Theorem grppncan

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

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

Proof

Step	Hyp	Ref	Expression
1		grpsubadd.b	⊢ 𝐵 = ( Base ‘ 𝐺 )
2		grpsubadd.p	⊢ + = ( +g ‘ 𝐺 )
3		grpsubadd.m	⊢ − = ( -g ‘ 𝐺 )
4		simp1	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝐺 ∈ Grp )
5		simp2	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝑋 ∈ 𝐵 )
6		simp3	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → 𝑌 ∈ 𝐵 )
7	1 2 3	grpaddsubass	⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( 𝑋 + 𝑌 ) − 𝑌 ) = ( 𝑋 + ( 𝑌 − 𝑌 ) ) )
8	4 5 6 6 7	syl13anc	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 + 𝑌 ) − 𝑌 ) = ( 𝑋 + ( 𝑌 − 𝑌 ) ) )
9		eqid	⊢ ( 0g ‘ 𝐺 ) = ( 0g ‘ 𝐺 )
10	1 9 3	grpsubid	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( 𝑌 − 𝑌 ) = ( 0g ‘ 𝐺 ) )
11	10	oveq2d	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 + ( 𝑌 − 𝑌 ) ) = ( 𝑋 + ( 0g ‘ 𝐺 ) ) )
12	11	3adant2	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 + ( 𝑌 − 𝑌 ) ) = ( 𝑋 + ( 0g ‘ 𝐺 ) ) )
13	1 2 9	grprid	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑋 + ( 0g ‘ 𝐺 ) ) = 𝑋 )
14	13	3adant3	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 + ( 0g ‘ 𝐺 ) ) = 𝑋 )
15	8 12 14	3eqtrd	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 + 𝑌 ) − 𝑌 ) = 𝑋 )