Metamath Proof Explorer

Theorem grpcominv1

Description: If two elements commute, then they commute with each other's inverses (case of the first element commuting with the inverse of the second element). (Contributed by SN, 29-Jan-2025)

Ref Expression
Hypotheses grpcominv.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpcominv.p ⊢ + = ( +g ‘ 𝐺 )
grpcominv.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
grpcominv.g ⊢ ( 𝜑 → 𝐺 ∈ Grp )
grpcominv.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
grpcominv.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
grpcominv.1 ⊢ ( 𝜑 → ( 𝑋 + 𝑌 ) = ( 𝑌 + 𝑋 ) )
Assertion grpcominv1 ( 𝜑 → ( 𝑋 + ( 𝑁 ‘ 𝑌 ) ) = ( ( 𝑁 ‘ 𝑌 ) + 𝑋 ) )

Proof

Step Hyp Ref Expression
1 grpcominv.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 grpcominv.p ⊢ + = ( +g ‘ 𝐺 )
3 grpcominv.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
4 grpcominv.g ⊢ ( 𝜑 → 𝐺 ∈ Grp )
5 grpcominv.x ⊢ ( 𝜑 → 𝑋 ∈ 𝐵 )
6 grpcominv.y ⊢ ( 𝜑 → 𝑌 ∈ 𝐵 )
7 grpcominv.1 ⊢ ( 𝜑 → ( 𝑋 + 𝑌 ) = ( 𝑌 + 𝑋 ) )
8 1 3 4 6 grpinvcld ⊢ ( 𝜑 → ( 𝑁 ‘ 𝑌 ) ∈ 𝐵 )
9 1 2 4 8 6 5 grpassd ⊢ ( 𝜑 → ( ( ( 𝑁 ‘ 𝑌 ) + 𝑌 ) + 𝑋 ) = ( ( 𝑁 ‘ 𝑌 ) + ( 𝑌 + 𝑋 ) ) )
10 eqid ⊢ ( 0g ‘ 𝐺 ) = ( 0g ‘ 𝐺 )
11 1 2 10 3 4 6 grplinvd ⊢ ( 𝜑 → ( ( 𝑁 ‘ 𝑌 ) + 𝑌 ) = ( 0g ‘ 𝐺 ) )
12 11 oveq1d ⊢ ( 𝜑 → ( ( ( 𝑁 ‘ 𝑌 ) + 𝑌 ) + 𝑋 ) = ( ( 0g ‘ 𝐺 ) + 𝑋 ) )
13 1 2 10 4 5 grplidd ⊢ ( 𝜑 → ( ( 0g ‘ 𝐺 ) + 𝑋 ) = 𝑋 )
14 12 13 eqtr2d ⊢ ( 𝜑 → 𝑋 = ( ( ( 𝑁 ‘ 𝑌 ) + 𝑌 ) + 𝑋 ) )
15 7 oveq2d ⊢ ( 𝜑 → ( ( 𝑁 ‘ 𝑌 ) + ( 𝑋 + 𝑌 ) ) = ( ( 𝑁 ‘ 𝑌 ) + ( 𝑌 + 𝑋 ) ) )
16 9 14 15 3eqtr4rd ⊢ ( 𝜑 → ( ( 𝑁 ‘ 𝑌 ) + ( 𝑋 + 𝑌 ) ) = 𝑋 )
17 1 2 4 8 5 6 grpassd ⊢ ( 𝜑 → ( ( ( 𝑁 ‘ 𝑌 ) + 𝑋 ) + 𝑌 ) = ( ( 𝑁 ‘ 𝑌 ) + ( 𝑋 + 𝑌 ) ) )
18 1 2 3 4 5 6 grpasscan2d ⊢ ( 𝜑 → ( ( 𝑋 + ( 𝑁 ‘ 𝑌 ) ) + 𝑌 ) = 𝑋 )
19 16 17 18 3eqtr4rd ⊢ ( 𝜑 → ( ( 𝑋 + ( 𝑁 ‘ 𝑌 ) ) + 𝑌 ) = ( ( ( 𝑁 ‘ 𝑌 ) + 𝑋 ) + 𝑌 ) )
20 1 2 4 5 8 grpcld ⊢ ( 𝜑 → ( 𝑋 + ( 𝑁 ‘ 𝑌 ) ) ∈ 𝐵 )
21 1 2 4 8 5 grpcld ⊢ ( 𝜑 → ( ( 𝑁 ‘ 𝑌 ) + 𝑋 ) ∈ 𝐵 )
22 1 2 grprcan ⊢ ( ( 𝐺 ∈ Grp ∧ ( ( 𝑋 + ( 𝑁 ‘ 𝑌 ) ) ∈ 𝐵 ∧ ( ( 𝑁 ‘ 𝑌 ) + 𝑋 ) ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( ( 𝑋 + ( 𝑁 ‘ 𝑌 ) ) + 𝑌 ) = ( ( ( 𝑁 ‘ 𝑌 ) + 𝑋 ) + 𝑌 ) ↔ ( 𝑋 + ( 𝑁 ‘ 𝑌 ) ) = ( ( 𝑁 ‘ 𝑌 ) + 𝑋 ) ) )
23 4 20 21 6 22 syl13anc ⊢ ( 𝜑 → ( ( ( 𝑋 + ( 𝑁 ‘ 𝑌 ) ) + 𝑌 ) = ( ( ( 𝑁 ‘ 𝑌 ) + 𝑋 ) + 𝑌 ) ↔ ( 𝑋 + ( 𝑁 ‘ 𝑌 ) ) = ( ( 𝑁 ‘ 𝑌 ) + 𝑋 ) ) )
24 19 23 mpbid ⊢ ( 𝜑 → ( 𝑋 + ( 𝑁 ‘ 𝑌 ) ) = ( ( 𝑁 ‘ 𝑌 ) + 𝑋 ) )