Metamath Proof Explorer


Theorem subsub

Description: Law for double subtraction. (Contributed by NM, 13-May-2004)

Ref Expression
Assertion subsub ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( 𝐵𝐶 ) ) = ( ( 𝐴𝐵 ) + 𝐶 ) )

Proof

Step Hyp Ref Expression
1 subsub2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( 𝐵𝐶 ) ) = ( 𝐴 + ( 𝐶𝐵 ) ) )
2 addsubass ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − 𝐵 ) = ( 𝐴 + ( 𝐶𝐵 ) ) )
3 addsub ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − 𝐵 ) = ( ( 𝐴𝐵 ) + 𝐶 ) )
4 2 3 eqtr3d ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + ( 𝐶𝐵 ) ) = ( ( 𝐴𝐵 ) + 𝐶 ) )
5 4 3com23 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + ( 𝐶𝐵 ) ) = ( ( 𝐴𝐵 ) + 𝐶 ) )
6 1 5 eqtrd ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( 𝐵𝐶 ) ) = ( ( 𝐴𝐵 ) + 𝐶 ) )