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Metamath Proof Explorer

Theorem lsmval

Description: Subgroup sum value (for a left module or left vector space). (Contributed by NM, 4-Feb-2014) (Revised by Mario Carneiro, 19-Apr-2016)

		Ref	Expression
	Hypotheses	lsmval.v	$⊢ B = {Base}_{G}$
		lsmval.a	$⊢ \underset{˙}{+} = +_{G}$
		lsmval.p	$⊢ \underset{˙}{\oplus} = LSSum (G)$
	Assertion	lsmval	$⊢ (T \in SubGrp (G) \land U \in SubGrp (G)) \to T \underset{˙}{\oplus} U = ran (x \in T, y \in U ⟼ x \underset{˙}{+} y)$

Proof

Step	Hyp	Ref	Expression
1		lsmval.v	$⊢ B = {Base}_{G}$
2		lsmval.a	$⊢ \underset{˙}{+} = +_{G}$
3		lsmval.p	$⊢ \underset{˙}{\oplus} = LSSum (G)$
4		subgrcl	$⊢ T \in SubGrp (G) \to G \in Grp$
5	1	subgss	$⊢ T \in SubGrp (G) \to T \subseteq B$
6	1	subgss	$⊢ U \in SubGrp (G) \to U \subseteq B$
7	1 2 3	lsmvalx	$⊢ (G \in Grp \land T \subseteq B \land U \subseteq B) \to T \underset{˙}{\oplus} U = ran (x \in T, y \in U ⟼ x \underset{˙}{+} y)$
8	4 5 6 7	syl2an3an	$⊢ (T \in SubGrp (G) \land U \in SubGrp (G)) \to T \underset{˙}{\oplus} U = ran (x \in T, y \in U ⟼ x \underset{˙}{+} y)$