Metamath Proof Explorer


Theorem sn-mul02

Description: mul02 without ax-mulcom . See https://github.com/icecream17/Stuff/blob/main/math/0A%3D0.md for an outline. (Contributed by SN, 30-Jun-2024)

Ref Expression
Assertion sn-mul02
|- ( A e. CC -> ( 0 x. A ) = 0 )

Proof

Step Hyp Ref Expression
1 cnre
 |-  ( A e. CC -> E. x e. RR E. y e. RR A = ( x + ( _i x. y ) ) )
2 0cnd
 |-  ( ( x e. RR /\ y e. RR ) -> 0 e. CC )
3 recn
 |-  ( x e. RR -> x e. CC )
4 3 adantr
 |-  ( ( x e. RR /\ y e. RR ) -> x e. CC )
5 ax-icn
 |-  _i e. CC
6 5 a1i
 |-  ( ( x e. RR /\ y e. RR ) -> _i e. CC )
7 recn
 |-  ( y e. RR -> y e. CC )
8 7 adantl
 |-  ( ( x e. RR /\ y e. RR ) -> y e. CC )
9 6 8 mulcld
 |-  ( ( x e. RR /\ y e. RR ) -> ( _i x. y ) e. CC )
10 2 4 9 adddid
 |-  ( ( x e. RR /\ y e. RR ) -> ( 0 x. ( x + ( _i x. y ) ) ) = ( ( 0 x. x ) + ( 0 x. ( _i x. y ) ) ) )
11 remul02
 |-  ( x e. RR -> ( 0 x. x ) = 0 )
12 11 adantr
 |-  ( ( x e. RR /\ y e. RR ) -> ( 0 x. x ) = 0 )
13 sn-0tie0
 |-  ( 0 x. _i ) = 0
14 13 oveq1i
 |-  ( ( 0 x. _i ) x. y ) = ( 0 x. y )
15 2 6 8 mulassd
 |-  ( ( x e. RR /\ y e. RR ) -> ( ( 0 x. _i ) x. y ) = ( 0 x. ( _i x. y ) ) )
16 remul02
 |-  ( y e. RR -> ( 0 x. y ) = 0 )
17 16 adantl
 |-  ( ( x e. RR /\ y e. RR ) -> ( 0 x. y ) = 0 )
18 14 15 17 3eqtr3a
 |-  ( ( x e. RR /\ y e. RR ) -> ( 0 x. ( _i x. y ) ) = 0 )
19 12 18 oveq12d
 |-  ( ( x e. RR /\ y e. RR ) -> ( ( 0 x. x ) + ( 0 x. ( _i x. y ) ) ) = ( 0 + 0 ) )
20 sn-00id
 |-  ( 0 + 0 ) = 0
21 19 20 eqtrdi
 |-  ( ( x e. RR /\ y e. RR ) -> ( ( 0 x. x ) + ( 0 x. ( _i x. y ) ) ) = 0 )
22 10 21 eqtrd
 |-  ( ( x e. RR /\ y e. RR ) -> ( 0 x. ( x + ( _i x. y ) ) ) = 0 )
23 oveq2
 |-  ( A = ( x + ( _i x. y ) ) -> ( 0 x. A ) = ( 0 x. ( x + ( _i x. y ) ) ) )
24 23 eqeq1d
 |-  ( A = ( x + ( _i x. y ) ) -> ( ( 0 x. A ) = 0 <-> ( 0 x. ( x + ( _i x. y ) ) ) = 0 ) )
25 22 24 syl5ibrcom
 |-  ( ( x e. RR /\ y e. RR ) -> ( A = ( x + ( _i x. y ) ) -> ( 0 x. A ) = 0 ) )
26 25 rexlimivv
 |-  ( E. x e. RR E. y e. RR A = ( x + ( _i x. y ) ) -> ( 0 x. A ) = 0 )
27 1 26 syl
 |-  ( A e. CC -> ( 0 x. A ) = 0 )