Theorem shless

Description: Subset implies subset of subspace sum. (Contributed by Mario Carneiro, 15-May-2014) (New usage is discouraged.)

	Ref	Expression
Assertion	shless	⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ∧ 𝐶 ∈ Sℋ ) ∧ 𝐴 ⊆ 𝐵 ) → ( 𝐴 +ℋ 𝐶 ) ⊆ ( 𝐵 +ℋ 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		ssrexv	⊢ ( 𝐴 ⊆ 𝐵 → ( ∃ 𝑦 ∈ 𝐴 ∃ 𝑧 ∈ 𝐶 𝑥 = ( 𝑦 +ℎ 𝑧 ) → ∃ 𝑦 ∈ 𝐵 ∃ 𝑧 ∈ 𝐶 𝑥 = ( 𝑦 +ℎ 𝑧 ) ) )
2	1	adantl	⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ∧ 𝐶 ∈ Sℋ ) ∧ 𝐴 ⊆ 𝐵 ) → ( ∃ 𝑦 ∈ 𝐴 ∃ 𝑧 ∈ 𝐶 𝑥 = ( 𝑦 +ℎ 𝑧 ) → ∃ 𝑦 ∈ 𝐵 ∃ 𝑧 ∈ 𝐶 𝑥 = ( 𝑦 +ℎ 𝑧 ) ) )
3		simpl1	⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ∧ 𝐶 ∈ Sℋ ) ∧ 𝐴 ⊆ 𝐵 ) → 𝐴 ∈ Sℋ )
4		simpl3	⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ∧ 𝐶 ∈ Sℋ ) ∧ 𝐴 ⊆ 𝐵 ) → 𝐶 ∈ Sℋ )
5		shsel	⊢ ( ( 𝐴 ∈ Sℋ ∧ 𝐶 ∈ Sℋ ) → ( 𝑥 ∈ ( 𝐴 +ℋ 𝐶 ) ↔ ∃ 𝑦 ∈ 𝐴 ∃ 𝑧 ∈ 𝐶 𝑥 = ( 𝑦 +ℎ 𝑧 ) ) )
6	3 4 5	syl2anc	⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ∧ 𝐶 ∈ Sℋ ) ∧ 𝐴 ⊆ 𝐵 ) → ( 𝑥 ∈ ( 𝐴 +ℋ 𝐶 ) ↔ ∃ 𝑦 ∈ 𝐴 ∃ 𝑧 ∈ 𝐶 𝑥 = ( 𝑦 +ℎ 𝑧 ) ) )
7		simpl2	⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ∧ 𝐶 ∈ Sℋ ) ∧ 𝐴 ⊆ 𝐵 ) → 𝐵 ∈ Sℋ )
8		shsel	⊢ ( ( 𝐵 ∈ Sℋ ∧ 𝐶 ∈ Sℋ ) → ( 𝑥 ∈ ( 𝐵 +ℋ 𝐶 ) ↔ ∃ 𝑦 ∈ 𝐵 ∃ 𝑧 ∈ 𝐶 𝑥 = ( 𝑦 +ℎ 𝑧 ) ) )
9	7 4 8	syl2anc	⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ∧ 𝐶 ∈ Sℋ ) ∧ 𝐴 ⊆ 𝐵 ) → ( 𝑥 ∈ ( 𝐵 +ℋ 𝐶 ) ↔ ∃ 𝑦 ∈ 𝐵 ∃ 𝑧 ∈ 𝐶 𝑥 = ( 𝑦 +ℎ 𝑧 ) ) )
10	2 6 9	3imtr4d	⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ∧ 𝐶 ∈ Sℋ ) ∧ 𝐴 ⊆ 𝐵 ) → ( 𝑥 ∈ ( 𝐴 +ℋ 𝐶 ) → 𝑥 ∈ ( 𝐵 +ℋ 𝐶 ) ) )
11	10	ssrdv	⊢ ( ( ( 𝐴 ∈ Sℋ ∧ 𝐵 ∈ Sℋ ∧ 𝐶 ∈ Sℋ ) ∧ 𝐴 ⊆ 𝐵 ) → ( 𝐴 +ℋ 𝐶 ) ⊆ ( 𝐵 +ℋ 𝐶 ) )