Metamath Proof Explorer


Theorem shs00i

Description: Two subspaces are zero iff their join is zero. (Contributed by NM, 7-Aug-2004) (New usage is discouraged.)

Ref Expression
Hypotheses shne0.1 𝐴S
shs00.2 𝐵S
Assertion shs00i ( ( 𝐴 = 0𝐵 = 0 ) ↔ ( 𝐴 + 𝐵 ) = 0 )

Proof

Step Hyp Ref Expression
1 shne0.1 𝐴S
2 shs00.2 𝐵S
3 oveq12 ( ( 𝐴 = 0𝐵 = 0 ) → ( 𝐴 + 𝐵 ) = ( 0 + 0 ) )
4 h0elsh 0S
5 4 shs0i ( 0 + 0 ) = 0
6 3 5 eqtrdi ( ( 𝐴 = 0𝐵 = 0 ) → ( 𝐴 + 𝐵 ) = 0 )
7 1 2 shsub1i 𝐴 ⊆ ( 𝐴 + 𝐵 )
8 sseq2 ( ( 𝐴 + 𝐵 ) = 0 → ( 𝐴 ⊆ ( 𝐴 + 𝐵 ) ↔ 𝐴 ⊆ 0 ) )
9 7 8 mpbii ( ( 𝐴 + 𝐵 ) = 0𝐴 ⊆ 0 )
10 shle0 ( 𝐴S → ( 𝐴 ⊆ 0𝐴 = 0 ) )
11 1 10 ax-mp ( 𝐴 ⊆ 0𝐴 = 0 )
12 9 11 sylib ( ( 𝐴 + 𝐵 ) = 0𝐴 = 0 )
13 2 1 shsub2i 𝐵 ⊆ ( 𝐴 + 𝐵 )
14 sseq2 ( ( 𝐴 + 𝐵 ) = 0 → ( 𝐵 ⊆ ( 𝐴 + 𝐵 ) ↔ 𝐵 ⊆ 0 ) )
15 13 14 mpbii ( ( 𝐴 + 𝐵 ) = 0𝐵 ⊆ 0 )
16 shle0 ( 𝐵S → ( 𝐵 ⊆ 0𝐵 = 0 ) )
17 2 16 ax-mp ( 𝐵 ⊆ 0𝐵 = 0 )
18 15 17 sylib ( ( 𝐴 + 𝐵 ) = 0𝐵 = 0 )
19 12 18 jca ( ( 𝐴 + 𝐵 ) = 0 → ( 𝐴 = 0𝐵 = 0 ) )
20 6 19 impbii ( ( 𝐴 = 0𝐵 = 0 ) ↔ ( 𝐴 + 𝐵 ) = 0 )