Theorem subcan

Description: Cancellation law for subtraction. (Contributed by NM, 8-Feb-2005) (Revised by Mario Carneiro, 27-May-2016)

	Ref	Expression
Assertion	subcan	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		simp2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐵 ∈ ℂ )
2		simp1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐴 ∈ ℂ )
3	1 2	addcomd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐴 ) = ( 𝐴 + 𝐵 ) )
4	3	eqeq1d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐴 ) = ( 𝐴 + 𝐶 ) ↔ ( 𝐴 + 𝐵 ) = ( 𝐴 + 𝐶 ) ) )
5		simp3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐶 ∈ ℂ )
6		addsubeq4	⊢ ( ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) ∧ ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) ) → ( ( 𝐵 + 𝐴 ) = ( 𝐴 + 𝐶 ) ↔ ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ) )
7	1 2 2 5 6	syl22anc	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐴 ) = ( 𝐴 + 𝐶 ) ↔ ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ) )
8		addcan	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) = ( 𝐴 + 𝐶 ) ↔ 𝐵 = 𝐶 ) )
9	4 7 8	3bitr3d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) = ( 𝐴 − 𝐶 ) ↔ 𝐵 = 𝐶 ) )