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Metamath Proof Explorer

Theorem shsvs

Description: Vector subtraction belongs to subspace sum. (Contributed by NM, 15-Dec-2004) (New usage is discouraged.)

Ref Expression
Assertion shsvs ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( ( 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐵 ) → ( 𝐶 −ℎ 𝐷 ) ∈ ( 𝐴 +ℋ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 shscl ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( 𝐴 +ℋ 𝐵 ) ∈ Sℋ )
2 1 a1d ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( ( 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐵 ) → ( 𝐴 +ℋ 𝐵 ) ∈ Sℋ ) )
3 shsel1 ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( 𝐶 ∈ 𝐴 → 𝐶 ∈ ( 𝐴 +ℋ 𝐵 ) ) )
4 3 adantrd ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( ( 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐵 ) → 𝐶 ∈ ( 𝐴 +ℋ 𝐵 ) ) )
5 shsel2 ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( 𝐷 ∈ 𝐵 → 𝐷 ∈ ( 𝐴 +ℋ 𝐵 ) ) )
6 5 adantld ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( ( 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐵 ) → 𝐷 ∈ ( 𝐴 +ℋ 𝐵 ) ) )
7 2 4 6 3jcad ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( ( 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐵 ) → ( ( 𝐴 +ℋ 𝐵 ) ∈ Sℋ ∧ 𝐶 ∈ ( 𝐴 +ℋ 𝐵 ) ∧ 𝐷 ∈ ( 𝐴 +ℋ 𝐵 ) ) ) )
8 shsubcl ⊢ ( ( ( 𝐴 +ℋ 𝐵 ) ∈ Sℋ ∧ 𝐶 ∈ ( 𝐴 +ℋ 𝐵 ) ∧ 𝐷 ∈ ( 𝐴 +ℋ 𝐵 ) ) → ( 𝐶 −ℎ 𝐷 ) ∈ ( 𝐴 +ℋ 𝐵 ) )
9 7 8 syl6 ⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ) → ( ( 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐵 ) → ( 𝐶 −ℎ 𝐷 ) ∈ ( 𝐴 +ℋ 𝐵 ) ) )