Metamath Proof Explorer


Theorem eq0rdv

Description: Deduction for equality to the empty set. (Contributed by NM, 11-Jul-2014)

Ref Expression
Hypothesis eq0rdv.1 φ ¬ x A
Assertion eq0rdv φ A =

Proof

Step Hyp Ref Expression
1 eq0rdv.1 φ ¬ x A
2 1 pm2.21d φ x A x
3 2 ssrdv φ A
4 ss0 A A =
5 3 4 syl φ A =