Take the following function:

CONSTANTS parity
PROPERTIES
 parity : (NATURAL --> {0,1}) & 
 parity(0) = 0 & 
 !x.(x:NATURAL => parity(x+1) = 1 - parity(x))

Here, ProB will complain that it cannot find a solution for parity. The reason is that parity is a function over an infinite domain, but ProB tries to represent the function as a finite set of maplets.

There are basically three solutions to this problem:

• Write a finite function:

parity : (NAT --> {0,1}) & 
parity(0) = 0 & 
!x.(x:NAT & x<MAXINT => parity(x+1) = 1 - parity(x))

• Rewrite your function so that in the forall you do not refer to values of parity greater than x:

parity : (NATURAL --> {0,1}) & 
parity(0) = 0 & 
!x.(x:NATURAL1 => parity(x) = 1 - parity(x-1))

• Write your function constructively using a single recursive equation using set comprehensions, lambda abstractions, finite sets and set union. This requires ProB 1.3.5-beta7 or higher and you need to declare parity as ABSTRACT_CONSTANT. Here is a possible equation:

parity : INTEGER <-> INTEGER & 
parity = {0|->0} \/ %x.(x:NATURAL1|1-parity(x-1))

Note, you have to remove the check parity : (NATURAL --> {0,1}), as this will currently cause expansion of the recursive function. We describe this new scheme in more detail below.

• Another solution is try and write your function constructively and non-recursively, ideally using a lambda abstraction:

  parity : (NATURAL --> {0,1}) & 
  parity = %x.(x:NATURAL|x mod 2)

• Here ProB detects that parity is an infinite function and will keep it symbolic (if possible). With such an infinite function you can:
  • apply the function, e.g., parity(10001) is the value 1
  • compute the image of the function, e.g., parity[10..20] is {0,1}
  • compose the function with a finite relation, e.g., (id(1..10) ; parity) gives the value [1,0,1,0,1,0,1,0,1,0]
  • sometimes compute the domain of the function, here, dom(parity) is determined to be NATURAL. But this only works for simple infinite functions represented.
  • sometimes check that you have a total function, e.g., parity: NATURAL --> INTEGER can be checked by ProB. However, if you change the range (say from INTEGER to 0..1), then ProB will try to expand the function.

You can experiment with those by using the Eval console of ProB, experimenting for example with the following complete machine. Note, you should use ProB 1.3.5-beta2 or higher.

MACHINE InfiniteParityFunction
CONSTANTS parity
PROPERTIES
  parity : NATURAL --> INTEGER & 
  parity = %x.(x:NATURAL|x mod 2)
VARIABLES c
INVARIANT
c: NATURAL
INITIALISATION c:=0
OPERATIONS
 Inc = BEGIN c:=c+1 END;
 r <-- Parity = BEGIN r:= parity(c) END;
 r <-- ParityImage = BEGIN r:= parity[0..c] END;
 r <-- ParityHistory = BEGIN r:= (%i.(i:1..c+1|i-1) ; parity) END
END

You may also want to look at the tutorial page on modeling infinite datatypes.

Recursive Function Definitions in ProB

As of version 1.3.5-beta7 ProB now accepts recursively defined functions. For this:

• the function has to be declared an ABSTRACT_CONSTANT.
• the function has to be defined using a single recursive equation with the name of the function at the left of the equation
• the right-hand side of the equation can make use of lambda abstractions, set comprehensions, set union and other finite sets

Here is a full example:

MACHINE Parity
ABSTRACT_CONSTANTS parity
PROPERTIES
 parity : INTEGER <-> INTEGER & 
 parity = {0|->0} \/ %x.(x:NATURAL1|1-parity(x-1))
END

With such a recursive function you can:

* apply the function to a given argument, e.g., parity(100) will give you 0

Computing the image of the function (parity[1..10]) or composing it with another function (([1,2] ; parity)) does not work yet. These work for non-recursive infinite functions, as described above. Future versions of ProB will probably allow these constructs for recursive functions as well.

There are the following further restrictions:

* ProB does not support mutual recursion yet
* the function is not allowed to depend on other constants, unless those other constants can be valued in a deterministic way (i.e., ProB finds only one possible solution for them)