Metamath Proof Explorer


Theorem subsub2

Description: Law for double subtraction. (Contributed by NM, 30-Jun-2005) (Revised by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion subsub2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( 𝐵𝐶 ) ) = ( 𝐴 + ( 𝐶𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 subcl ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵𝐶 ) ∈ ℂ )
2 1 3adant1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵𝐶 ) ∈ ℂ )
3 simp1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐴 ∈ ℂ )
4 simp3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐶 ∈ ℂ )
5 simp2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐵 ∈ ℂ )
6 subcl ( ( 𝐶 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐶𝐵 ) ∈ ℂ )
7 4 5 6 syl2anc ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐶𝐵 ) ∈ ℂ )
8 2 3 7 add12d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵𝐶 ) + ( 𝐴 + ( 𝐶𝐵 ) ) ) = ( 𝐴 + ( ( 𝐵𝐶 ) + ( 𝐶𝐵 ) ) ) )
9 npncan2 ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵𝐶 ) + ( 𝐶𝐵 ) ) = 0 )
10 9 3adant1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵𝐶 ) + ( 𝐶𝐵 ) ) = 0 )
11 10 oveq2d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + ( ( 𝐵𝐶 ) + ( 𝐶𝐵 ) ) ) = ( 𝐴 + 0 ) )
12 3 addid1d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 0 ) = 𝐴 )
13 8 11 12 3eqtrd ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵𝐶 ) + ( 𝐴 + ( 𝐶𝐵 ) ) ) = 𝐴 )
14 3 7 addcld ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + ( 𝐶𝐵 ) ) ∈ ℂ )
15 subadd ( ( 𝐴 ∈ ℂ ∧ ( 𝐵𝐶 ) ∈ ℂ ∧ ( 𝐴 + ( 𝐶𝐵 ) ) ∈ ℂ ) → ( ( 𝐴 − ( 𝐵𝐶 ) ) = ( 𝐴 + ( 𝐶𝐵 ) ) ↔ ( ( 𝐵𝐶 ) + ( 𝐴 + ( 𝐶𝐵 ) ) ) = 𝐴 ) )
16 3 2 14 15 syl3anc ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵𝐶 ) ) = ( 𝐴 + ( 𝐶𝐵 ) ) ↔ ( ( 𝐵𝐶 ) + ( 𝐴 + ( 𝐶𝐵 ) ) ) = 𝐴 ) )
17 13 16 mpbird ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( 𝐵𝐶 ) ) = ( 𝐴 + ( 𝐶𝐵 ) ) )