Metamath Proof Explorer


Theorem abssi

Description: Inference of abstraction subclass from implication. (Contributed by NM, 20-Jan-2006)

Ref Expression
Hypothesis abssi.1
|- ( ph -> x e. A )
Assertion abssi
|- { x | ph } C_ A

Proof

Step Hyp Ref Expression
1 abssi.1
 |-  ( ph -> x e. A )
2 1 ss2abi
 |-  { x | ph } C_ { x | x e. A }
3 abid2
 |-  { x | x e. A } = A
4 2 3 sseqtri
 |-  { x | ph } C_ A