Theorem lsmub1x

Description: Subgroup sum is an upper bound of its arguments. (Contributed by Mario Carneiro, 19-Apr-2016)

	Ref	Expression
Hypotheses	lsmless2.v	⊢ 𝐵 = ( Base ‘ 𝐺 )
	lsmless2.s	⊢ ⊕ = ( LSSum ‘ 𝐺 )
Assertion	lsmub1x	⊢ ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) → 𝑇 ⊆ ( 𝑇 ⊕ 𝑈 ) )

Proof

Step	Hyp	Ref	Expression
1		lsmless2.v	⊢ 𝐵 = ( Base ‘ 𝐺 )
2		lsmless2.s	⊢ ⊕ = ( LSSum ‘ 𝐺 )
3		submrcl	⊢ ( 𝑈 ∈ ( SubMnd ‘ 𝐺 ) → 𝐺 ∈ Mnd )
4	3	ad2antlr	⊢ ( ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) ∧ 𝑥 ∈ 𝑇 ) → 𝐺 ∈ Mnd )
5		simpll	⊢ ( ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) ∧ 𝑥 ∈ 𝑇 ) → 𝑇 ⊆ 𝐵 )
6		simpr	⊢ ( ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) ∧ 𝑥 ∈ 𝑇 ) → 𝑥 ∈ 𝑇 )
7	5 6	sseldd	⊢ ( ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) ∧ 𝑥 ∈ 𝑇 ) → 𝑥 ∈ 𝐵 )
8		eqid	⊢ ( +g ‘ 𝐺 ) = ( +g ‘ 𝐺 )
9		eqid	⊢ ( 0g ‘ 𝐺 ) = ( 0g ‘ 𝐺 )
10	1 8 9	mndrid	⊢ ( ( 𝐺 ∈ Mnd ∧ 𝑥 ∈ 𝐵 ) → ( 𝑥 ( +g ‘ 𝐺 ) ( 0g ‘ 𝐺 ) ) = 𝑥 )
11	4 7 10	syl2anc	⊢ ( ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) ∧ 𝑥 ∈ 𝑇 ) → ( 𝑥 ( +g ‘ 𝐺 ) ( 0g ‘ 𝐺 ) ) = 𝑥 )
12	1	submss	⊢ ( 𝑈 ∈ ( SubMnd ‘ 𝐺 ) → 𝑈 ⊆ 𝐵 )
13	12	ad2antlr	⊢ ( ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) ∧ 𝑥 ∈ 𝑇 ) → 𝑈 ⊆ 𝐵 )
14	9	subm0cl	⊢ ( 𝑈 ∈ ( SubMnd ‘ 𝐺 ) → ( 0g ‘ 𝐺 ) ∈ 𝑈 )
15	14	ad2antlr	⊢ ( ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) ∧ 𝑥 ∈ 𝑇 ) → ( 0g ‘ 𝐺 ) ∈ 𝑈 )
16	1 8 2	lsmelvalix	⊢ ( ( ( 𝐺 ∈ Mnd ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵 ) ∧ ( 𝑥 ∈ 𝑇 ∧ ( 0g ‘ 𝐺 ) ∈ 𝑈 ) ) → ( 𝑥 ( +g ‘ 𝐺 ) ( 0g ‘ 𝐺 ) ) ∈ ( 𝑇 ⊕ 𝑈 ) )
17	4 5 13 6 15 16	syl32anc	⊢ ( ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) ∧ 𝑥 ∈ 𝑇 ) → ( 𝑥 ( +g ‘ 𝐺 ) ( 0g ‘ 𝐺 ) ) ∈ ( 𝑇 ⊕ 𝑈 ) )
18	11 17	eqeltrrd	⊢ ( ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) ∧ 𝑥 ∈ 𝑇 ) → 𝑥 ∈ ( 𝑇 ⊕ 𝑈 ) )
19	18	ex	⊢ ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) → ( 𝑥 ∈ 𝑇 → 𝑥 ∈ ( 𝑇 ⊕ 𝑈 ) ) )
20	19	ssrdv	⊢ ( ( 𝑇 ⊆ 𝐵 ∧ 𝑈 ∈ ( SubMnd ‘ 𝐺 ) ) → 𝑇 ⊆ ( 𝑇 ⊕ 𝑈 ) )