Metamath Proof Explorer


Theorem nfss

Description: If x is not free in A and B , it is not free in A C_ B . (Contributed by NM, 27-Dec-1996)

Ref Expression
Hypotheses dfss2f.1
|- F/_ x A
dfss2f.2
|- F/_ x B
Assertion nfss
|- F/ x A C_ B

Proof

Step Hyp Ref Expression
1 dfss2f.1
 |-  F/_ x A
2 dfss2f.2
 |-  F/_ x B
3 1 2 dfss3f
 |-  ( A C_ B <-> A. x e. A x e. B )
4 nfra1
 |-  F/ x A. x e. A x e. B
5 3 4 nfxfr
 |-  F/ x A C_ B