Metamath Proof Explorer


Theorem obs2ss

Description: A basis has no proper subsets that are also bases. (Contributed by Mario Carneiro, 23-Oct-2015)

Ref Expression
Assertion obs2ss ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) → 𝐶 = 𝐵 )

Proof

Step Hyp Ref Expression
1 simp3 ⊢ ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) → 𝐶 ⊆ 𝐵 )
2 eqid ⊢ ( 0g ‘ 𝑊 ) = ( 0g ‘ 𝑊 )
3 2 obsne0 ⊢ ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝑥 ∈ 𝐵 ) → 𝑥 ≠ ( 0g ‘ 𝑊 ) )
4 3 3ad2antl1 ⊢ ( ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) ∧ 𝑥 ∈ 𝐵 ) → 𝑥 ≠ ( 0g ‘ 𝑊 ) )
5 eqid ⊢ ( ocv ‘ 𝑊 ) = ( ocv ‘ 𝑊 )
6 5 obselocv ⊢ ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ∧ 𝑥 ∈ 𝐵 ) → ( 𝑥 ∈ ( ( ocv ‘ 𝑊 ) ‘ 𝐶 ) ↔ ¬ 𝑥 ∈ 𝐶 ) )
7 6 3expa ⊢ ( ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) ∧ 𝑥 ∈ 𝐵 ) → ( 𝑥 ∈ ( ( ocv ‘ 𝑊 ) ‘ 𝐶 ) ↔ ¬ 𝑥 ∈ 𝐶 ) )
8 7 3adantl2 ⊢ ( ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) ∧ 𝑥 ∈ 𝐵 ) → ( 𝑥 ∈ ( ( ocv ‘ 𝑊 ) ‘ 𝐶 ) ↔ ¬ 𝑥 ∈ 𝐶 ) )
9 simpl2 ⊢ ( ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) ∧ 𝑥 ∈ 𝐵 ) → 𝐶 ∈ ( OBasis ‘ 𝑊 ) )
10 2 5 obsocv ⊢ ( 𝐶 ∈ ( OBasis ‘ 𝑊 ) → ( ( ocv ‘ 𝑊 ) ‘ 𝐶 ) = { ( 0g ‘ 𝑊 ) } )
11 9 10 syl ⊢ ( ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) ∧ 𝑥 ∈ 𝐵 ) → ( ( ocv ‘ 𝑊 ) ‘ 𝐶 ) = { ( 0g ‘ 𝑊 ) } )
12 11 eleq2d ⊢ ( ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) ∧ 𝑥 ∈ 𝐵 ) → ( 𝑥 ∈ ( ( ocv ‘ 𝑊 ) ‘ 𝐶 ) ↔ 𝑥 ∈ { ( 0g ‘ 𝑊 ) } ) )
13 elsni ⊢ ( 𝑥 ∈ { ( 0g ‘ 𝑊 ) } → 𝑥 = ( 0g ‘ 𝑊 ) )
14 12 13 biimtrdi ⊢ ( ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) ∧ 𝑥 ∈ 𝐵 ) → ( 𝑥 ∈ ( ( ocv ‘ 𝑊 ) ‘ 𝐶 ) → 𝑥 = ( 0g ‘ 𝑊 ) ) )
15 8 14 sylbird ⊢ ( ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) ∧ 𝑥 ∈ 𝐵 ) → ( ¬ 𝑥 ∈ 𝐶 → 𝑥 = ( 0g ‘ 𝑊 ) ) )
16 15 necon1ad ⊢ ( ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) ∧ 𝑥 ∈ 𝐵 ) → ( 𝑥 ≠ ( 0g ‘ 𝑊 ) → 𝑥 ∈ 𝐶 ) )
17 4 16 mpd ⊢ ( ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) ∧ 𝑥 ∈ 𝐵 ) → 𝑥 ∈ 𝐶 )
18 1 17 eqelssd ⊢ ( ( 𝐵 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ∈ ( OBasis ‘ 𝑊 ) ∧ 𝐶 ⊆ 𝐵 ) → 𝐶 = 𝐵 )