Metamath Proof Explorer


Theorem ablnncan

Description: Cancellation law for group subtraction. ( nncan analog.) (Contributed by NM, 7-Apr-2015)

Ref Expression
Hypotheses ablnncan.b 𝐵 = ( Base ‘ 𝐺 )
ablnncan.m = ( -g𝐺 )
ablnncan.g ( 𝜑𝐺 ∈ Abel )
ablnncan.x ( 𝜑𝑋𝐵 )
ablnncan.y ( 𝜑𝑌𝐵 )
Assertion ablnncan ( 𝜑 → ( 𝑋 ( 𝑋 𝑌 ) ) = 𝑌 )

Proof

Step Hyp Ref Expression
1 ablnncan.b 𝐵 = ( Base ‘ 𝐺 )
2 ablnncan.m = ( -g𝐺 )
3 ablnncan.g ( 𝜑𝐺 ∈ Abel )
4 ablnncan.x ( 𝜑𝑋𝐵 )
5 ablnncan.y ( 𝜑𝑌𝐵 )
6 eqid ( +g𝐺 ) = ( +g𝐺 )
7 1 6 2 3 4 4 5 ablsubsub ( 𝜑 → ( 𝑋 ( 𝑋 𝑌 ) ) = ( ( 𝑋 𝑋 ) ( +g𝐺 ) 𝑌 ) )
8 ablgrp ( 𝐺 ∈ Abel → 𝐺 ∈ Grp )
9 3 8 syl ( 𝜑𝐺 ∈ Grp )
10 eqid ( 0g𝐺 ) = ( 0g𝐺 )
11 1 10 2 grpsubid ( ( 𝐺 ∈ Grp ∧ 𝑋𝐵 ) → ( 𝑋 𝑋 ) = ( 0g𝐺 ) )
12 9 4 11 syl2anc ( 𝜑 → ( 𝑋 𝑋 ) = ( 0g𝐺 ) )
13 12 oveq1d ( 𝜑 → ( ( 𝑋 𝑋 ) ( +g𝐺 ) 𝑌 ) = ( ( 0g𝐺 ) ( +g𝐺 ) 𝑌 ) )
14 1 6 10 grplid ( ( 𝐺 ∈ Grp ∧ 𝑌𝐵 ) → ( ( 0g𝐺 ) ( +g𝐺 ) 𝑌 ) = 𝑌 )
15 9 5 14 syl2anc ( 𝜑 → ( ( 0g𝐺 ) ( +g𝐺 ) 𝑌 ) = 𝑌 )
16 7 13 15 3eqtrd ( 𝜑 → ( 𝑋 ( 𝑋 𝑌 ) ) = 𝑌 )