Metamath Proof Explorer


Theorem subcan2

Description: Cancellation law for subtraction. (Contributed by NM, 8-Feb-2005)

Ref Expression
Assertion subcan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐶 ) = ( 𝐵𝐶 ) ↔ 𝐴 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 simp1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐴 ∈ ℂ )
2 simp3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐶 ∈ ℂ )
3 subcl ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵𝐶 ) ∈ ℂ )
4 3 3adant1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵𝐶 ) ∈ ℂ )
5 subadd2 ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ ( 𝐵𝐶 ) ∈ ℂ ) → ( ( 𝐴𝐶 ) = ( 𝐵𝐶 ) ↔ ( ( 𝐵𝐶 ) + 𝐶 ) = 𝐴 ) )
6 1 2 4 5 syl3anc ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐶 ) = ( 𝐵𝐶 ) ↔ ( ( 𝐵𝐶 ) + 𝐶 ) = 𝐴 ) )
7 npcan ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵𝐶 ) + 𝐶 ) = 𝐵 )
8 7 3adant1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵𝐶 ) + 𝐶 ) = 𝐵 )
9 8 eqeq1d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐵𝐶 ) + 𝐶 ) = 𝐴𝐵 = 𝐴 ) )
10 eqcom ( 𝐵 = 𝐴𝐴 = 𝐵 )
11 9 10 bitrdi ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐵𝐶 ) + 𝐶 ) = 𝐴𝐴 = 𝐵 ) )
12 6 11 bitrd ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐶 ) = ( 𝐵𝐶 ) ↔ 𝐴 = 𝐵 ) )