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Metamath Proof Explorer

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 ) )