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Metamath Proof Explorer

Theorem isdomn5

Description: The equivalence between the right conjuncts in the right hand sides of isdomn and isdomn2 , in predicate calculus form. (Contributed by SN, 16-Sep-2024)

Ref Expression
Assertion isdomn5 
|- ( A. a e. B A. b e. B ( ( a .x. b ) = .0. -> ( a = .0. \/ b = .0. ) ) <-> A. a e. ( B \ { .0. } ) A. b e. B ( ( a .x. b ) = .0. -> b = .0. ) )

Proof

Step Hyp Ref Expression
1 bi2.04 
 |-  ( ( -. a = .0. -> ( ( a .x. b ) = .0. -> b = .0. ) ) <-> ( ( a .x. b ) = .0. -> ( -. a = .0. -> b = .0. ) ) )
2 df-ne 
 |-  ( a =/= .0. <-> -. a = .0. )
3 2 imbi1i 
 |-  ( ( a =/= .0. -> ( ( a .x. b ) = .0. -> b = .0. ) ) <-> ( -. a = .0. -> ( ( a .x. b ) = .0. -> b = .0. ) ) )
4 df-or 
 |-  ( ( a = .0. \/ b = .0. ) <-> ( -. a = .0. -> b = .0. ) )
5 4 imbi2i 
 |-  ( ( ( a .x. b ) = .0. -> ( a = .0. \/ b = .0. ) ) <-> ( ( a .x. b ) = .0. -> ( -. a = .0. -> b = .0. ) ) )
6 1 3 5 3bitr4ri 
 |-  ( ( ( a .x. b ) = .0. -> ( a = .0. \/ b = .0. ) ) <-> ( a =/= .0. -> ( ( a .x. b ) = .0. -> b = .0. ) ) )
7 6 2ralbii 
 |-  ( A. a e. B A. b e. B ( ( a .x. b ) = .0. -> ( a = .0. \/ b = .0. ) ) <-> A. a e. B A. b e. B ( a =/= .0. -> ( ( a .x. b ) = .0. -> b = .0. ) ) )
8 r19.21v 
 |-  ( A. b e. B ( a =/= .0. -> ( ( a .x. b ) = .0. -> b = .0. ) ) <-> ( a =/= .0. -> A. b e. B ( ( a .x. b ) = .0. -> b = .0. ) ) )
9 8 ralbii 
 |-  ( A. a e. B A. b e. B ( a =/= .0. -> ( ( a .x. b ) = .0. -> b = .0. ) ) <-> A. a e. B ( a =/= .0. -> A. b e. B ( ( a .x. b ) = .0. -> b = .0. ) ) )
10 raldifsnb 
 |-  ( A. a e. B ( a =/= .0. -> A. b e. B ( ( a .x. b ) = .0. -> b = .0. ) ) <-> A. a e. ( B \ { .0. } ) A. b e. B ( ( a .x. b ) = .0. -> b = .0. ) )
11 7 9 10 3bitri 
 |-  ( A. a e. B A. b e. B ( ( a .x. b ) = .0. -> ( a = .0. \/ b = .0. ) ) <-> A. a e. ( B \ { .0. } ) A. b e. B ( ( a .x. b ) = .0. -> b = .0. ) )