Theorem ablpncan2

Description: Cancellation law for subtraction. (Contributed by NM, 2-Oct-2014)

	Ref	Expression
Hypotheses	ablsubadd.b	⊢ 𝐵 = ( Base ‘ 𝐺 )
	ablsubadd.p	⊢ + = ( +g ‘ 𝐺 )
	ablsubadd.m	⊢ − = ( -g ‘ 𝐺 )
Assertion	ablpncan2	⊢ ( ( 𝐺 ∈ Abel ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 + 𝑌 ) − 𝑋 ) = 𝑌 )

Proof

Step	Hyp	Ref	Expression
1		ablsubadd.b	⊢ 𝐵 = ( Base ‘ 𝐺 )
2		ablsubadd.p	⊢ + = ( +g ‘ 𝐺 )
3		ablsubadd.m	⊢ − = ( -g ‘ 𝐺 )
4		simp1	⊢ ( ( 𝐺 ∈ Abel ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝐺 ∈ Abel )
5		simp2	⊢ ( ( 𝐺 ∈ Abel ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝑋 ∈ 𝐵 )
6		simp3	⊢ ( ( 𝐺 ∈ Abel ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝑌 ∈ 𝐵 )
7	1 2 3	abladdsub	⊢ ( ( 𝐺 ∈ Abel ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑋 ∈ 𝐵 ) ) → ( ( 𝑋 + 𝑌 ) − 𝑋 ) = ( ( 𝑋 − 𝑋 ) + 𝑌 ) )
8	4 5 6 5 7	syl13anc	⊢ ( ( 𝐺 ∈ Abel ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 + 𝑌 ) − 𝑋 ) = ( ( 𝑋 − 𝑋 ) + 𝑌 ) )
9		ablgrp	⊢ ( 𝐺 ∈ Abel → 𝐺 ∈ Grp )
10	4 9	syl	⊢ ( ( 𝐺 ∈ Abel ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝐺 ∈ Grp )
11		eqid	⊢ ( 0g ‘ 𝐺 ) = ( 0g ‘ 𝐺 )
12	1 11 3	grpsubid	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑋 − 𝑋 ) = ( 0g ‘ 𝐺 ) )
13	10 5 12	syl2anc	⊢ ( ( 𝐺 ∈ Abel ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 − 𝑋 ) = ( 0g ‘ 𝐺 ) )
14	13	oveq1d	⊢ ( ( 𝐺 ∈ Abel ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 − 𝑋 ) + 𝑌 ) = ( ( 0g ‘ 𝐺 ) + 𝑌 ) )
15	1 2 11	grplid	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( ( 0g ‘ 𝐺 ) + 𝑌 ) = 𝑌 )
16	10 6 15	syl2anc	⊢ ( ( 𝐺 ∈ Abel ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 0g ‘ 𝐺 ) + 𝑌 ) = 𝑌 )
17	8 14 16	3eqtrd	⊢ ( ( 𝐺 ∈ Abel ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 + 𝑌 ) − 𝑋 ) = 𝑌 )