Metamath Proof Explorer


Theorem difidALT

Description: Alternate proof of difid . (Contributed by David Abernethy, 17-Jun-2012) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion difidALT
|- ( A \ A ) = (/)

Proof

Step Hyp Ref Expression
1 dfdif2
 |-  ( A \ A ) = { x e. A | -. x e. A }
2 dfnul3
 |-  (/) = { x e. A | -. x e. A }
3 1 2 eqtr4i
 |-  ( A \ A ) = (/)