Metamath Proof Explorer


Theorem ehl0

Description: The Euclidean space of dimension 0 consists of the neutral element only. (Contributed by AV, 12-Feb-2023)

Ref Expression
Hypotheses ehl0base.e 𝐸 = ( 𝔼hil ‘ 0 )
ehl0base.0 0 = ( 0g𝐸 )
Assertion ehl0 ( Base ‘ 𝐸 ) = { 0 }

Proof

Step Hyp Ref Expression
1 ehl0base.e 𝐸 = ( 𝔼hil ‘ 0 )
2 ehl0base.0 0 = ( 0g𝐸 )
3 1 ehl0base ( Base ‘ 𝐸 ) = { ∅ }
4 ovex ( 1 ... 0 ) ∈ V
5 0nn0 0 ∈ ℕ0
6 1 ehlval ( 0 ∈ ℕ0𝐸 = ( ℝ^ ‘ ( 1 ... 0 ) ) )
7 5 6 ax-mp 𝐸 = ( ℝ^ ‘ ( 1 ... 0 ) )
8 fz10 ( 1 ... 0 ) = ∅
9 8 xpeq1i ( ( 1 ... 0 ) × { 0 } ) = ( ∅ × { 0 } )
10 9 eqcomi ( ∅ × { 0 } ) = ( ( 1 ... 0 ) × { 0 } )
11 7 10 rrx0 ( ( 1 ... 0 ) ∈ V → ( 0g𝐸 ) = ( ∅ × { 0 } ) )
12 4 11 ax-mp ( 0g𝐸 ) = ( ∅ × { 0 } )
13 2 12 eqtri 0 = ( ∅ × { 0 } )
14 0xp ( ∅ × { 0 } ) = ∅
15 13 14 eqtri 0 = ∅
16 15 eqcomi ∅ = 0
17 16 sneqi { ∅ } = { 0 }
18 3 17 eqtri ( Base ‘ 𝐸 ) = { 0 }