Metamath Proof Explorer


Theorem elspansn4

Description: A span membership condition implying two vectors belong to the same subspace. (Contributed by NM, 17-Dec-2004) (New usage is discouraged.)

Ref Expression
Assertion elspansn4 ( ( ( 𝐴S𝐵 ∈ ℋ ) ∧ ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶 ≠ 0 ) ) → ( 𝐵𝐴𝐶𝐴 ) )

Proof

Step Hyp Ref Expression
1 elspansn3 ( ( 𝐴S𝐵𝐴𝐶 ∈ ( span ‘ { 𝐵 } ) ) → 𝐶𝐴 )
2 1 3exp ( 𝐴S → ( 𝐵𝐴 → ( 𝐶 ∈ ( span ‘ { 𝐵 } ) → 𝐶𝐴 ) ) )
3 2 com23 ( 𝐴S → ( 𝐶 ∈ ( span ‘ { 𝐵 } ) → ( 𝐵𝐴𝐶𝐴 ) ) )
4 3 imp ( ( 𝐴S𝐶 ∈ ( span ‘ { 𝐵 } ) ) → ( 𝐵𝐴𝐶𝐴 ) )
5 4 ad2ant2r ( ( ( 𝐴S𝐵 ∈ ℋ ) ∧ ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶 ≠ 0 ) ) → ( 𝐵𝐴𝐶𝐴 ) )
6 spansnid ( 𝐵 ∈ ℋ → 𝐵 ∈ ( span ‘ { 𝐵 } ) )
7 6 ad2antrr ( ( ( 𝐵 ∈ ℋ ∧ 𝐶 ≠ 0 ) ∧ 𝐶 ∈ ( span ‘ { 𝐵 } ) ) → 𝐵 ∈ ( span ‘ { 𝐵 } ) )
8 spansneleq ( ( 𝐵 ∈ ℋ ∧ 𝐶 ≠ 0 ) → ( 𝐶 ∈ ( span ‘ { 𝐵 } ) → ( span ‘ { 𝐶 } ) = ( span ‘ { 𝐵 } ) ) )
9 8 imp ( ( ( 𝐵 ∈ ℋ ∧ 𝐶 ≠ 0 ) ∧ 𝐶 ∈ ( span ‘ { 𝐵 } ) ) → ( span ‘ { 𝐶 } ) = ( span ‘ { 𝐵 } ) )
10 7 9 eleqtrrd ( ( ( 𝐵 ∈ ℋ ∧ 𝐶 ≠ 0 ) ∧ 𝐶 ∈ ( span ‘ { 𝐵 } ) ) → 𝐵 ∈ ( span ‘ { 𝐶 } ) )
11 elspansn3 ( ( 𝐴S𝐶𝐴𝐵 ∈ ( span ‘ { 𝐶 } ) ) → 𝐵𝐴 )
12 11 3expia ( ( 𝐴S𝐶𝐴 ) → ( 𝐵 ∈ ( span ‘ { 𝐶 } ) → 𝐵𝐴 ) )
13 10 12 syl5 ( ( 𝐴S𝐶𝐴 ) → ( ( ( 𝐵 ∈ ℋ ∧ 𝐶 ≠ 0 ) ∧ 𝐶 ∈ ( span ‘ { 𝐵 } ) ) → 𝐵𝐴 ) )
14 13 exp4b ( 𝐴S → ( 𝐶𝐴 → ( ( 𝐵 ∈ ℋ ∧ 𝐶 ≠ 0 ) → ( 𝐶 ∈ ( span ‘ { 𝐵 } ) → 𝐵𝐴 ) ) ) )
15 14 com24 ( 𝐴S → ( 𝐶 ∈ ( span ‘ { 𝐵 } ) → ( ( 𝐵 ∈ ℋ ∧ 𝐶 ≠ 0 ) → ( 𝐶𝐴𝐵𝐴 ) ) ) )
16 15 exp4a ( 𝐴S → ( 𝐶 ∈ ( span ‘ { 𝐵 } ) → ( 𝐵 ∈ ℋ → ( 𝐶 ≠ 0 → ( 𝐶𝐴𝐵𝐴 ) ) ) ) )
17 16 com23 ( 𝐴S → ( 𝐵 ∈ ℋ → ( 𝐶 ∈ ( span ‘ { 𝐵 } ) → ( 𝐶 ≠ 0 → ( 𝐶𝐴𝐵𝐴 ) ) ) ) )
18 17 imp43 ( ( ( 𝐴S𝐵 ∈ ℋ ) ∧ ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶 ≠ 0 ) ) → ( 𝐶𝐴𝐵𝐴 ) )
19 5 18 impbid ( ( ( 𝐴S𝐵 ∈ ℋ ) ∧ ( 𝐶 ∈ ( span ‘ { 𝐵 } ) ∧ 𝐶 ≠ 0 ) ) → ( 𝐵𝐴𝐶𝐴 ) )