Theorem grpsubsub

Description: Double group subtraction. (Contributed by NM, 24-Feb-2008) (Revised by Mario Carneiro, 2-Dec-2014)

	Ref	Expression
Hypotheses	grpsubadd.b	⊢ 𝐵 = ( Base ‘ 𝐺 )
	grpsubadd.p	⊢ + = ( +g ‘ 𝐺 )
	grpsubadd.m	⊢ − = ( -g ‘ 𝐺 )
Assertion	grpsubsub	⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 − ( 𝑌 − 𝑍 ) ) = ( 𝑋 + ( 𝑍 − 𝑌 ) ) )

Proof

Step	Hyp	Ref	Expression
1		grpsubadd.b	⊢ 𝐵 = ( Base ‘ 𝐺 )
2		grpsubadd.p	⊢ + = ( +g ‘ 𝐺 )
3		grpsubadd.m	⊢ − = ( -g ‘ 𝐺 )
4		simpr1	⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → 𝑋 ∈ 𝐵 )
5	1 3	grpsubcl	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( 𝑌 − 𝑍 ) ∈ 𝐵 )
6	5	3adant3r1	⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑌 − 𝑍 ) ∈ 𝐵 )
7		eqid	⊢ ( invg ‘ 𝐺 ) = ( invg ‘ 𝐺 )
8	1 2 7 3	grpsubval	⊢ ( ( 𝑋 ∈ 𝐵 ∧ ( 𝑌 − 𝑍 ) ∈ 𝐵 ) → ( 𝑋 − ( 𝑌 − 𝑍 ) ) = ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ ( 𝑌 − 𝑍 ) ) ) )
9	4 6 8	syl2anc	⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 − ( 𝑌 − 𝑍 ) ) = ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ ( 𝑌 − 𝑍 ) ) ) )
10	1 3 7	grpinvsub	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) → ( ( invg ‘ 𝐺 ) ‘ ( 𝑌 − 𝑍 ) ) = ( 𝑍 − 𝑌 ) )
11	10	3adant3r1	⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( ( invg ‘ 𝐺 ) ‘ ( 𝑌 − 𝑍 ) ) = ( 𝑍 − 𝑌 ) )
12	11	oveq2d	⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 + ( ( invg ‘ 𝐺 ) ‘ ( 𝑌 − 𝑍 ) ) ) = ( 𝑋 + ( 𝑍 − 𝑌 ) ) )
13	9 12	eqtrd	⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ) ) → ( 𝑋 − ( 𝑌 − 𝑍 ) ) = ( 𝑋 + ( 𝑍 − 𝑌 ) ) )