NF+swf-Separation is inconsistent. 

Let P1(X) be the set of one element subsets of X. Let S={{P1(A),Ø}| A is infinite}. 

Let T be the set of subsets of S. 


(1) Suppose A is infinite. Then {P1(A),Ø}∩P1(A)=Ø and therefore swf({P1(A),Ø}).

(2) Suppose A is infinite. Then {P1(A),Ø}∩S=Ø. and therefore swf(S).

    Proof: P1(A) is not in S because every element of S is a set with 2 elements. 
           
           Ø  is not in S because every element of S is a set with 2 elements.

(3) S∩T=Ø.

    Proof: Suppose A is infinite. Then {P1(A),Ø} is not  a subset of S.


Let P={{a,b}| a∈T and b∈S}.

(4) Suppose a∈T and b∈S. Then {a,b}∩P=Ø and therefore swf(P).
  
    Proof: By (3), a∩T=Ø an so a∉P. P1(A)∈b for some infinite set A. By (2), P1(A)∉S.   
           
           P1(A)∉T because T is a set of 2 element sets. Therefore b∉P.

(5) There is a 1-1 function from T to S.

    Proof: Let N be the set of natural numbers and let O be the set of odd numbers.

           Let d be the 1-1 function from sets to infinite sets defined by 

             da={2n|n∈N∩a}U{x|x∈(a-N)}UO.

           Now define f:T-->S by ft={P1(d{A|{P1(A),Ø}∈t}),Ø}. Then f is 1-1.


Let f be a 1-1 function from T to S. Let F={{a,b}|a∈T and b∈S and fa=b}. By (4), swf(F).

Let φ(x) be the formula ∃t(t∈T∧∃p(p∈F∧t∈p∧x∈p∧x∉t)). Then for x∈S, φ(x) is equivalent to its 
 
relativization to swf. By swf-Separation, there is a C such that x∈C<-->x∈S∧φ(x). 

Suppose fC=c. Then c∈C<-->c∉C.