Theorem lsmub2

Description: Subgroup sum is an upper bound of its arguments. (Contributed by NM, 6-Feb-2014) (Revised by Mario Carneiro, 19-Apr-2016)

	Ref	Expression
Hypothesis	lsmub1.p	$$⊢\underset{\dot{}}{\oplus }=\mathrm{LSSum}\left(G\right)$$
Assertion	lsmub2	$$⊢\left(T\in \mathrm{SubGrp}\left(G\right)\wedge U\in \mathrm{SubGrp}\left(G\right)\right)\to U\subseteq T\underset{\dot{}}{\oplus }U$$

Proof

Step	Hyp	Ref	Expression
1		lsmub1.p	$$⊢\underset{\dot{}}{\oplus }=\mathrm{LSSum}\left(G\right)$$
2		subgsubm	$$⊢T\in \mathrm{SubGrp}\left(G\right)\to T\in \mathrm{SubMnd}\left(G\right)$$
3		eqid	$$⊢{\mathrm{Base}}_G={\mathrm{Base}}_G$$
4	3	subgss	$$⊢U\in \mathrm{SubGrp}\left(G\right)\to U\subseteq {\mathrm{Base}}_G$$
5	3 1	lsmub2x	$$⊢\left(T\in \mathrm{SubMnd}\left(G\right)\wedge U\subseteq {\mathrm{Base}}_G\right)\to U\subseteq T\underset{\dot{}}{\oplus }U$$
6	2 4 5	syl2an	$$⊢\left(T\in \mathrm{SubGrp}\left(G\right)\wedge U\in \mathrm{SubGrp}\left(G\right)\right)\to U\subseteq T\underset{\dot{}}{\oplus }U$$