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Metamath Proof Explorer

Theorem grpinvsub

Description: Inverse of a group subtraction. (Contributed by NM, 9-Sep-2014)

Ref Expression
Hypotheses grpsubcl.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpsubcl.m ⊢ − = ( -g ‘ 𝐺 )
grpinvsub.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
Assertion grpinvsub ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ ( 𝑋 − 𝑌 ) ) = ( 𝑌 − 𝑋 ) )

Proof

Step Hyp Ref Expression
1 grpsubcl.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 grpsubcl.m ⊢ − = ( -g ‘ 𝐺 )
3 grpinvsub.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
4 1 3 grpinvcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑌 ) ∈ 𝐵 )
5 4 3adant2 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑌 ) ∈ 𝐵 )
6 eqid ⊢ ( +g ‘ 𝐺 ) = ( +g ‘ 𝐺 )
7 1 6 3 grpinvadd ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ ( 𝑁 ‘ 𝑌 ) ∈ 𝐵 ) → ( 𝑁 ‘ ( 𝑋 ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑌 ) ) ) = ( ( 𝑁 ‘ ( 𝑁 ‘ 𝑌 ) ) ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑋 ) ) )
8 5 7 syld3an3 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ ( 𝑋 ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑌 ) ) ) = ( ( 𝑁 ‘ ( 𝑁 ‘ 𝑌 ) ) ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑋 ) ) )
9 1 3 grpinvinv ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ ( 𝑁 ‘ 𝑌 ) ) = 𝑌 )
10 9 3adant2 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ ( 𝑁 ‘ 𝑌 ) ) = 𝑌 )
11 10 oveq1d ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑁 ‘ ( 𝑁 ‘ 𝑌 ) ) ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑋 ) ) = ( 𝑌 ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑋 ) ) )
12 8 11 eqtrd ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ ( 𝑋 ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑌 ) ) ) = ( 𝑌 ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑋 ) ) )
13 1 6 3 2 grpsubval ⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 − 𝑌 ) = ( 𝑋 ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑌 ) ) )
14 13 3adant1 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 − 𝑌 ) = ( 𝑋 ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑌 ) ) )
15 14 fveq2d ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ ( 𝑋 − 𝑌 ) ) = ( 𝑁 ‘ ( 𝑋 ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑌 ) ) ) )
16 1 6 3 2 grpsubval ⊢ ( ( 𝑌 ∈ 𝐵 ∧ 𝑋 ∈ 𝐵 ) → ( 𝑌 − 𝑋 ) = ( 𝑌 ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑋 ) ) )
17 16 ancoms ⊢ ( ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑌 − 𝑋 ) = ( 𝑌 ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑋 ) ) )
18 17 3adant1 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑌 − 𝑋 ) = ( 𝑌 ( +g ‘ 𝐺 ) ( 𝑁 ‘ 𝑋 ) ) )
19 12 15 18 3eqtr4d ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ ( 𝑋 − 𝑌 ) ) = ( 𝑌 − 𝑋 ) )