Theorem ablodiv32

Description: Swap the second and third terms in a double division. (Contributed by NM, 29-Feb-2008) (New usage is discouraged.)

	Ref	Expression
Hypotheses	abldiv.1	⊢ 𝑋 = ran 𝐺
	abldiv.3	⊢ 𝐷 = ( /𝑔 ‘ 𝐺 )
Assertion	ablodiv32	⊢ ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( ( 𝐴 𝐷 𝐵 ) 𝐷 𝐶 ) = ( ( 𝐴 𝐷 𝐶 ) 𝐷 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		abldiv.1	⊢ 𝑋 = ran 𝐺
2		abldiv.3	⊢ 𝐷 = ( /𝑔 ‘ 𝐺 )
3	1	ablocom	⊢ ( ( 𝐺 ∈ AbelOp ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) → ( 𝐵 𝐺 𝐶 ) = ( 𝐶 𝐺 𝐵 ) )
4	3	3adant3r1	⊢ ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( 𝐵 𝐺 𝐶 ) = ( 𝐶 𝐺 𝐵 ) )
5	4	oveq2d	⊢ ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( 𝐴 𝐷 ( 𝐵 𝐺 𝐶 ) ) = ( 𝐴 𝐷 ( 𝐶 𝐺 𝐵 ) ) )
6	1 2	ablodivdiv4	⊢ ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( ( 𝐴 𝐷 𝐵 ) 𝐷 𝐶 ) = ( 𝐴 𝐷 ( 𝐵 𝐺 𝐶 ) ) )
7		3ancomb	⊢ ( ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ↔ ( 𝐴 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) )
8	1 2	ablodivdiv4	⊢ ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) ) → ( ( 𝐴 𝐷 𝐶 ) 𝐷 𝐵 ) = ( 𝐴 𝐷 ( 𝐶 𝐺 𝐵 ) ) )
9	7 8	sylan2b	⊢ ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( ( 𝐴 𝐷 𝐶 ) 𝐷 𝐵 ) = ( 𝐴 𝐷 ( 𝐶 𝐺 𝐵 ) ) )
10	5 6 9	3eqtr4d	⊢ ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ 𝐶 ∈ 𝑋 ) ) → ( ( 𝐴 𝐷 𝐵 ) 𝐷 𝐶 ) = ( ( 𝐴 𝐷 𝐶 ) 𝐷 𝐵 ) )