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Metamath Proof Explorer

Theorem nzrpropd

Description: If two structures have the same components (properties), one is a nonzero ring iff the other one is. (Contributed by SN, 21-Jun-2025)

Ref Expression
Hypotheses nzrpropd.1 
|- ( ph -> B = ( Base ` K ) )
nzrpropd.2 
|- ( ph -> B = ( Base ` L ) )
nzrpropd.3 
|- ( ( ph /\ ( x e. B /\ y e. B ) ) -> ( x ( +g ` K ) y ) = ( x ( +g ` L ) y ) )
nzrpropd.4 
|- ( ( ph /\ ( x e. B /\ y e. B ) ) -> ( x ( .r ` K ) y ) = ( x ( .r ` L ) y ) )
Assertion nzrpropd 
|- ( ph -> ( K e. NzRing <-> L e. NzRing ) )

Proof

Step Hyp Ref Expression
1 nzrpropd.1 
 |-  ( ph -> B = ( Base ` K ) )
2 nzrpropd.2 
 |-  ( ph -> B = ( Base ` L ) )
3 nzrpropd.3 
 |-  ( ( ph /\ ( x e. B /\ y e. B ) ) -> ( x ( +g ` K ) y ) = ( x ( +g ` L ) y ) )
4 nzrpropd.4 
 |-  ( ( ph /\ ( x e. B /\ y e. B ) ) -> ( x ( .r ` K ) y ) = ( x ( .r ` L ) y ) )
5 1 2 3 4 ringpropd 
 |-  ( ph -> ( K e. Ring <-> L e. Ring ) )
6 1 2 4 rngidpropd 
 |-  ( ph -> ( 1r ` K ) = ( 1r ` L ) )
7 1 2 3 grpidpropd 
 |-  ( ph -> ( 0g ` K ) = ( 0g ` L ) )
8 6 7 neeq12d 
 |-  ( ph -> ( ( 1r ` K ) =/= ( 0g ` K ) <-> ( 1r ` L ) =/= ( 0g ` L ) ) )
9 5 8 anbi12d 
 |-  ( ph -> ( ( K e. Ring /\ ( 1r ` K ) =/= ( 0g ` K ) ) <-> ( L e. Ring /\ ( 1r ` L ) =/= ( 0g ` L ) ) ) )
10 eqid 
 |-  ( 1r ` K ) = ( 1r ` K )
11 eqid 
 |-  ( 0g ` K ) = ( 0g ` K )
12 10 11 isnzr 
 |-  ( K e. NzRing <-> ( K e. Ring /\ ( 1r ` K ) =/= ( 0g ` K ) ) )
13 eqid 
 |-  ( 1r ` L ) = ( 1r ` L )
14 eqid 
 |-  ( 0g ` L ) = ( 0g ` L )
15 13 14 isnzr 
 |-  ( L e. NzRing <-> ( L e. Ring /\ ( 1r ` L ) =/= ( 0g ` L ) ) )
16 9 12 15 3bitr4g 
 |-  ( ph -> ( K e. NzRing <-> L e. NzRing ) )