Metamath Proof Explorer


Theorem grpsubinv

Description: Subtraction of an inverse. (Contributed by NM, 7-Apr-2015)

Ref Expression
Hypotheses grpsubinv.b 𝐵 = ( Base ‘ 𝐺 )
grpsubinv.p + = ( +g𝐺 )
grpsubinv.m = ( -g𝐺 )
grpsubinv.n 𝑁 = ( invg𝐺 )
grpsubinv.g ( 𝜑𝐺 ∈ Grp )
grpsubinv.x ( 𝜑𝑋𝐵 )
grpsubinv.y ( 𝜑𝑌𝐵 )
Assertion grpsubinv ( 𝜑 → ( 𝑋 ( 𝑁𝑌 ) ) = ( 𝑋 + 𝑌 ) )

Proof

Step Hyp Ref Expression
1 grpsubinv.b 𝐵 = ( Base ‘ 𝐺 )
2 grpsubinv.p + = ( +g𝐺 )
3 grpsubinv.m = ( -g𝐺 )
4 grpsubinv.n 𝑁 = ( invg𝐺 )
5 grpsubinv.g ( 𝜑𝐺 ∈ Grp )
6 grpsubinv.x ( 𝜑𝑋𝐵 )
7 grpsubinv.y ( 𝜑𝑌𝐵 )
8 1 4 grpinvcl ( ( 𝐺 ∈ Grp ∧ 𝑌𝐵 ) → ( 𝑁𝑌 ) ∈ 𝐵 )
9 5 7 8 syl2anc ( 𝜑 → ( 𝑁𝑌 ) ∈ 𝐵 )
10 1 2 4 3 grpsubval ( ( 𝑋𝐵 ∧ ( 𝑁𝑌 ) ∈ 𝐵 ) → ( 𝑋 ( 𝑁𝑌 ) ) = ( 𝑋 + ( 𝑁 ‘ ( 𝑁𝑌 ) ) ) )
11 6 9 10 syl2anc ( 𝜑 → ( 𝑋 ( 𝑁𝑌 ) ) = ( 𝑋 + ( 𝑁 ‘ ( 𝑁𝑌 ) ) ) )
12 1 4 grpinvinv ( ( 𝐺 ∈ Grp ∧ 𝑌𝐵 ) → ( 𝑁 ‘ ( 𝑁𝑌 ) ) = 𝑌 )
13 5 7 12 syl2anc ( 𝜑 → ( 𝑁 ‘ ( 𝑁𝑌 ) ) = 𝑌 )
14 13 oveq2d ( 𝜑 → ( 𝑋 + ( 𝑁 ‘ ( 𝑁𝑌 ) ) ) = ( 𝑋 + 𝑌 ) )
15 11 14 eqtrd ( 𝜑 → ( 𝑋 ( 𝑁𝑌 ) ) = ( 𝑋 + 𝑌 ) )