Metamath Proof Explorer


Theorem lazval

Description: An adjunction of a complex number to the ring of integers. (Contributed by AV, 24-Oct-2026)

Ref Expression
Assertion lazval
|- ( S e. CC -> ( Z[]l ` S ) = { x e. CC | E. a e. ZZ E. b e. ZZ x = ( a + ( b x. S ) ) } )

Proof

Step Hyp Ref Expression
1 oveq2
 |-  ( s = S -> ( b x. s ) = ( b x. S ) )
2 1 oveq2d
 |-  ( s = S -> ( a + ( b x. s ) ) = ( a + ( b x. S ) ) )
3 2 eqeq2d
 |-  ( s = S -> ( x = ( a + ( b x. s ) ) <-> x = ( a + ( b x. S ) ) ) )
4 3 2rexbidv
 |-  ( s = S -> ( E. a e. ZZ E. b e. ZZ x = ( a + ( b x. s ) ) <-> E. a e. ZZ E. b e. ZZ x = ( a + ( b x. S ) ) ) )
5 4 rabbidv
 |-  ( s = S -> { x e. CC | E. a e. ZZ E. b e. ZZ x = ( a + ( b x. s ) ) } = { x e. CC | E. a e. ZZ E. b e. ZZ x = ( a + ( b x. S ) ) } )
6 df-laz
 |-  Z[]l = ( s e. CC |-> { x e. CC | E. a e. ZZ E. b e. ZZ x = ( a + ( b x. s ) ) } )
7 cnex
 |-  CC e. _V
8 7 rabex
 |-  { x e. CC | E. a e. ZZ E. b e. ZZ x = ( a + ( b x. S ) ) } e. _V
9 5 6 8 fvmpt
 |-  ( S e. CC -> ( Z[]l ` S ) = { x e. CC | E. a e. ZZ E. b e. ZZ x = ( a + ( b x. S ) ) } )