grppncan

Description: Cancellation law for subtraction ( pncan analog). (Contributed by NM, 16-Apr-2014)

	Ref	Expression
Hypotheses	grpsubadd.b	$⊢ B = {Base}_{G}$
	grpsubadd.p	$⊢ \underset{˙}{+} = +_{G}$
	grpsubadd.m	$⊢ \underset{˙}{-} = -_{G}$
Assertion	grppncan	$⊢ (G \in Grp \land X \in B \land Y \in B) \to (X \underset{˙}{+} Y) \underset{˙}{-} Y = X$

Step	Hyp	Ref	Expression
1		grpsubadd.b	$⊢ B = {Base}_{G}$
2		grpsubadd.p	$⊢ \underset{˙}{+} = +_{G}$
3		grpsubadd.m	$⊢ \underset{˙}{-} = -_{G}$
4		simp1	$⊢ (G \in Grp \land X \in B \land Y \in B) \to G \in Grp$
5		simp2	$⊢ (G \in Grp \land X \in B \land Y \in B) \to X \in B$
6		simp3	$⊢ (G \in Grp \land X \in B \land Y \in B) \to Y \in B$
7	1 2 3	grpaddsubass	$⊢ (G \in Grp \land (X \in B \land Y \in B \land Y \in B)) \to (X \underset{˙}{+} Y) \underset{˙}{-} Y = X \underset{˙}{+} (Y \underset{˙}{-} Y)$
8	4 5 6 6 7	syl13anc	$⊢ (G \in Grp \land X \in B \land Y \in B) \to (X \underset{˙}{+} Y) \underset{˙}{-} Y = X \underset{˙}{+} (Y \underset{˙}{-} Y)$
9		eqid	$⊢ 0_{G} = 0_{G}$
10	1 9 3	grpsubid	$⊢ (G \in Grp \land Y \in B) \to Y \underset{˙}{-} Y = 0_{G}$
11	10	oveq2d	$⊢ (G \in Grp \land Y \in B) \to X \underset{˙}{+} (Y \underset{˙}{-} Y) = X \underset{˙}{+} 0_{G}$
12	11	3adant2	$⊢ (G \in Grp \land X \in B \land Y \in B) \to X \underset{˙}{+} (Y \underset{˙}{-} Y) = X \underset{˙}{+} 0_{G}$
13	1 2 9	grprid	$⊢ (G \in Grp \land X \in B) \to X \underset{˙}{+} 0_{G} = X$
14	13	3adant3	$⊢ (G \in Grp \land X \in B \land Y \in B) \to X \underset{˙}{+} 0_{G} = X$
15	8 12 14	3eqtrd	$⊢ (G \in Grp \land X \in B \land Y \in B) \to (X \underset{˙}{+} Y) \underset{˙}{-} Y = X$

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