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Metamath Proof Explorer

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 ) )