Metamath Proof Explorer


Theorem elspansn5

Description: A vector belonging to both a subspace and the span of the singleton of a vector not in it must be zero. (Contributed by NM, 17-Dec-2004) (New usage is discouraged.)

Ref Expression
Assertion elspansn5 ( 𝐴S → ( ( ( 𝐵 ∈ ℋ ∧ ¬ 𝐵𝐴 ) ∧ ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶𝐴 ) ) → 𝐶 = 0 ) )

Proof

Step Hyp Ref Expression
1 elspansn4 ( ( ( 𝐴S𝐵 ∈ ℋ ) ∧ ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶 ≠ 0 ) ) → ( 𝐵𝐴𝐶𝐴 ) )
2 1 biimprd ( ( ( 𝐴S𝐵 ∈ ℋ ) ∧ ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶 ≠ 0 ) ) → ( 𝐶𝐴𝐵𝐴 ) )
3 2 exp32 ( ( 𝐴S𝐵 ∈ ℋ ) → ( 𝐶 ∈ ( span ‘ { 𝐵 } ) → ( 𝐶 ≠ 0 → ( 𝐶𝐴𝐵𝐴 ) ) ) )
4 3 com34 ( ( 𝐴S𝐵 ∈ ℋ ) → ( 𝐶 ∈ ( span ‘ { 𝐵 } ) → ( 𝐶𝐴 → ( 𝐶 ≠ 0𝐵𝐴 ) ) ) )
5 4 imp32 ( ( ( 𝐴S𝐵 ∈ ℋ ) ∧ ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶𝐴 ) ) → ( 𝐶 ≠ 0𝐵𝐴 ) )
6 5 necon1bd ( ( ( 𝐴S𝐵 ∈ ℋ ) ∧ ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶𝐴 ) ) → ( ¬ 𝐵𝐴𝐶 = 0 ) )
7 6 exp31 ( 𝐴S → ( 𝐵 ∈ ℋ → ( ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶𝐴 ) → ( ¬ 𝐵𝐴𝐶 = 0 ) ) ) )
8 7 com34 ( 𝐴S → ( 𝐵 ∈ ℋ → ( ¬ 𝐵𝐴 → ( ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶𝐴 ) → 𝐶 = 0 ) ) ) )
9 8 imp4c ( 𝐴S → ( ( ( 𝐵 ∈ ℋ ∧ ¬ 𝐵𝐴 ) ∧ ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶𝐴 ) ) → 𝐶 = 0 ) )