a:SSN," you would have to either infer that "123-45-6789"^^xsd:string is an SSN by the usage of a:hasSSN or by asserting that "123-45-6789"^^xsd:string is an instance of a:SSN.

From the axiom forall A, ~~A -> A that you were given, you can prove the law of the excluded middle, forall A, A \/ ~A. They are actually logically equivalent. If you do that, then it is easy for you to solve your assignment.

I don't give the full solution here because it is not good to give solutions to course exercises, and I promise you will learn from doing it.

But as a small hint, when trying to prove

After some research and help the solution has been found:

There appeared a breaking change in Axis2 1.7.0 (https://issues.apache.org/jira/browse/AXIS2-5340)

Adding the following lines to the axis2.xml fixed the problem for me: