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Metamath Proof Explorer

Theorem ablpncan2

Description: Cancellation law for subtraction in an Abelian group. (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 )