Metamath Proof Explorer


Theorem ablodivdiv

Description: Law for double group division. (Contributed by NM, 29-Feb-2008) (New usage is discouraged.)

Ref Expression
Hypotheses abldiv.1 𝑋 = ran 𝐺
abldiv.3 𝐷 = ( /𝑔𝐺 )
Assertion ablodivdiv ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴 𝐷 ( 𝐵 𝐷 𝐶 ) ) = ( ( 𝐴 𝐷 𝐵 ) 𝐺 𝐶 ) )

Proof

Step Hyp Ref Expression
1 abldiv.1 𝑋 = ran 𝐺
2 abldiv.3 𝐷 = ( /𝑔𝐺 )
3 ablogrpo ( 𝐺 ∈ AbelOp → 𝐺 ∈ GrpOp )
4 1 2 grpodivdiv ( ( 𝐺 ∈ GrpOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴 𝐷 ( 𝐵 𝐷 𝐶 ) ) = ( 𝐴 𝐺 ( 𝐶 𝐷 𝐵 ) ) )
5 3 4 sylan ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴 𝐷 ( 𝐵 𝐷 𝐶 ) ) = ( 𝐴 𝐺 ( 𝐶 𝐷 𝐵 ) ) )
6 3ancomb ( ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ↔ ( 𝐴𝑋𝐶𝑋𝐵𝑋 ) )
7 1 2 grpomuldivass ( ( 𝐺 ∈ GrpOp ∧ ( 𝐴𝑋𝐶𝑋𝐵𝑋 ) ) → ( ( 𝐴 𝐺 𝐶 ) 𝐷 𝐵 ) = ( 𝐴 𝐺 ( 𝐶 𝐷 𝐵 ) ) )
8 3 7 sylan ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐶𝑋𝐵𝑋 ) ) → ( ( 𝐴 𝐺 𝐶 ) 𝐷 𝐵 ) = ( 𝐴 𝐺 ( 𝐶 𝐷 𝐵 ) ) )
9 1 2 ablomuldiv ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐶𝑋𝐵𝑋 ) ) → ( ( 𝐴 𝐺 𝐶 ) 𝐷 𝐵 ) = ( ( 𝐴 𝐷 𝐵 ) 𝐺 𝐶 ) )
10 8 9 eqtr3d ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐶𝑋𝐵𝑋 ) ) → ( 𝐴 𝐺 ( 𝐶 𝐷 𝐵 ) ) = ( ( 𝐴 𝐷 𝐵 ) 𝐺 𝐶 ) )
11 6 10 sylan2b ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴 𝐺 ( 𝐶 𝐷 𝐵 ) ) = ( ( 𝐴 𝐷 𝐵 ) 𝐺 𝐶 ) )
12 5 11 eqtrd ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴 𝐷 ( 𝐵 𝐷 𝐶 ) ) = ( ( 𝐴 𝐷 𝐵 ) 𝐺 𝐶 ) )