Metamath Proof Explorer


Theorem dih0bN

Description: A lattice element is zero iff its isomorphism is the zero subspace. (Contributed by NM, 16-Aug-2014) (New usage is discouraged.)

Ref Expression
Hypotheses dih0b.b 𝐵 = ( Base ‘ 𝐾 )
dih0b.h 𝐻 = ( LHyp ‘ 𝐾 )
dih0b.o 0 = ( 0. ‘ 𝐾 )
dih0b.i 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
dih0b.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
dih0b.z 𝑍 = ( 0g𝑈 )
dih0b.k ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) )
dih0b.x ( 𝜑𝑋𝐵 )
Assertion dih0bN ( 𝜑 → ( 𝑋 = 0 ↔ ( 𝐼𝑋 ) = { 𝑍 } ) )

Proof

Step Hyp Ref Expression
1 dih0b.b 𝐵 = ( Base ‘ 𝐾 )
2 dih0b.h 𝐻 = ( LHyp ‘ 𝐾 )
3 dih0b.o 0 = ( 0. ‘ 𝐾 )
4 dih0b.i 𝐼 = ( ( DIsoH ‘ 𝐾 ) ‘ 𝑊 )
5 dih0b.u 𝑈 = ( ( DVecH ‘ 𝐾 ) ‘ 𝑊 )
6 dih0b.z 𝑍 = ( 0g𝑈 )
7 dih0b.k ( 𝜑 → ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) )
8 dih0b.x ( 𝜑𝑋𝐵 )
9 7 simpld ( 𝜑𝐾 ∈ HL )
10 hlop ( 𝐾 ∈ HL → 𝐾 ∈ OP )
11 1 3 op0cl ( 𝐾 ∈ OP → 0𝐵 )
12 9 10 11 3syl ( 𝜑0𝐵 )
13 1 2 4 dih11 ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑋𝐵0𝐵 ) → ( ( 𝐼𝑋 ) = ( 𝐼0 ) ↔ 𝑋 = 0 ) )
14 7 8 12 13 syl3anc ( 𝜑 → ( ( 𝐼𝑋 ) = ( 𝐼0 ) ↔ 𝑋 = 0 ) )
15 3 2 4 5 6 dih0 ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝐼0 ) = { 𝑍 } )
16 7 15 syl ( 𝜑 → ( 𝐼0 ) = { 𝑍 } )
17 16 eqeq2d ( 𝜑 → ( ( 𝐼𝑋 ) = ( 𝐼0 ) ↔ ( 𝐼𝑋 ) = { 𝑍 } ) )
18 14 17 bitr3d ( 𝜑 → ( 𝑋 = 0 ↔ ( 𝐼𝑋 ) = { 𝑍 } ) )