Theorem lsmub1

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	lsmub1	$$⊢\left(T\in \mathrm{SubGrp}\left(G\right)\wedge U\in \mathrm{SubGrp}\left(G\right)\right)\to T\subseteq T\underset{\dot{}}{\oplus }U$$

Proof

Step	Hyp	Ref	Expression
1		lsmub1.p	$$⊢\underset{\dot{}}{\oplus }=\mathrm{LSSum}\left(G\right)$$
2		eqid	$$⊢{\mathrm{Base}}_G={\mathrm{Base}}_G$$
3	2	subgss	$$⊢T\in \mathrm{SubGrp}\left(G\right)\to T\subseteq {\mathrm{Base}}_G$$
4		subgsubm	$$⊢U\in \mathrm{SubGrp}\left(G\right)\to U\in \mathrm{SubMnd}\left(G\right)$$
5	2 1	lsmub1x	$$⊢\left(T\subseteq {\mathrm{Base}}_G\wedge U\in \mathrm{SubMnd}\left(G\right)\right)\to T\subseteq T\underset{\dot{}}{\oplus }U$$
6	3 4 5	syl2an	$$⊢\left(T\in \mathrm{SubGrp}\left(G\right)\wedge U\in \mathrm{SubGrp}\left(G\right)\right)\to T\subseteq T\underset{\dot{}}{\oplus }U$$