Metamath Proof Explorer


Theorem grpodivdiv

Description: Double group division. (Contributed by NM, 24-Feb-2008) (New usage is discouraged.)

Ref Expression
Hypotheses grpdivf.1 𝑋 = ran 𝐺
grpdivf.3 𝐷 = ( /𝑔𝐺 )
Assertion grpodivdiv ( ( 𝐺 ∈ GrpOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴 𝐷 ( 𝐵 𝐷 𝐶 ) ) = ( 𝐴 𝐺 ( 𝐶 𝐷 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 grpdivf.1 𝑋 = ran 𝐺
2 grpdivf.3 𝐷 = ( /𝑔𝐺 )
3 simpl ( ( 𝐺 ∈ GrpOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → 𝐺 ∈ GrpOp )
4 simpr1 ( ( 𝐺 ∈ GrpOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → 𝐴𝑋 )
5 1 2 grpodivcl ( ( 𝐺 ∈ GrpOp ∧ 𝐵𝑋𝐶𝑋 ) → ( 𝐵 𝐷 𝐶 ) ∈ 𝑋 )
6 5 3adant3r1 ( ( 𝐺 ∈ GrpOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐵 𝐷 𝐶 ) ∈ 𝑋 )
7 eqid ( inv ‘ 𝐺 ) = ( inv ‘ 𝐺 )
8 1 7 2 grpodivval ( ( 𝐺 ∈ GrpOp ∧ 𝐴𝑋 ∧ ( 𝐵 𝐷 𝐶 ) ∈ 𝑋 ) → ( 𝐴 𝐷 ( 𝐵 𝐷 𝐶 ) ) = ( 𝐴 𝐺 ( ( inv ‘ 𝐺 ) ‘ ( 𝐵 𝐷 𝐶 ) ) ) )
9 3 4 6 8 syl3anc ( ( 𝐺 ∈ GrpOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴 𝐷 ( 𝐵 𝐷 𝐶 ) ) = ( 𝐴 𝐺 ( ( inv ‘ 𝐺 ) ‘ ( 𝐵 𝐷 𝐶 ) ) ) )
10 1 7 2 grpoinvdiv ( ( 𝐺 ∈ GrpOp ∧ 𝐵𝑋𝐶𝑋 ) → ( ( inv ‘ 𝐺 ) ‘ ( 𝐵 𝐷 𝐶 ) ) = ( 𝐶 𝐷 𝐵 ) )
11 10 3adant3r1 ( ( 𝐺 ∈ GrpOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( ( inv ‘ 𝐺 ) ‘ ( 𝐵 𝐷 𝐶 ) ) = ( 𝐶 𝐷 𝐵 ) )
12 11 oveq2d ( ( 𝐺 ∈ GrpOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴 𝐺 ( ( inv ‘ 𝐺 ) ‘ ( 𝐵 𝐷 𝐶 ) ) ) = ( 𝐴 𝐺 ( 𝐶 𝐷 𝐵 ) ) )
13 9 12 eqtrd ( ( 𝐺 ∈ GrpOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴 𝐷 ( 𝐵 𝐷 𝐶 ) ) = ( 𝐴 𝐺 ( 𝐶 𝐷 𝐵 ) ) )