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Metamath Proof Explorer

Theorem grpodivinv

Description: Group division by an inverse. (Contributed by NM, 15-Feb-2008) (New usage is discouraged.)

Ref Expression
Hypotheses grpdiv.1 ⊢ 𝑋 = ran 𝐺
grpdiv.2 ⊢ 𝑁 = ( inv ‘ 𝐺 )
grpdiv.3 ⊢ 𝐷 = ( /𝑔 ‘ 𝐺 )
Assertion grpodivinv ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( 𝐴 𝐷 ( 𝑁 ‘ 𝐵 ) ) = ( 𝐴 𝐺 𝐵 ) )

Proof

Step Hyp Ref Expression
1 grpdiv.1 ⊢ 𝑋 = ran 𝐺
2 grpdiv.2 ⊢ 𝑁 = ( inv ‘ 𝐺 )
3 grpdiv.3 ⊢ 𝐷 = ( /𝑔 ‘ 𝐺 )
4 1 2 grpoinvcl ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐵 ∈ 𝑋 ) → ( 𝑁 ‘ 𝐵 ) ∈ 𝑋 )
5 4 3adant2 ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( 𝑁 ‘ 𝐵 ) ∈ 𝑋 )
6 1 2 3 grpodivval ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ∧ ( 𝑁 ‘ 𝐵 ) ∈ 𝑋 ) → ( 𝐴 𝐷 ( 𝑁 ‘ 𝐵 ) ) = ( 𝐴 𝐺 ( 𝑁 ‘ ( 𝑁 ‘ 𝐵 ) ) ) )
7 5 6 syld3an3 ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( 𝐴 𝐷 ( 𝑁 ‘ 𝐵 ) ) = ( 𝐴 𝐺 ( 𝑁 ‘ ( 𝑁 ‘ 𝐵 ) ) ) )
8 1 2 grpo2inv ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐵 ∈ 𝑋 ) → ( 𝑁 ‘ ( 𝑁 ‘ 𝐵 ) ) = 𝐵 )
9 8 3adant2 ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( 𝑁 ‘ ( 𝑁 ‘ 𝐵 ) ) = 𝐵 )
10 9 oveq2d ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( 𝐴 𝐺 ( 𝑁 ‘ ( 𝑁 ‘ 𝐵 ) ) ) = ( 𝐴 𝐺 𝐵 ) )
11 7 10 eqtrd ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( 𝐴 𝐷 ( 𝑁 ‘ 𝐵 ) ) = ( 𝐴 𝐺 𝐵 ) )