Metamath Proof Explorer


Theorem equequ2

Description: An equivalence law for equality. (Contributed by NM, 21-Jun-1993) (Proof shortened by Wolf Lammen, 4-Aug-2017) (Proof shortened by BJ, 12-Apr-2021)

Ref Expression
Assertion equequ2
|- ( x = y -> ( z = x <-> z = y ) )

Proof

Step Hyp Ref Expression
1 equtrr
 |-  ( x = y -> ( z = x -> z = y ) )
2 equeuclr
 |-  ( x = y -> ( z = y -> z = x ) )
3 1 2 impbid
 |-  ( x = y -> ( z = x <-> z = y ) )