Metamath Proof Explorer


Theorem ablonnncan1

Description: Cancellation law for group division. ( nnncan1 analog.) (Contributed by NM, 7-Mar-2008) (New usage is discouraged.)

Ref Expression
Hypotheses abldiv.1 𝑋 = ran 𝐺
abldiv.3 𝐷 = ( /𝑔𝐺 )
Assertion ablonnncan1 ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( ( 𝐴 𝐷 𝐵 ) 𝐷 ( 𝐴 𝐷 𝐶 ) ) = ( 𝐶 𝐷 𝐵 ) )

Proof

Step Hyp Ref Expression
1 abldiv.1 𝑋 = ran 𝐺
2 abldiv.3 𝐷 = ( /𝑔𝐺 )
3 simpr1 ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → 𝐴𝑋 )
4 simpr2 ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → 𝐵𝑋 )
5 ablogrpo ( 𝐺 ∈ AbelOp → 𝐺 ∈ GrpOp )
6 1 2 grpodivcl ( ( 𝐺 ∈ GrpOp ∧ 𝐴𝑋𝐶𝑋 ) → ( 𝐴 𝐷 𝐶 ) ∈ 𝑋 )
7 5 6 syl3an1 ( ( 𝐺 ∈ AbelOp ∧ 𝐴𝑋𝐶𝑋 ) → ( 𝐴 𝐷 𝐶 ) ∈ 𝑋 )
8 7 3adant3r2 ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴 𝐷 𝐶 ) ∈ 𝑋 )
9 3 4 8 3jca ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴𝑋𝐵𝑋 ∧ ( 𝐴 𝐷 𝐶 ) ∈ 𝑋 ) )
10 1 2 ablodiv32 ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋 ∧ ( 𝐴 𝐷 𝐶 ) ∈ 𝑋 ) ) → ( ( 𝐴 𝐷 𝐵 ) 𝐷 ( 𝐴 𝐷 𝐶 ) ) = ( ( 𝐴 𝐷 ( 𝐴 𝐷 𝐶 ) ) 𝐷 𝐵 ) )
11 9 10 syldan ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( ( 𝐴 𝐷 𝐵 ) 𝐷 ( 𝐴 𝐷 𝐶 ) ) = ( ( 𝐴 𝐷 ( 𝐴 𝐷 𝐶 ) ) 𝐷 𝐵 ) )
12 1 2 ablonncan ( ( 𝐺 ∈ AbelOp ∧ 𝐴𝑋𝐶𝑋 ) → ( 𝐴 𝐷 ( 𝐴 𝐷 𝐶 ) ) = 𝐶 )
13 12 3adant3r2 ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( 𝐴 𝐷 ( 𝐴 𝐷 𝐶 ) ) = 𝐶 )
14 13 oveq1d ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( ( 𝐴 𝐷 ( 𝐴 𝐷 𝐶 ) ) 𝐷 𝐵 ) = ( 𝐶 𝐷 𝐵 ) )
15 11 14 eqtrd ( ( 𝐺 ∈ AbelOp ∧ ( 𝐴𝑋𝐵𝑋𝐶𝑋 ) ) → ( ( 𝐴 𝐷 𝐵 ) 𝐷 ( 𝐴 𝐷 𝐶 ) ) = ( 𝐶 𝐷 𝐵 ) )